mechanicsModules - User contributions [en]

Module 6: Circular Motion

2019-10-21T18:54:27Z

Kevin:

''Karen Tennenhouse and Kevin Lenton''

==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand that a change in the velocity vector could indicate a change in the magnitude of velocity (speed), a change in the direction, or a change in both magnitude and direction.
* Understand why uniform circular motion requires centripetal acceleration.
* Know the direction of the centripetal acceleration.
* Know the magnitude of the centripetal acceleration.
* Apply the concepts of centripetal acceleration to some circular motion problems
 

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/CircularMotion/Module_CircularMotion_Worksheet.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/CircularMotion/Module_CircularMotion_Worksheet.pdf pdfWorksheet] 
[https://cnx.org/contents/Ax2o07Ul@17.22:lsUL0z9f@18/6-2-Centripetal-Acceleration Openstax] 

==Can a particle move at a constant speed and yet be accelerating?==

The answer is '''YES'''. 

The point is that the acceleration is the rate of change of the velocity '''vector''' with respect to time. A vector has both a value (magnitude) and a direction. You can change the velocity vector by changing the magnitude ''and/or'' by changing the direction. 
An example of this is an object moving around a circle. The particle can move at ''constant speed'' around a circular or any other curved path and have an acceleration at the same time. Since the direction of the velocity changes, the ''velocity vector'' is not constant and therefore the motion is an accelerated motion.

 

==A Special Case: Uniform Circular Motion==

Consider an object travelling in a circle (radius R) at some instantaneous speed (v). 

If the speed (magnitude of velocity) is constant, we say that this is '''Uniform Circular Motion'''. 
Such an object does not have constant velocity; its direction is changing. Therefore, it is accelerating! This acceleration is called '''centripetal acceleration''' <math> \vec a_c </math>. 

This '''centripetal acceleration''' <math> \vec a_c </math> is also a vector and therefore has a magnitude and direction.

===Direction of the '''centripetal acceleration''' <math> \vec a_c </math>===
Because the definition of acceleration is 
<math>
\vec{a_{average}} = \frac{\vec{\Delta v}}{\Delta t} = \frac{\vec{v_{f}}-\vec{v_{i}}}{\Delta t} \ \ \ or \frac{\vec{v_{2}}-\vec{v_{1}}}{\Delta t}
</math> 
The direction of the acceleration will be in the same direction as
<math>
\vec{\Delta v}
</math>. 
Watch this video to find out the direction of the centripetal acceleration. 
[http://www.youtube.com/watch?v=WKvhgkg2_dw Direction of the centripetal acceleration]

:::::<youtube>WKvhgkg2_dw</youtube> 

===Magnitude of the '''centripetal acceleration''' <math>| \vec a_c |</math>===
The magnitude of the centripetal acceleration is equal to: 

<math>| \vec a_c | = \frac{v^2}{R }
</math> 
Where v is the magnitude of the velocity and R is the radius of the circle. 
Here is a proof of this equation for the magnitude of the centripetal acceleration. 
[http://www.youtube.com/watch?v=TNX-Z6XR3gA Magnitude of the centripetal acceleration]

:::::<youtube>TNX-Z6XR3gA</youtube> 

What causes the object to accelerate towards the center? 
You will be seeing that what causes this acceleration is the same thing that ever causes any object to accelerate in any way: The real forces acting on it (gravity, friction, tension or whatever) are adding up to some nonzero net force.
 

===What happens for non-uniform Circular Motion?===
<li>All the above notes deal with uniform circular motion, in other words the object is going in a circle but at constant speed. 
If the object has any sort of change in the velocity vector, there will be an acceleration. For any motion in any circle, there must be a centripetal acceleration with direction towards the centre of the circle and magnitude <math>| \vec a_c | = \frac{v^2}{R }
</math> 
However, as you know points on a rotating object can be speeding up or slowing down, that is to say, they can have an angular acceleration <math>\alpha</math>. That means that a point on the rotating object can also have a linear (or tangential) acceleration. We say tangential because this acceleration is a tangent to the circle. 
Therefore the point is travelling in a circle (thus changing direction) and is also changing its speed. 
In such a case, the object’s acceleration will have two (perpendicular) components: 
*One component, along the direction of motion, is called the tangential acceleration, aT . The tangential acceleration is related to the change in speed: it is in the same direction as velocity if the object is speeding up, or opposite to velocity if object is slowing down.) 
*The second component, called the radial acceleration aR , is exactly the centripetal acceleration: it points towards the center and has magnitude aR = (v2 / R) 
*The point’s total acceleration is the vector sum of its two components. 
<math> | \vec a_{total} |=\sqrt{a_{Radial} ^{2}+a_{Tangential} ^{2}}</math>
 
[http://www.youtube.com/watch?v=zW3KCKOXe0k The first 5 mins of this video explain this]

:::::<youtube>zW3KCKOXe0k</youtube> 

===Period of the Circular Motion===
The period of the circular motion is the time for the object to complete one revolution, one circle. 
The distance travelled in this time is the circumference of the circle <math> 2\pi r</math>. 
Therefore <math> v =\frac{2\pi r}{T_{period}}</math>
 

==Exercises==
''Helena Dedic''

===Exercise 1===

True or False: When a particle moves in uniform circular motion its acceleration is constant.

'''Solution:'''
 

'''False.''' Trick Question! The radial acceleration is still a vector. The radial acceleration has a constant magnitude but its direction changes as the particle moves around the circle. The acceleration always pointing towards the centre, and is therefore constantly changing.
 

===Exercise 2===

(a)The electron in a hydrogen atom has a speed of <math>2.2 \times 10^6 m/s</math> and orbits the proton at a distance of <math>5.3 \times 10^{-11} m</math>. What is its centripetal acceleration? 

(b) A neutron star of radius 20 km is found to rotate once per second. What is the centripetal acceleration of a point on its equator?

'''Solution:''' 

a. It is given that the electron has speed <math> v = 2 \times 10^6 m/s</math> and orbits the proton with radius <math>r = 5.3 \times 10^{-11} m</math>. You are asked to find its centripetal or radial acceleration:

<math>a = \frac{(2 \times 10^6 m/s)^2 }{ (5.3 \times 10^{-11} m)} = 9.1 \times 10^{22} m/s^2</math> 

It is always interesting to think about one's results. It is particularly intriguiging in this case because the magnitude of the acceleration is such an awesome number (compared to <math>9.8 m/s^2</math> on Earth). Now, suppose the Earth were to shrink and become a Black Hole; then at a distance of 3 cm from the centre, acceleration would still be only <math> 4 \times 10^{17} m/s^2</math>. Thus the large radial acceleration of the electron is testimony to the strength of the interaction between the electron and the proton in the atom.
 

b. You are told that a neutron star rotates once per second (i.e that it has period T = 1 s) and that it has a radius of <math>r = 2 \times 10^4</math> m. You are asked to find the centripetal acceleration of a point on its equator:

<math>v = \frac{(2\pi) }{ T} = \frac{(2\pi \times 2 \times 10^4 m) }{ 1 s} = 4\pi \times 10^4 m/s</math> 

<math>a = v^2 / r = \frac{(4\pi \times 10^4 m/s)^2 }{ (2 \times 10^4 m)} = 7.9 \times 10^5 m/s^2</math>

Main Page

2016-11-26T17:52:21Z

Kevin: /* Study Guide */

'''Kevin Lenton's Mechanics Wiki.'''

Consult the [//meta.wikimedia.org/wiki/Help:Contents User's Guide] for information on using the wiki software.

== Study Guide ==
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-VECT-20160913-SG.pdf Vectors] [ Solutions in progress] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-Graphs_1and2DKinematics.pdf Graphs, 1D and 2D Kinematics] [http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-TorqueandStaticEquilibrium_Solutions.pdf Solutions] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-TorqueandStaticEquilibrium.pdf Torque and Static Equilibrium] [http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-TorqueandStaticEquilibrium_Solutions.pdf Solutions] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-STATDYN-20160913-SG.pdf Statics and Dynamics : Includes Friction and Circular Dynamics] 


[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-STATDYN-20160913-SG-Sol-A.pdf Solutions A]
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-STATDYN-20160913-SG-Sol-B.pdf Solutions B]
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-STATDYN-20160913-SG-Sol-C.pdf Solutions C]
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-STATDYN-20160913-SG-Sol-D.pdf Solutions D]
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-ENERGY-20160913-SG.docx Work and Energy]

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-ENERGY-20160913-SG-Sol-ABCE.pdf Solutions ABCE]
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/NYAStudyGuide_KJL/NYA-ENERGY-20160913-SG-Sol-D.pdf Solutions D]

 

== Modules ==
* [[Module 1: Units and Measurements]]
* [[Module 2: Intro to Vectors]]
* [[Module 3: Intro to 1D Motion]]
* [[Module 3b: Area under Kinematic Graphs]]
* [[Module 4: Equations of Motion for Constant Acceleration]]
* [[Module 5: 2D Projectile Motion]]
* [[Module 6: Circular Motion]]
* [[Module : Rotational Kinematics]]
* [[Module 8: Newton's Laws of Motion]]
* [[Module : Applications of Newton' Laws : Friction ]]
* [[Module : Applications of Newton' Laws : Circular Motion ]]
* [[Module : Torque]]
* [[Module : Work and Kinetic Energy]]
* [[Module : Conservation of Energy]]
* [[Module : Conservation of Momentum]]
* [[Module : Rotational Dynamics]]
* [[ClassNotes and Compiled Class Problems: 203NYA H14]]
* [[ClassNotes and Compiled Class Problems: 203NYA A14]]

== Electricity and Magnetism Modules ==
* [[Charge and Electric Force]]
* [[The Electric Field]]
* [[Electric Potential Energy and Electric Potential]]

* [[Capacitors]]
* [[Electric Current]]

* [[Electromotive Force and Kirchoff's Rules]]
* [[The Magnetic Field]]
* [[Motion of a Charged Particle in a Magnetic Field]]
* [[Sources of Magnetic Field]]
* [[Magnetic Induction]]
* [[ClassNotes and Compiled Class Problems: 203NYB A13]]
* [[ClassNotes and Compiled Class Problems: 203NYB A14]]

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYB_Notes.pdf A set of Notes from a Student] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/203NYB_AdditionalProblems.docx Compiled Additional Problems]

Module : Applications of Newton' Laws : Friction

2015-04-09T21:52:46Z

Kevin: /* Misconception #1: Fs= mu N */

''Kreshnik Angoni, and Kevin Lenton''
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Friction/Module%20Friction%20Worksheet.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Friction/Module%20Friction%20Worksheet.pdf PDF Worksheet] 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 6.1-4
*[http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@7.61:32/College_Physics Openstax]
==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand what causes friction forces.
* Know the formulas for static and kinetic friction.
* Know to avoid some common misconceptions.
* Apply the concepts to some dynamics problems.
 

==Basic Definition==
Friction is a resistive force, which resists the sliding (kinetic friction) or attempted sliding (static friction) of two surfaces. 

==Tribology==
Tribology is the science of 'interacting surfaces in relative motion'. It includes the study of friction, wear, lubrication, etc. 
Any surface that appears smooth, is really quite rough when looked at very closely. See this image below from the [https://www.mcgill.ca/materials/research/surfacescienceandengineering McGill Materials Engineering Group]: 
https://www.mcgill.ca/materials/files/materials/sem-images.png 
Imagine two of these surfaces pushed together, and then made to slide over top of each other. 
Friction is caused by forces between points where surfaces meet. If two glass plates are made really flat and polished and cleaned and made to touch in a vacuum, they bond together. It would look as if the glass was just one piece. The bonds are formed as in a normal glass piece. This is called cold welding. And this is the main cause of friction. 
The detailed microscopics of friction are complicated. However, by focusing on practical details we can arrive at useful Models of Friction.

Some observations:

*Heavier objects have more friction than lighter ones
*On surfaces with friction it usually takes more force to get an object moving than keep it moving
*It is harder to move objects on rough surfaces than smooth ones

Friction should be proportional to the force the surface exerts on another surface, which is of course the normal force <math> \vec N</math> . As an equation this means that the magnitude of the frictional force can be expressed as 

<math> |\vec F_{friction}|=\mu |\vec N|</math> 

and we can also suppose we will need different constants for a stationary object compared to a moving one. i.e we have different coefficients of static friction (μs) and kinetic friction (μk). 
Remember that the normal force is perpendicular to the surface. Frictional forces are always parallel to the surface. 
===The Coefficient of Friction <math>\mu</math>===
*The coefficient is always positive. A negative coefficient would imply the force of friction is assistive (adds energy) rather than resistive (removes energy from the system). 
*The coefficient has no units. It is just a number.
*Coefficients of friction usually have values between 0.3 and 0.6. However, teflon can have a coefficient as low as 0.04. Rubber-coated surfaces can have a coefficient of friction substantially larger than 1. 
 
[http://www.youtube.com/watch?v=jSu0Tvlm6LY How much force does it take to separate two interleaved phone books?]

:::::<youtube>jSu0Tvlm6LY</youtube> 

[http://www.youtube.com/watch?v=QMW_uYWwHWQ&NR Second part of video: Worth Watching]

==Pushing a book on the Table==
Perform a short experiment by pushing a book on the table with an ever increasing horizontal force. The book does not move initially, even though you are pushing the book. The force of static friction is matching the force with which you are pushing on it. Eventually you overcome the force of static friction and the book starts to slide. The sliding is opposed by the force of kinetic friction, which then remains roughly constant. 
The maximum static friction is always greater than the kinetic friction. This means the coefficient of static friction is always greater than the coefficient of kinetic friction. 
This is shown in this figure showing the frictional forces as a function of the applied force: 
[[File:statickinetic.jpeg]] 
This is explained in this video. 
[http://www.youtube.com/watch?v=J9BWNiOSGlc Static and kinetic Friction]

:::::<youtube>J9BWNiOSGlc</youtube> 

==Force of Static Friction==
For static friction, the force of friction matches the force applied, keeping the surfaces from sliding. When the applied force exceeds the maximum static frictional force, the surfaces will begin to slide. The maximum frictional force is calculated as follows: 
<math> |\vec F_{static}|\leq \mu_{static}|\vec N|</math> 
The static frictional force is less than or equal to the coefficient of static friction times the normal force. Once the frictional force exceeds this value, the surfaces will break away and begin to slide.

==Force of Kinetic Friction==
Once it is moving, the frictional force then obeys: 
<math> |\vec F_{kinetic}|= \mu_{kinetic}|\vec N|</math> 
Note that once moving the force of kinetic friction is constant; that is to say it has the same value whatever velocity the object is moving at.

==Misconception #1: Fs= mu N ==
Students can sometimes be heard saying: 
"The force of static friction is equal to <math> \mu_{s}|\vec N|</math>
" 
Be very careful with this because the formula gives the ''limit'' or ''max'' for the force of static friction. 
<math> |\vec F_{static}|\leq \mu_{static}|\vec N|</math> 
"The force of static friction is ''less than or'' equal to <math> \mu_{s}|\vec N|</math>" 
If the applied horizontal force on the above book is less than this limit, the force of static friction will be equal in magnitude, but opposite in direction, to the applied force. 
For instance, let's say the maximum the static friction can be is 30N (Ok, ok it's a big book). 
You apply 10N horizontally, the static friction matches it to 10N. 
You apply 20N, the static friction matches it to 20N. 
You apply 30N, the static friction matches it to 30N. 
You apply 31N, the maximum static friction is exceeded and the surface starts to slide. The force of kinetic friction is now in operation. 
This misconception is not helped by the fact that many textbook problems ask questions at this limit, just so you can use the equation.
 
This is explained in this video. 
[http://www.youtube.com/watch?v=P0XbjUVES8Q Static Friction]

:::::<youtube>P0XbjUVES8Q</youtube> 

==Misconception #2: Friction always opposes motion ==
You will often see "friction always opposes the motion" on websites and even in textbooks. This is not true, if you just say it like that. What is true is that friction always opposes the sliding or the attempted sliding. 
The classic case for this is the tires of a car on the road, moving forward. First off, assume the wheels are not spinning: the friction is static friction, the rubber is not sliding on the road. 
If the road were covered in ice (no friction), how would the wheels spin? They would push back on the road. 
This means the road pushes forward on the wheels of the car (Newton 3). This is because friction always involves two surfaces and you have to 'frictionize' both surfaces. 
So what direction is the force of static friction on the car? It pushes the car forward. What direction is the car moving? Also forward. 
You can use similar arguments to show that the force of static friction pushes you forward when you walk forward.
 

==Case of Friction on an Inclined Plane==
Consider a block simply sitting on an inclined plane. If the plane were covered in ice, the block would slide down. Therefore, it is the force of friction, static friction, which is keeping it there. The acceleration is zero, so the net force on the block must also be zero. 
We can use a free body diagram and Newton's second law to determine the value of the force. 
 [[File:FrictiononanInclinedPlane.jpg|left|500px]] 
{| class="wikitable"
|-
! X Axis !! Y Axis
|-
| <math> \sum F_x = ma_x </math> 
<math> F_{staticfriction}-mgsin\theta = ma_x </math> 
But the acceleration is zero 
<math> F_{staticfriction}-mgsin\theta = 0 </math> 
<math> F_{staticfriction}=mgsin\theta </math> 

||
<math> \sum F_y = ma_y </math> 
<math> n-mgcos\theta = ma_y </math> 
But the acceleration is zero 
<math> n-mgcos\theta = 0 </math> 
<math> n=mgcos\theta </math> 

|}
 

Therefore the required force of static friction is <math> F_{staticfriction}=mgsin\theta </math> 
and the normal force is 
<math> n=mgcos\theta </math> 
You can see that these are dependent on the angle. The bigger the angle, the bigger the required force of static friction, and the smaller the normal force.
As shown in this diagram: 
 [[File:FrictiononanInclinedPlaneBigAngle.jpg|center|400px]] 
What happens if the angle gets too big? 
If the x component of the weight exceeds the maximum possible force of static friction, the block will start to slide, and the force of kinetic friction will be in play. 
If the block wants to slide up (e.g. because of an applied force or a previously applied force has made it move that way), the force of friction will be down the slope, if the block wants to slide down, the force of friction will be up the slope. 
Remember, the force of friction always opposes the sliding motion, or the attempted sliding motion.

==Examples==
Example 1: 
A block is sliding down an incline with angle <math> \theta</math>. The mass of the block is m. The coefficient of kinetic friction is equal to <math> \mu_k</math>. Find the acceleration of the block. (Ans:<math> a=gsin\theta-\mu_kgcos\theta</math>) 
Bonus: At what angle will the block slide with constant velocity?
 
[http://www.youtube.com/watch?v=FA3H0WKB3eo Video Solution]

:::::<youtube>FA3H0WKB3eo</youtube>

The Magnetic Field

2014-12-10T20:11:29Z

Kevin:

ClassNotes and Compiled Class Problems: 203NYB A14

2014-12-09T20:36:41Z

Kevin:

These are files used in class containing some notes, problems worked on in class. 
 
Smart Notebook files can be viewed and modified using [http://express.smarttech.com/# SmartNotebook Express] 

 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYB_A14_ElectricFieldWebwork.PDF 203NYB_A14_ElectricForceWebwork.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/Electric%20Field%20Webwork.docx 203NYB_A14_ElectricFieldWebwork.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYBA14_CapacitorStudentSolutions.pdf 203NYB_A14_CapacitorWebwork.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYB_RCCircuits.pdf 203NYB_RCCircuits.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYB_Kirchoff.pdf 203NYB_Kirchoff.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYBA14/203NYBA14MagneticInductionWebwork.pdf 203NYB_MagneticInduction.pdf]

Sources of Magnetic Field

2014-11-26T17:49:46Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/SourcesofMagneticFields/SourcesofMagneticFieldWorksheet_wclickerquestions.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/SourcesofMagneticFields/SourcesofMagneticFieldWorksheet_wclickerquestions.pdf Worksheet.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/11.pdf Additional Problems] 
 
 
Other Resources:
*Haliday & Resnick, Fundamentals of Physics 29.1-4
*[http://cnx.org/content/m42382/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_9.pdf?FCItemID=S0027D8BB Printable version]
 
We saw the way a magnetic field acts on charged particles in movement and on a current-currying wire or loop . But, we have not explained the origins of the magnetic field itself, yet. It’s important to remember that we cannot have a unique magnetic pole and the quantitative definition of a magnetic field makes use of “an electric charge in movement”. The first experiments were published by Oersted. Also, he was the first to prove experimentally that a carrying current wire exerts a force on another carrying current wire as if it were a magnet itself.
===CURRENTS PRODUCE MAGNETIC FIELD: A LONG, STRAIGHT WIRE===
You may verify easily, by use of iron particles, that a straight long wire carrying current creates a magnetic field with circular field lines around it. Also, by using a compass needle (fig.1), one may find that the field direction obeys the right hand curl rule (or bottle-cap rule): If the thumb is in the direction of the current , the curled fingers of the right hand indicate the direction of the magnetic field (see Fig.1 and the video below). The magnitude of the field vector is constant around a circle centered at wire and it depends only on the current in the wire I and the distance from it R. It can be shown that the magnetic field is given by the expression 
<math> B =\frac{\mu_0I}{2\pi R} </math> (1) 
where the constant <math>\mu_0 = 4\pi\times 10^{-7} Tm/A</math> is known as the permeability constant of free space. (free space =vacuum) 
Note: As the magnitude of the field decreases with distance it turns out that the density of field lines decreases with increase of distance, too.
 [[File:SourcesofMagneticField1.png|center|500x375px|alt=Figure 1]] 
:::::<youtube>jCcUochTbVo</youtube>

===MAGNETIC FORCE BETWEEN TWO PARALLEL WIRES===
Wires carrying current produce magnetic fields, but those magnetic fields can apply force on other current carrying wires. 
Consider two long straight wires carrying the currents <math>I_1</math> and <math>I_2</math> at a distance d from each other. The first wire creates a magnetic field with magnitude <math>B_1</math> at distance d ; it is perpendicular to the direction of current <math>I_2</math> passing thought the second wire. Referring to a length <math>l_2</math> on the second wire, you can show that the magnitude of the magnetic force exerted on this length is
 
<math>F_{21}=I_2l_2B_1= I_2l_2 \times\frac{\mu_0 I_1}{2\pi d} = l_2 \times\frac{\mu_0 I_1I_2}{2\pi d} </math>(2)
 
When considering the magnetic force exerted on the first wire by the magnetic field of the current in the second wire, you get 
<math>F_{12}=I_1l_1B_2= I_1l_1 \times\frac{\mu_0 I_2}{2\pi d} = l_1 \times\frac{\mu_0 I_1I_2}{2\pi d} </math> (3) 
Note that, if the currents have the same direction (fig. 2) the exerted forces are attractive and if the currents have opposite directions the exerted forces are repulsive. 
 [[File:SourcesofMagneticField2.png|center|500x525px|alt=Figure 2]] 
By comparing the expressions (2) and (3) you can see that the same force magnitude is exerted on the unit length of each wire 
<math>\frac{F}{l} = \frac{\mu_0 I_1I_2}{2\pi d} </math> (4) 
It is very interesting to mention that the definition of the current unit (the Ampere) is based precisely on this expression: The current through two parallel wires at a distance 1m is 1A when the magnitude of the magnetic force exerted on 1m length of each of them is equal to <math>2\times 10^{-7}</math> N.

==Examples==
:::::<youtube>NsIeoG_TbEI</youtube>
:::::<youtube>epLI3u9mJqw</youtube>

===BIOT-SAVART LAW FOR A CURRENT ELEMENT ===
The definition of magnetic field due to current in a long straight wire opened the way for finding a basic expression in calculation of magnetic fields created by different forms of wires (not only straight). As we know from the similar situations in case of electric field calculations, at first, one starts by the definition of field due to an infinitesimal element and then you follow up with an integral calculation. 
To make our lives easier we can define <math> \frac{\mu_0}{2\pi} \equiv k'</math>
to expression (1) and get it in form 
<math> B = \frac{2k'I}{R}</math> (5) 
Which, you will note, is a similar expression to that of the magnitude of the electric field from an electric charge uniformly distributed along a long straight wire <math> E = \frac{2k\lambda}{R}</math> (6) 
Without going into detailed explanations, we note that the symmetry between the magnetic and electric fields expressions is much more profound than it first appears. The electric and magnetic forces must fundamentally be the same! 
As a second step, note that the elementary field vector from an elementary charge dq = λdl is given by the expression 
<math>d\vec{E}= k\frac{dq}{r^2}\hat{r} = k\frac{\lambda dl}{r^2}\hat{r} </math> (7) 
<math> \hat{r} </math> is the unit vector directed from the charge “dq” to the point where the electric field is calculated. Then, by using integration along the wire you get equation(6) for the magnitude of the field. 
Following the same method, you can derive equation (5) for the magnitude of the magnetic field as the integral of the elementary magnetic fields vectors <math>d\vec{B}</math> due to elementary currents dI = Idl in the wire. When comparing the expressions (6) and (7) one observes that the distance appears as <math>1/R</math> in expression (6) and <math>1/R^2</math> at expression (7). 
So, due to symmetries between (5) and (6), one should expect a factor “ <math>1/R^2</math> “ for the elementary magnetic field expression as well. Also, similarly to “dq” in (7) you might expect proportionality to “ dI ”. 
Besides these similarities, you should remember that the directions of the E and B fields depend on different rules; <math> \vec{ E} \parallel \hat{r}</math> (see fig.3.a) but <math>\vec{B}</math> obeys the vector product rule (see fig.3.b).
 [[File:SourcesofMagneticField3.png|center|750x325px|alt=Figure 3]] 
Using these ideas you can get the general expression: 
<math> d\vec{B}= k'\frac{I\vec{dl}}{r^2}\times \hat{r} = \frac{\mu_0}{4\pi}\frac{I\vec{dl}}{r^2}\times \hat{r} </math>(8) 
In this expression, <math>\vec{dl}</math> is the infinitesimal length vector whose direction is defined by the current and <math>\hat{r}</math> is the unit vector directed from the elementary current to the point where the magnetic field is calculated. 
Equation (8) is known as the Bio-Savart law for the magnetic field due to an elementary current dI. 
From equation (8), you can see that the magnitude of the elementary magnetic field is 
<math> dB= \frac{\mu_0}{4\pi}\frac{Idl}{r^2} sin\theta </math> (9) 
Where <math>\theta</math> is the angle between the <math>\vec{dl}</math> direction and <math>\hat{r}</math> direction at the infinitesimal length location. 
Important note: The elementary current only exists in the physics world where it is very helpful for calculations. There is no way to isolate or measure an elementary current in the laboratory.
==== Calculating the magnetic field from a finite length wire carrying a current I====
[[File:SourcesofMagneticField4.png|left|500x325px|alt=Figure 4]] 
The set-up is shown in Figure 4. A short length of wire (Length=L1+L2) is carrying a current I. All the infinitesimal contributions of dB, at the point in question point P, are directed perpendicularly into the page at P, as determined by the right hand rule. The net magnetic field is the sum of all these contributions and therefore will also be directed into the page. To solve the problem you can divide up the wire into upper and lower parts. The division is defined by the perpendicular distance R. Using this artificial division, the net magnetic field at point P from the wire (fig.4) is constituted by two contributions; one due to the part of the wire below R and one due to the part above R. 
We will calculate each of these contributions and sum them up.That is to say, integrate them! 
So, let’s start by calculating the contribution from the lower half (see fig.3(b). The magnitude of dB (contribution of current element at distance “<math>l</math>” from the level of considered point) is 
<math> dB= \frac{\mu_0}{4\pi}\frac{Idl}{r^2} sin\theta </math> (8) 

From the figure 3.b you can see that <math>sin\theta = R / r</math> and <math>r = [l^2 +R^2]^{1/2}</math> 
So: 
<math> dB= \frac{\mu_0}{4\pi}\frac{Idl}{r^2} \frac{R}{r} = \frac{\mu_0}{4\pi}\frac{IRdl}{r^3}=\frac{\mu_0}{4\pi}\frac{IRdl}{[l^2+R^2]^{\frac{3}{2}}} </math> (10) 
Then, the contribution from the lower part of wire with length L1 is 

<math> B_{L_1}= \frac{\mu_0IR}{4\pi} \int^0_{-L_1}\frac{dl}{[l^2+R^2]^{\frac{3}{2}}} = \frac{\mu_0IR}{4\pi} \frac{1}{R^2} | \frac{l}{[l^2+R^2]^{\frac{1}{2}}}]^0_{-L_1} = -\frac{\mu_0IR}{4\pi R} \frac{L_1}{[L_1^2+R^2]^{\frac{1}{2}}} = -\frac{\mu_0I}{4\pi R} sin\alpha_1</math>(11) 
which means a magnitude equal to 
<math> |B_{L_1}|= \frac{\mu_0I}{4\pi R} sin\alpha_1</math> (12) 
You can get the direction from the right hand rule. 
Similarly, the contribution to the field from all elements in the upper part of wire is: 
<math>
B_{L_2} = \frac{\mu_0IR}{4\pi} \int_0^{L_2} \frac{dl}{[l^2+R^2]^{ 3/2 } }=
\frac{\mu_0I}{4\pi R} \frac{L_2}{[L_2^2+R^2]^{\frac{1}{2}}} =
\frac{\mu_0I}{4\pi R} sin\alpha_2
</math> (13) 
Finally, the net magnitude of magnetic field due to a finite length wire (<math>L=L_1+L_2</math>) is given by the expression 
<math> B_{total}= B_{L_1} +B_{L_2} = \frac{\mu_0I}{4\pi R} (sin\alpha_1 +sin\alpha_2 )</math> (14) 

====Magnetic Field from an Infinitely Long Wire====
Note that, for an infinite long wire, <math>sin\theta_1= sin\theta_2= sin90=1</math> and the expression (14) gives 
<math> B_{infinitely\ long}= \frac{\mu_0I}{4\pi R} ( 1+1 ) =\frac{\mu_0I}{2\pi R} </math> <div style="text-align: right; direction: ltr">(15)</div> 
Which is the same expression as equation (1).
 
:::::<youtube>WUpDMi50zPs</youtube>

==Example==
:::::<youtube>9nWTiPGjbjY</youtube>

Module : Conservation of Momentum

2014-11-26T17:42:02Z

Kevin:

''Kreshnik Angoni and Kevin Lenton''
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Impulse_Momentum/Module%20Momentum%20Worksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Impulse_Momentum/Module%20Momentum%20Worksheet.pdf Worksheet.pdf] 

Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 9.4-11
*[http://cnx.org/content/m42155/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00042F8A/FOV1-0006425E/FOV1-0006425F/LECTURE_11.pdf?FCItemID=S00282F7E Printable version]
 
===How to Save the Earth from Catastrophe===
http://static.ddmcdn.com/gif/asteroid-hits-earth-2.jpg 
Asteroids hit the earth every year, and it is just a matter of time before ‘the big one’ has a path close enough to the earth to cause concern. 
How would you save Earth from such an asteroid? 
The most obvious choice is to send a nuclear bomb up to the asteroid and explode it (as seen in the movie Armageddon). You hope that this big force applied for a short amount of time will be enough to change the orbit of the asteroid such that it misses the earth.
This probably will not work very well because the bomb will probably just split the asteroid up into smaller fragments – all still heading for Earth! 
Experts recommend, rather, shining a powerful laser onto the asteroid. The force from the laser onto the asteroid is small, but by applying it for a long time you can get the same effect of changing the path of the asteroid. 
'''The point is that applying a small force for a long time can have the same effect as applying a big force for a short time. The Force times Time is the same in the two cases.''' 
This value, the product of Force and Time is called the Impulse J. More correctly, it is the integral of force times time.
<math> \vec{ J } =\int \vec{ F }dt </math> 
Note that the direction of the Impulse <math> \vec{ J }</math> is in the same direction as the force <math> \vec{ F }</math>. 
If the force is constant, or assumed to be constant then: 
<math> \vec{ J } = \vec{ F }\times t</math> 
(Also just a quick note to point out that Work is the product of Force and Displacement, whereas Impulse is the product of Force and Time.) 

===Impulse is equal to the Change in Momentum, momentum is mass times velocity===
Starting with this definition we can use Newton’s Second Law to derive an expression for momentum: 
<math> \vec{ F_{net} } =m \vec{ a } </math> 
Multiply both sides by dt 
<math> \vec{ F_{net} }dt =m \vec{ a }dt </math> 
Take the integral of both sides, from the initial to the final condition 
<math> \int\vec{ F_{net} }dt =\int^{final}_{initial} m \vec{ a }dt </math> 
Rewrite using the definition of J: 
<math> \vec{ J_{net} } =\int ^{f}_{i}m \vec{ a }dt </math> 
Rewrite the acceleration in terms of velocity: 
<math> \vec{ J_{net} } =\int^{f}_{i} m \frac{d\vec{ v }}{dt}dt= \int^{f}_{i} m d\vec{ v } </math> 
therefore 
<math> \vec{ J_{net} } =m\vec{v_{f}}-m\vec{v_{i}} </math> 
This value <math> m\vec{v}</math> is called the Momentum and is given the symbol <math> \vec{p}=m\vec{v}</math> 
Therefore: 
<math> \vec{ J_{net} } =\vec{p_{f}}-\vec{p_{i}} </math> or <math> \vec{ J_{net} } =\Delta\vec{p} </math> 
That is to say Impulse is equal to the change in Momentum. 
The units of impulse and momentum are either: <math> Ns </math> (from the definition of Impulse), or <math> kg\frac{m}{s} </math> from the definition of momentum.
:::::<youtube>7lD0IveNnNQ</youtube>

===The Conservation of Momentum depends on the System===
Consider the universe as a closed, isolated system. Any force in the universe is internal to the universe and has an equal opposite reaction on some other object (Newton's Third Law). The force imparts Impulse to the object that it is acting on, but the reaction to this force (which is equal and opposite)produces an equal and opposite Impulse. The vector sum of these two impulses sum to zero. That is to say for every impulse a force gives to the universe, there is an equal and opposite impulse somewhere else. In other words all you are doing is transferring impulse from one object in one part of the universe to another object in another part of the universe. 
The net impulse is zero, and therefore the change in momentum is zero.
<math> \vec{ J_{net} } =\vec{p_{f}}-\vec{p_{i}} =0</math> 
or 
<math> \vec{p_{f}}=\vec{p_{i}} </math> 
That is to say Momentum is conserved for a closed system with no external Impulses. 
 
For a systme with external Impulses, the general equation, for a system smaller than the universe is 
<math> \vec{ J_{net} } =\vec{p_{f}}-\vec{p_{i}} \neq 0</math> 
That is to say the change in momentum is equal to any impulse coming in, or leaving the system.

===Momentum: What is it good for?===
Momentum is a very useful way of describing interactions, particularly collisions between objects. As we have seen, Conservation of Momentum is simply another way of describing Newton's Second and Third Law. 
Normally the forces between objects colliding, for example, are quite complicated, being dynamic and non-linear. You would need sophisticated computer models to predict how objects would move during the collision, if you just looked at the forces. 
However, we know that the momentum is conserved in an isolated system, so that the momentum before and after the collision is conserved. This means that you can use momentum to calculate the motion of the system components after the collision, without having to know the exact forces involved. Very useful!
 
Ultimately however, momentum is simply another way of describing the effect of force.

===Misconception #1: "Momentum is Inertia"===

Momentum is not inertia. The concept of inertia is really the concept of mass. Mass describes how an object accelerates given a net force. Therefore inertia describes how easy it is to ''change'' the velocity of an object. 
Momentum is the mass times the velocity, something quite different. Momentum is really only useful when looking at the change in momentum of a system. But, if it helps you, the momentum of an object can be thought of as a measure of how much impulse (a force applied for a length of time)is required to bring the system to rest.

===Misconception #2: "Momentum is always conserved"===
Momentum is only conserved when: 
<math> \vec{ J_{net} } = 0</math> 
That is to say the net force on the system is zero, and therefore no net impulses are acting from outside the system, the system is closed. 
This depends on the definition of your system! 
If the universe is the system, then yes, momentum is always conserved. Anything less, and you have to determine if the system is closed or not. If there ''are'' external forces to the system, adding impulse, you will have to use: 
<math> \vec{ J_{net} } =\vec{p_{f}}-\vec{p_{i}} \neq 0</math>

===Types of Collisions: Elastic and Inelastic===
The principle of linear momentum conservation is very general, it can be used to apply to any sort of force interaction. It applies in all fields of physics,
mechanical collisions, explosions, reactive motion, light emission/absorption, nuclear decay and nuclear
reactions. So, it is important to clarify some basic issues related to term “collisions”. 
Collisions can be classified by the energy transfer into or out of the system. 
A collision where no kinetic energy is lost during the collision is called a '''completely elastic collision'''. An example of an elastic collision would be two electrons colliding. The electrons do not actually touch and no energy is lost due to e.g. friction etc. 
A collision where maximum kinetic energy is lost during the collision is called a '''completely inelastic collision''' (Note: maximum is not the same as ''all''). An example of a completely inelastic collision is two cars colliding and '''sticking together'''. The cars lose energy due to the crumpling of the car body, sound etc. Any collision when the objects stick together is completely inelastic. 
In reality, most collisions are somewhere in between: some energy is lost, but the objects do not stick together. 
You can know whether a collision is elastic or inelastic by comparing the total kinetic energy of the system before and after the collision.

===Examples===
Example Impulse and Momentum 
An 800g baseball travelling with a velocity equal to 5m/s hits the ground at an angle of 40 degrees to the horizontal. The ball then rebounds at an angle of 40 degrees with the same speed. The ball is in contact with the ground for 0.006s. 
Find the impulse of the ground on the ball. 
Find the average force of the ground on the ball. 
:::::<youtube>qWvSdw170ik</youtube>
 
Example: 1D Conservation of Momentum: Elastic Collision 
Two identical 1.50-kg masses are pressed against oppo- site ends of a light spring of force constant 1.75 N/cm, compress- ing the spring by 20.0 cm from its normal length. Find the speed of each mass when it has moved free of the spring on a frictionless horizontal table.
:::::<youtube>MK8l47daHLk</youtube>
Example 2D Momentum: Inelastic Collision 
At the intersection of Ste.Croix Avenue and Cote-Vertu Blvd, a yellow subcompact car with mass 950 kg traveling east on Cote-Vertu collides with a red pickup truck with mass 1900 kg that is traveling north on Ste. Croix and has run a red light. The two vehicles stick together as a result of the collision, and the wreckage slides at 16.0 m/s in the direction 24.0° east of north. Calculate the speed of each vehicle before the collision. The collision occurs during a heavy rainstorm; you can ignore friction forces between the vehicles and the wet road.
:::::<youtube>2yq9tvgfYwc</youtube>

Magnetic Induction

2014-11-22T21:01:23Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.pdf Worksheet.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/12.pdf Additional Problems] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/13.pdf Additional Problems2] 
 
 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 30.1-5
*[http://cnx.org/content/m42390/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_10.pdf?FCItemID=S0027DBAB Printable version]
 
===Learning Objectives===
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand the conditions required for magnetic induction.
* Know the equations to calculated induced potential, and how to apply them.
* Know to avoid some common misconceptions.
* Apply the concepts to some induction problems.
 
===Some Experimental Observations===
Just as currents produce magnetic fields, changing magnetic fields can produce currents in conductors. 
Look at these videos to see some of the amazing effects of magnetic induction. 
<youtube>txmKr69jGBk</youtube> <youtube>NqdOyxJZj0U</youtube>

===Beginning to Build a Model of Magnetic Induction===
A conductor rod that slides at constant speed
on two conducting rails terminating in a galvanometer is part of a closed passive circuit (without a potential difference); generally, there is no current in the circuit. But, if the rod moves inside a magnetic field, the galvanometer G shows a current passing through the circuit (fig.1) 
[[File:MagneticInduction1.png|left|500x375px|alt=Figure 1]]
So, “a current appears in a conductor moving inside a magnetic field ”. Actually, this is quite expected from the effects of magnetic force. The free electrons inside the conductor move with velocity v with respect to the magnetic field B . 
The magnetic force <math> \vec{F_B} = -e \vec{v} \times \vec{B}
</math> (1) sets the electrons in movement “down” and a current (in the opposite direction) passes through the circuit as long as the wire moves. 

Consider now the rod at rest and a current <math>I_2</math> passing through a second wire located close by as shown in figure 1. We know that this current must have a magnetic field around it. This field will also be in the area marked out by the circuit. Observations show that, starting from the moment t=0 (when the current starts in the second wire) and during a very short time interval ( <math>\Delta t</math> ) the galvanometer needle moves; so “a current appears in the circuit”. Also, for longer times (steady state) t > t+<math>\Delta t</math>, the current in the circuit returns to zero. This was a new effect observed for the first time by [http://en.wikipedia.org/wiki/Joseph_Henry Joseph Henry]. 
It is clear that, in the two cases, the appearance of an electric current in a circuit is due to “some kind of action from the magnetic field”. Before going into a better understanding of these phenomena, keep in mind that the current means charges in movement and this requires the presence of an electric field. 
So, this action of the magnetic field is consistent with the creation of an electric field inside the circuit wires. Electromagnetic induction is the induction of a current inside a circuit due to a magnetic field. 
=====Definition of Magnetic Flux=====
From an electrical point of view, the fact that a current passes through the circuit means that an emf (<math>\mathcal{E}</math> ElectroMotiveForce=Voltage=potential difference) is operating on the circuit. Let’s define the magnitude and the direction of this emf. 
Consistent observations show that the magnitude of this induced emf is related to the change of magnetic flux <math> \Phi_B</math> inside the closed circuit. The flux <math> \Phi_B</math> of a magnetic field B through a loop,(or a closed circuit) is defined as 
<math> \Phi_B= \vec{B} \cdot \vec{n} A= \vec{B} \cdot \vec{A} =BAcos\theta</math> (2) 
 [[File:MagneticInduction2.png|left|250x188px|alt=Figure 1]]
<math>\vec{A} =A\hat{n} </math> is the area vector of the loop. A is the magnitude of the area of loop and <math>\hat{n} </math> is the unit vector of the loop direction, defined as being perpendicular to the loop surface. Then, after aligning the thumb with the <math>\hat{n} </math> direction, the right hand rule gives the positive sense of the circulation <math> \vec{B} </math> in the loop along the direction of your curled fingers (=right hand curl rule). 
Basically Flux is a measure of the magnetic field going through a certain area. 

====Different Ways of Changing the Flux====
Note that for a given field strength <math> \vec{B} </math> , the flux <math> \Phi_B</math> can be changed by changing the loop area (fig 3a; the loop goes from a circle to an elipse) or by changing the angle (fig3.b). The rotation of the loop (changing the angle) is the situation generally met in practice (AC generators, electric motors,).
[[File:MagneticInduction3.png|center|1000x750px|alt=Figure 1]]

Also, you can change the flux <math>\Phi_B</math> by changing the magnitude of the <math> \vec{B}</math>vector. In fig.4, the magnetic field produced by the primary circuit passes through the secondary circuit. You can change the magnitude of the <math> \vec{B}</math> field inside the area of the secondary circuit by changing the current <math>I_{primary}</math> in the primary circuit.
 [[File:MagneticInduction4.png|left|250x188px|alt=Figure 1]]

=====Units of Magnetic Flux=====
The unit of magnetic flux in the SI system is the Weber (Wb). From
expression (2), you can find that 
<math> 1Wb =1T \cdot1m^2 </math> (3) 

In our introduction for flux, we were referring to a loop inside a

uniform magnetic field. In a more general situation, the magnetic field may change from one location to another inside the loop and the flux is defined as the sum of all local contributions. This means that the general formula for the magnetic flux through a real loop must be calculated as 
<math> \Phi_B= \underbrace{\oint\oint}_\textrm{CircuitSurface}\vec{B} d \vec{A} </math> (4) 
That is to the say the integral of all the contributions to flux across the area in question. Most problems however will have a uniform B field inside the loop. 

===Faraday’s Law: The Induced EMF is equal to the Rate of Change of Magnetic Flux===
Faraday introduced the mathematical definition for the induced emf. He found that the magnitude of induced emf is equal to the rate of change of the magnetic flux through the loop: 
<math>| \mathcal{E}| = |\frac{ d\Phi_B }{dt}</math>| (5) 
Applying the chain rule: 

<math> |\frac{ d\Phi_B }{dt} |= \frac{ d(BAcos\theta) }{dt} = \frac{ d(B) }{dt} Acos\theta + B\frac{ dA }{dt} cos\theta - BAsin\theta \frac{ d\theta }{dt} </math>(6) 
Faraday’s Law shows that there are three ways to induce an emf ( change B, change A, or change angle ). 
Important: In practice, usually you deal mainly with achange of the B field or the angle. 

===Lenz’s Law: The Induced EMF always Opposes Changes to Flux===
The direction of the induced emf is defined by Lenz's law. Lenz's law is a particular expression of action-reaction behaviour . It says that: ''While the exterior action is changing the flux, the induced emf in a circuit has a direction such that the related current produces a magnetic field which tends to keep the previous value of magnetic flux into the circuit.'' 
 [[File:MagneticInduction4a.png|center|1000x750px|alt=Figure 1]] 

In fig4a the external action is increasing the flux; the induced current has a direction such that the related field <math>B_{ind}</math> (directed upside) works against the exterior action (by decreasing the net field). In fig4.b the external action is decreasing the flux; the induced current has a direction such that the related field <math>B_{ind}</math> (directed downside) works against the exterior action (by increasing the net field). 
In fig4.c the system is in a stationary state; there is no flux change through the circuit (e = 0). In fig4.d the external field magnitude is decreasing (<math>\Phi_B</math> < 0) and the induced current direction is tending to compensate for external action by creating an “up” directed Bind. In fig4.e the external field is increasing (<math>\Phi_B</math> > 0) and the induced current direction is tenting to compensate for external action, by creating a
“down” directed induced magnetic field <math>B_{ind}</math> which tends to decrease the net magnetic flux. 
Now, let’s use the figures 4.c,d,e to include Lenz’s law into expression (5). At t = 0, there is a field <math> \vec{B} </math> through the circuit area. We chose the area vector A directed parallel to <math> \vec{B} </math> so that at t = 0 FB is positive. This way we fix the selection for the positive direction (fig4.c) of the current and emf around the circuit. Then, at a moment t = dt, the figures 4.d,e show that the sign of induced emf is the opposite of sign at flux change .FB. So, the Faraday-Lenz law has the form

<math> \mathcal{E} =- \frac{ d\Phi_B }{dt}
</math>
(7) 
In the case of a coil with N loops Expression (7) becomes
<math> \mathcal{E} =- N\frac{ d\Phi_B }{dt}
</math> (8) 
====Additional Points====

a) Where is the source of the induced emf located? In the case of electromagnetic induction, the emf source is the magnetic field itself. It is distributed along the whole of the circuit and its action is distributed around the whole circuit. 
b) Lines of a field are used to visualize the field direction and its magnitude (their density at a given location is proportional to the magnitude of the B field at this point) but they do not have any physical meaning. That is to say they are merely a tool to help describe the physics. 

Whereas the magnetic flux ''is'' a basic physical quantity at the origin of a natural phenomenon. 

c) The change of magnetic field B induces an electric field E which produces a current into the circuit. 
This current induces an additional magnetic field in the surrounding space which brings a change of magnetic flux and so on…. This short comment shows that there is a close relationship between the changes of magnetic and electric fields. In fact, electromagnetic induction has implications far beyond: It is at the origin of all types of electromagnetic waves (light, TV, radio, X-rays,etc.). 

===GENERATORS OF ELECTRIC CURRENT ===
 [[File:MagneticInduction5.png|left|500x375px|alt=Figure 1]]
Essentially, an electric generator is a coil with N turns that rotates at (constant) angular velocity. 
(<math>\omega</math> ; <math>\theta =\omega t</math>) inside a uniform magnetic field (Fig.5) and its function is based on electromagnetic induction. Assume that at t = 0 the plane of the coil is perpendicular to <math>\vec{B}</math> and also that the Area vector <math>\vec{A}</math> parallel (<math>\theta = 0</math>) to <math>\vec{B}</math> . Then, at t = 0 the flux through the coil <math> \Phi_B = N (\vec{B} \cdot \vec{A}) = NBAcos\theta =NBA</math> is positive and maximum. 

If the coil starts to rotate as shown, the angle between th two vectors <math> \vec{A} </math> and <math> \vec{B} </math> increases, the magnetic flux changes. At a moment <math> t\neq 0</math> the angle
between the vectors is <math>\theta =\omega \times t </math> and the flux is decreased to 
<math> \Phi(t)_B = N \vec{B} \cdot \vec{A} = NBA cos\omega t </math>
(9) 
The equation (8) shows that if the coil is induced the emf is 
<math> \mathcal{E} = -N \frac{d\{\Phi_B}{dt} = NBA \omega sin\omega t </math> (10) 

This positive emf produces a current which flows in the positive direction of the circulation in the coil as shown in the figure. 
From expression (10) you can see that the induced emf is maximum <math> \mathcal{E}_0= NAB\omega </math> (11) 
when <math> \theta = 90</math> i.e. when <math>\vec{A} </math> perpendicular to <math>\vec{B} </math> which means the coil plane is parallel to the <math>\vec{B} </math> field. 
Now we can rewrite the expression (10) in the form <math>\mathcal{E}=\mathcal{E}_0 sin\omega t </math> (12) 

This expression shows that the induced emf and consequently the current in the coil change their direction sinusoidally. So, an alternating sinusoidal emf (fig.6) (and current) is induced in the coil.
In a real AC generator the current goes out of coil into the external circuit by the use of a set of two brush contacts that slide on two rings which rotate together with the coil : the [http://en.wikipedia.org/wiki/Commutator_(electric) commutators](fig.7).
 [[File:MagneticInduction67.png|center|1000x750px|alt=Figure 1]] 

The commutator is a set of two half-rings welded to coil wires (from inside, Fig.8) and sliding over two fixed brushes (from outside). The commutator rotates with the coil. Let’s consider the moment when <math> \theta=90^\circ</math> (i.e. when <math> \mathcal{E} =+\mathcal{E}_0 </math>) and the current is sent outside by the left side brush, as presented in fig.8.

 [[File:MagneticInduction8910.png|center|1000x750px|alt=Figure 1]] 

When the coil is rotated by <math> 90^\circ</math> (i.e. <math>\theta </math> becomes <math> 180^\circ</math>) the direction of the current in the coil is inverted but at that moment the “ex-right side half ring” contacts the “ex-left side brush”. So, the brush on the left always receives current in the same positive direction and sends it on to the external circuit. This device transforms an alternating current into a direct current pulse (Fig.9). 
[http://en.wikipedia.org/wiki/Charles_Wheatstone Wheatstone] improved the quality of DC current (by smoothing the pulses see Fig.10) by using a system of multiple coils and commutators oriented at different angles. This improvement gives us the DC generators we use today.

=== MOTIONAL EMF: Induction by changing the Area ===
Let’s consider a conducting rod with length l moving at constant velocity perpendicular to a uniform magnetic field (Fig.11). Due to magnetic force ( <math> \vec{ F_B } =-e \vec{v} \times \vec{ B } </math>) the free electrons move to the lower end and leave a net positive charge at the upper end of rod. This situation builds up an electrostatic field <math> E_0</math> directed downward. The related electrostatic force acting upon the electron <math> \vec{ F_E } =-e\vec{ E_0} </math> 
is directed in opposite sense of <math> \vec{ F_B } (\vec{ F_E }\uparrow \downarrow \vec{ F_B } </math> 
The magnitude of electric force <math> \vec{ F_E} </math> F increases with the increase of charge concentration at rod’s ends.

 [[File:MagneticInduction11.png|left|500x375px|alt=Figure 1]]

After a short interval of time it becomes equal to magnitude of <math>\vec{F_B}</math> and the electrons stop moving any more versus the rod’s ends because the exerted net force becomes zero. In this way a steady state equilibrium is set up. Starting from this moment
<math> BE_0=evB\ \ \ i.e.\ \ E_0=vB</math> (13) 
At steady state equilibrium there is a potential difference between rod’s ends <math> V_b-V_a=E_0l =Bvl </math> (14) 
If referring to terminology used for electric circuits, one might have to say that the charge separation and the potential difference are built by the presence of an emf inside the rod. This kind of emf is known as motional emf. Since no current is flowing through the rod one finds easily that <math> \mathcal{E}=V_b -V_a = Bvl </math> (15) 

What is the source of motional emf? To answer this question one must look closer at the work done moving the electrons. There are two forces that act on the rod; a) the magnetic field and b) the external force that moves the rod right side. We will consider the action of each of them separately. 
a) The Magnetic Field Force 
Each electron has two velocity components; <math> \vec{v} </math> along the rod motion and <math> \vec{v_D} </math> (drift velocity) directed downward (fig.12). The magnetic field interacts with both velocities. 

The magnetic force on the electrons from the velocity of the rod is 
<math> \vec{F_{1_B}}=-e\vec{v}\times \vec{B} </math> 
This force gives a power “the work
per 1sec” of (<math> P = \vec{F} \times \vec{v_{net}} =\vec{F} \times ( \vec{v} +\vec{v_{D}})
</math>
)
<math>P_1= \vec{F_{1_B}}\cdot(\vec{v}+\vec{v_D}) = (-e\vec{v}\times \vec{B} ) \cdot(\vec{v}+\vec{v_D}) </math> (16) 
The magnetic force on edue to its drift motion is 
<math> \vec{F_{2_B}}=-e\vec{v}\times\vec{B} </math> and it would
Provide a power equal to 
<math> P_2=\vec{F_{2_B}}\cdot(\vec{v}+\vec{v_D})=\vec{F_{2_B}}\times\vec{v}=-(-ev_DB)v=(evB)v_D </math>
(17) 
 [[File:MagneticInduction12.png|left|500x375px|alt=Figure 1]] 
So, the net power provided by magnetic field on each electron is <math> P = P_1 + P_2 = -(evB)v_D+ (evB)v_D= 0</math>
(18) 

This result does make sense because the magnetic forces are perpendicular to the velocity and therefore perpendicular to the displacement at any moment. 
Hence, at any moment, their dot product is zero; so a magnetic force can never do work. 

b) The Force from the Moving Rod. 
This force <math>\vec{F_{RodonElectron}} </math> is exerted by the rod on the electrons and it has the same direction as <math>\vec{v} </math>. 
From Fig.12, you can see that the net force acting on each electron in the direction of the rod movement has two components
<math> \vec{F_{Total_{electron}}} = \vec{F_{RodonElectron}} + (-e\vec{v_D} \times \vec{B}) </math>
(19) 
Since the electron is moving at constant velocity right side (as a particle of the rod) it comes out that the net force exerted on it is equal to zero. So, one gets: 
<math> F_{RodonElectron}-ev_DB=0\longrightarrow F_{RodonElectron}=ev_DB </math> (20) 

If the conductor has “n” (electron density) free electrons for m3, a section “A” and a length “l”, it contains in total “n*A*l” free electrons. Since the same “part of rod force” is applied on each of
them it comes out that the net force applied on all free electrons in the rod has the magnitude 
<math> F_{rod} =nAl\cdot F_{RodonElectron}=nAl(ev_DB)=(nAev_D)lB=IlB</math> (21) 

This result is not “unpredictable” if we remember that this is the magnitude of force exerted by magnetic field “B” on the wire with length “l” carrying the current “I”. The external force must balance the magnetic force action so that the “wire can move at constant velocity”. 

From the energetic point of view you can conclude that, the work done by the external force goes into moving the rod (motion at constant velocity) and to producing the motional emf (15). 
Note that, although the magnetic force does no work by itself, it acts as a kind of intermediary agent that converts a part of the external ''mechanical work'' into ''electrical energy''. 
====Examples====
1. A magnet is poked North Pole first into a solenoid. Determine the direction in which the induced current flows and whether the magnet and solenoid attract each other or repel each other. 
:::::<youtube>Q_7ZjZqrN0k</youtube>
2. A Square coil is made up of 10 turns and has a side equal to 10cm. The coil is flattened by pulling on the sides in 0.2 s. The coil's resistance is 10 <math>\Omega</math>. What is the magnitude and direction of the induced current? 

:::::<youtube>RJZ67-aBND8</youtube>
3. A circular 20 turn coil has a diameter of 7.5 cm. It lies in a uniform 0.1 T magnetic field that points out of the page. It is rotated a 1/4 turn, from perpendicular to parallel to the field. 
a)How quickly must this be done such that the induced voltage is 1.5V and in what direction will the current flow? 
b)You have a power supply that can deliver 5A. How would you produce the magnetic field used in this problem?

:::::<youtube>mubvkqPjWUA</youtube>
4. A 1.5m bar is pulled to the right at a steady 5m/s perpendicular to a uniform 0.75 T magnetic field directed out of the page. The parallel metal rails are connect to a 25 <math>\Omega</math> resistor. You may ignore the resistance of the bar and rails. 
a) What is the force vector required to keep the bar moving at 5m/s? 
b) At what rate is energy dissipated in the resistor? 
c) On the picture indicate the positive and negative ends of the metal bar, and the direction that current moved in the resistor? 
d) What is the current in the resistor if the magnetic field is now at an angle of 70 degrees to the vertical? (the bar is still moving at 5m/s)
:::::<youtube>Re_sWrOvQZg</youtube>

Motion of a Charged Particle in a Magnetic Field

2014-11-22T20:59:27Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MotioninMagneticFields/MotioninMagneticFieldsWorksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MotioninMagneticFields/MotioninMagneticFieldsWorksheet.pdf Worksheet.pdf] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/10.pdf Additional Problems] 
 
 
Other Resources:
*Haliday & Resnick, Fundamentals of Physics 28.4,6-7
*[http://cnx.org/content/m42375/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_8.pdf?FCItemID=S0027D7F5 Printable version]
 
This situation happens frequently in nature(ex: deflection of high-energy charged particles around earth) and
is widely used in technology (ex: Confinement of particles in a plasma gas, deflection and focusing of charged
particles in an accelerator or electrons in an old fashioned TV CTR tube.)

===MOTION IN A UNIFORM MAGNETIC FIELD (mainly met in technological applications) ===

We will consider the case when the velocity of the particle is perpendicular to B separately to when the velocity is not perpendicular to B . 
====A. Particle velocity perpendicular to B (Fig.1).====
[[File:MotioninMagneticFields14.png|left|250x175px|alt=Figure 1]] 
Let’s take, for simplicity, the case when q is positive. 

The magnetic force exerted on the charge has a direction perpendicular to both <math>\vec{v}</math> and <math>\vec{B}</math>. 
Remember that when any force is perpendicular to the direction of motion <math>\vec{v}</math> you always get some sort of circular motion: centripetal force provides a centripetal acceleration: direction towards the centre of the circle. This is exactly the situation here. That is to say, whenever a charged particle enters a magnetic field it performs some sort of circular motion.
Let’s assume that the magnetic force is the only acting force on the particle (neglecting gravity). 
This force is perpendicular to the particle movement and it will play the role of
a centripetal force. Its magnitude is 
<math>|F_B| =qvB </math>(1) 

We know that this particle will move in a circle 
and from the second
law of Newton we can write: 
<math>\sum F_{radial}= ma_{radial}</math> 
<math> F_B= m\frac{v^2}{r}</math> 
<math> qvB= m\frac{v^2}{r}</math> 
<math> qB= m\frac{v}{r}</math> 

From the equation (2), one finds the circle radius as <math> r= \frac{mv}{qB}</math> 
If the circumference of the circle is <math>C=2\pi r</math>
and the particle makes a full revolution in one period T then: 

<math>v=\frac{distance\ for\ 1rev}{time\ for\ 1rev}= \frac{ 2\pi r }{T} </math> 
<math>T=\frac{2\pi r }{v}= \frac{ 2\pi m }{qB} </math> 
The frequency is the inverse of the period. Therefore: 
<math>f_C = 1/T= \frac{ qB }{ 2\pi m } </math> 

Note that in this type of situation: 

a) The particle speed is not affected by the magnetic field.(only its direction changes, not its magnitude) 
b) The rotation period and frequency does not depend on the particle speed. 
c) Particles with the same ratio (q/m) rotate with equal frequency fC. 

====Application: the Cyclotron====
This type of movement is characteristic of a special type of charge particle accelerator called
cyclotrons and the frequency of one revolution fC is known as cyclotron frequency. In this situation
the magnetic field does not perform work and the energy of the charged particle is not affected by the magnetic field. Meanwhile, the main duty of an accelerator is the increase of particles’
energy. In cyclotron devices one uses the electric field to increase the particle energy.

A Cyclotron device is constituted by two half cylindrical parts separated
by a small gap. A uniform magnetic field is directed perpendicular to the
surface of cylinders and an alternating potential difference is applied
between the two parts of cyclotron. The charged particles are produced
by an ion source at the centre of this system. The electric field in the
space between the two parts at centre accelerates the charged particle
first and gives them an initial velocity perpendicular to the magnetic
field. Once inside the cyclotron cylinder, there is no more electric force.
Rather the magnetic force makes the particle bend around. Once the particle finishes the half circle, the potential switches sign and accelerates the particle across the gap between two cylinders in the other direction. 

[[File:MotioninMagneticFields2.png|left|250x175px|alt=Figure 2]] Fig.2

So, each time the particle passes through the gap its kinetic energy increases by

<math>\Delta KE=qV </math> (6) 

and the radius of the circle increases (because the particle is now moving faster). When the particle reaches the desired energy it is moving around the outside edge of the cyclotron and it can shot out of the cyclotron, and used, for example to bombard a target. 

The fact that the particle rotates at all times with a constant frequency is a big advantage for cyclotrons. You just have to calculate fC for a given type (q/m value) and apply an alternating voltage with
a frequency equal to 2*fC . A modern cyclotron has a radius ~ 2m and is used mainly to accelerate
protons instead of electrons. This is because electrons have such little mass their velocity quickly approaches the speed of light and special relativity applies. Particularly, the mass increases. The immediate consequence is a change in period of rotation that requires a continuous modification of frequency for alternating the potential. We have the technology: The synchrotron is a more complicated device that adjusts the frequency of alternating potential for relativistic particles. 

====B: Particle velocity not perpendicular to B ====
[[File:MotioninMagneticFields3.png|left|250x175px|alt=Figure 3]] 
We consider again only positive charge. For any motion, you can break the velocity vector into components: perpendicular to B <math>\vec{v_{\perp}}</math> and parallel to B <math>\vec{v_{||}}</math>. The magnetic
Force perpendicular to B ( <math>F_B= q v_{\perp} B</math> ) produces circular movement with period <math>T= \frac{ 2\pi m }{qB} </math>, as seen above. 
The parallel component velocity produces no magnetic force: the
particle simply moves parallel to the field lines with this constant velocity (see Figure 3). During one period T
Fig.3 the particle moves along this direction by: 
<math>d= v_{||}T = v_{||}\frac{2\pi m}{Bq} </math> (7) 
So, the net movement of the particle follows a helical path, orientated with the magnetic field. 

===Non-UNIFORM MAGNETIC FIELD (particles in a plasma gas) ===

A plasma gas is a medium with very high temperature and any material used for its confinement
would be melted. So, one uses this type of magnetic fields to keep particles inside a restricted space.
These fields are known as “magnetic bottles”. As the radius of circular motion is <math>R\propto 1/B</math>
, it turns out that, R decreases in any region of space where B increases (observe close to the coils in fig.4). Also, due to the B-field orientation (at any point the direction is always tangent to the field line at that point) the magnetic force is directed towards the region of weaker field no matter what the position of the particle (“seeking the bottle center” at fig. 4). This means that there exists a force component directed in the opposite direction to the velocity. This component slows down the motion of particle along field direction and, if the magnitude of the velocity is not too large, it reverses the direction of motion; the particle
returns back to the “center of magnetic bottle”. In this way, once the particle is inserted in a magnetic
bottle it remains trapped there inside.

[[File:MotioninMagneticFields4.png|center|500x350px|alt=Figure 4]] 
Our sun is a good example of a plasma, and plasma cork-screwing along magnetic field lines is known as a solar flare. The Northern Lights are caused by charged particles (from the sun) corkscrewing along the magnetic field of the earth and hitting the atmosphere at the North (and South) Poles. 
<youtube>u_V1IFZYz6o</youtube> <youtube>knwiWm4DpvQ</youtube>

=== CHARGED PARTICLE MOVING IN A MAGNETIC FIELD AND AN ELECTRIC FIELD SIMULTANEOUSLY===
When a charged particle enters a space region where an electric field E and a magnetic field B superimpose, two forces are exerted on the particle; 
the electric force <math>\vec{F}_{el}= q\vec{E}</math> and 
the magnetic <math>\vec{F}_{B}= q \vec{v}\times \vec{B}</math> . 
So, the net force (known as Lorentz force) acting on the particle is 
<math>\vec{F}_{net}= q(\vec{E}+ \vec{v}\times \vec{B})</math>(8)

In general, it is difficult to predict the path followed by a charged particle under the effect of Lorentz force. The simplest cases concern two fields either parallel or perpendicular to each other. We will consider the case of two fields perpendicular to each other. Let’s select a reference system where the electric field is directed along the y axis; the magnetic field is directed along the z axis and the charged particle +q enters this region at the origin with velocity v directed along the x axis (see Fig. 5).
So; <math>\vec{E} =
E\mathbf{\hat{ \jmath } }\ \ \ \vec{B} =
B\mathbf{\hat{ k } }\ \ \ \vec{v} =
v\mathbf{\hat{ \imath } }\ \ \ </math>

[[File:MotioninMagneticFields5.png|center|500x350px|alt=Figure 5]] 

One can easily find that the electric and magnetic forces lie along the Oy axis but in opposite directions. 

So, the magnitude of the net force acting on the particle is <math>F_{net} =qE-qvB </math>(9) 

The special case where the net force is zero occurs for only one velocity, this happens when 
<math>F_{net} =qE-qvB=0 </math>(9) 
The magnitude of this velocity is 
<math>v =\frac{E}{B} </math>(9) 

Note that, if a parallel beam of particles with exactly this velocity enters this region, all the particles follow
their original direction and are not deviated from their initial direction (Fig. 5). If the beam contains particles with different
velocities, only those that fulfill the condition (9) will go out in the same direction. The others particles
will deviate from their initial direction. 
This is how a velocity selector works. It uses this effect to select particles with a particular velocity, determined by the chosen values of E and B. 
====The Mass Spectrometer====
The mass spectrometer is a device that contains a velocity selector and analyses the abundance
of different isotopes of the same element in a sample. At first, the sample atoms are ionized (positively).
Then, these charged particles (ions) are directed through a set of slits S1, S2, which collimates the beam (shapes the beam to a nice and sharp line). 
This collimated beam is sent into a velocity selector ( <math>\vec{E} \perp
\vec{B}</math> ), which selects those particles with velocity <math>v=E/B_1</math> . Then, the ions enter a second section where a magnetic field
directed perpendicular to their velocity exerts a centripetal force 
<math>F_{B1} =qvB_1 = m\ a = m\frac{v^2}{R} </math>(11) 

These ions follow circular paths with radius 
<math>R = \frac{m}{q}\frac{v}{B_2}=\frac{m}{q}\frac{E}{B_1B_2}</math>(12) 
As the different isotopes have different masses (but q is the same for all ions) the radius is different for
each isotope and they hit the detector (photographic plate or electronic counter) at different positions.
So, based on density of traces or the records’ intensity one calculates the relative abundance of different

isotopes in the sample under study . Figure 6 shows a schematic of a mass spectrometer.
[[File:MotioninMagneticFields6.png|center|500x350px|alt=Figure 6]] 
=====Examples=====
:::::<youtube>c2GdOSFMNkg</youtube>
:::::<youtube>YCNi1J2Mdso</youtube>

Module : Conservation of Energy

2014-11-16T23:38:09Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.pdf Worksheet.pdf] 

 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 7.1-7,9
*[http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.28:45/College_Physics Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00042F8A/FOV1-0006425E/FOV1-0006425F/LECTURE%2010.pdf?FCItemID=S002829EF Printable version]
 
===What is the total energy of the Universe?===

Consider the universe as a closed, isolated system. Any force in the universe is internal to the universe and has an equal opposite reaction on some other object (Newton's Third Law). That is to say the amount of work that the force does on the universe has an equal and opposite work done somewhere else. In other words all you are doing is transferring work from one object in one part of the universe to another object in another part of the universe. You can see the concept of energy conservation beginning to develop. 
The positive mass-energy and kinetic energy of the universe is canceled by the negative gravitational potential energy (defined below). 
 
This implies that the total energy of the universe is perhaps [http://en.wikipedia.org/wiki/Zero-energy_universe zero], although that is debatable. 
But it also implies that whatever the total energy is, it is constant unless inputs of work either come in or leave the universe system. 
 
These concepts of ''system'', and ''potential energy'' are crucial to the concept of Energy Conservation. 

==CONSERVATIVE FORCES, A precursor to Potential Energy==

Some forces posses a specific and useful quality: the work done by these forces equals zero whenever the forces move along a closed
path ( the endpoint of path is the same as its initial point). That is to say the work done is ''path independent'', it depends on the final and initial positions, but not on the path taken between these points. 
These are known as conservative forces. 
Note that a
conservative force does ''non-zero work'' whenever the final position is different from the initial position. For instance, in Fig. 4 the force moves from point “1” to point “2” and does work <math>W_{1-2} </math>.
If the force then moves back from “2” to “1” <math>W_{1-2} </math> (fig.4) then 
<math>W_{2-1} = - W_{1-2} </math> 
so that <math>W_{1-1} = W_{1-2} + W_{2-1} = 0</math>. 
If you do a certain amount of work in one direction, and get that work back '''exactly''', going back in the opposite direction to the initial point, the force is essentially storing that energy for you. The energy is being conserved because the force is conservative. 

Three main examples of such conservative forces are the gravitational, restorative (elastic, or spring) and static electric forces.
 
The work done by these forces is described as a Potential Energy, and is treated differently than non-conservative forces in conservation of energy problems, because you can get that energy back. Every conservative force will have Potential Energy. 
====Take Home Message====
In the case of a conservative force applied on an object, it turns out that: 
The work done by a conservative force depends only on its
initial and final positions and not on the shape of the path from the initial to final location.

===What about non-conservative forces?===
Most forces are non-consevative. Any force that is directly pulling or pushing, e.g. friction, or the force of a hand is an example of a non-conservative force. The work done does depend on the path taken. 
The net work done by a force of kinetic friction force when an object slides on the floor is negative even when object is moved
along a closed path, because the friction removed energy from the object all along the path. 
The normal force is always perpendicular to displacement 1
and its work is always zero, no matter what is the path. Figure 4
Forces that depend on the velocity (the drag force in fluids, the magnetic force,..) are not conservative forces.

==INTRODUCTION TO POTENTIAL ENERGY ==

Consider an object at rest on the floor. As long as the application point of exerted forces ( FG , N ) does not move no work done. 
If you move the object to a new height ‘h’ and release it , the object
will fall, i.e. it will move vertically down back to h = 0. 
The gravity force will do (positive) work on the block because the force (the weight) is moved by a distance s (=h) (Fig.1) is shifted along weight direction by s . 
<math>W_G =
F_G s =
mgh \quad\quad\quad(1) </math>
 
You can place the object at the height ‘h’ by hand or by throwing it upward with the
right initial velocity, or any other method you want. No matter how the object gets to the height ‘h’, once there,
“it has the capacity to produce mechanical work”. This is entirely due to its height, because
the object now possesses a potential mechanical energy. This type of energy is due to the
weight force which is due to the gravitational attraction between the earth and object. When you move an object up, the earth also moves.
So, when talking about potential energy, the system has to include both the earth and object. 
In more general terms: 

Any kind of energy which is due to the configuration of a system, i.e. its position, is called potential energy.

==Elastic Potential Energy: The Block-spring system==
If you extend a spring by a distance “x”, the restoring force produced by the spring will obey Hooke's Law <math>F_{elastic}=-kx</math> and will

be directed towards the equilibrium position. When the block is

returned to the unstretched or equilibrium position, the elastic (or restoring)
force has done positive work (see previous wiki on work): 

<math>
W_{elastic} =
F_{ elastic} \cdot s =
(k / 2)(x^2 -
0^2) =
\frac{1}{2}kx^2 \quad\quad\quad(2)
</math>
 
If the spring is at rest, at its equilibrium(x = 0), it does not
Figure 2

s

possess any capacity to produce work. But, if it is extended

or compressed (x.0), the spring “is able” to produce work “on the block”; this means that it does possess
energy. Its capacity to produce work depending only on its configuration (extended or compressed); so this is a
potential energy. The two main types of mechanical potential energy are: gravitational and elastic.

In both cases, initially, an external force (its source is not part of system; like hand’s force) does a positive
work Wext on the system and shift it from an initial configuration ‘i’ to a final configuration ‘f ’. Then, just
because of being at “f ” configuration the system can provide work and the amount of work it can produce
(its capacity for work production) depends only on system configuration. This means that we may calculate
this capacity by use of a configuration function1; let’s call it U = Uconf. If we refer to values of this
function at initial-final configurations Ui and Uf and remember that the positive external work Wext brings
the system from state “i” to state “f ”, the simplest logical relation between U and Wext would be
W =Uf -U =.U (3) where Uf >Ui

ext i

E

This relation fits perfectly with the logic that the energy of the system increases when an
external force achieves positive external work (Wext > 0) on it (see figure 3). Actually,
this
definition requires that all the exterior work go only for configuration changes and not for
the change of kinetic energy of system parts. So, the system must be moved from the
initial configuration (Ui ) to the final configuration Uf by a constant speed (in practice

Figure 3

1 Mathematical function that depends only on the location (coordinates) not on velocities or accelerations

Uf
Ui
0
Wext

- To avoid the ambiguity related to the external force, one prefers to tie the definition of potential energy to
the work by internal forces. The third law tells that during system transfer from Ui to Uf , the work by
internal forces Wint = - Wext. From relation (3), one gets to the basic definition for potential energy as
-W =.U or .U =
(U -
U ) =-W (4)

intfi in

-Note that the equation (4) is based on the difference .U = Uf - Ui and not on the U- values.
This means that only .U has physical meaning (not U values). This definition for U leaves “free choice”
for the selection of configuration where U = 0. In practice, one fixes Ui = 0 to an initial configuration
which makes easier the solution of the considered problem. Then, considering the system shifted from
Ui = 0 to Uf , it comes out that (Uf -
0) =-Win . So, in practice, one starts by fixing u=0 to a given system
configuration and the calculate the potential energy function as
Uf = U = - Wint (5)

Example: One selects Ui = Ufloor = 0 when studying the displacement by “h” of an object from the floor up
and Ui = Uground = 0 when shifting it from the ground level up. In both cases, the system is the same “earth

object”, the internal force is the weight and W =
W =
F
.
*
.
s =-mgh . So, UG = - (-mgh) = mgh

int GG

In case of the “spring- block” system, one selects Ui = 0 for spring at equilibrium (x = 0) and the Hook’s

2

22

fx x

force is the internal force. Then Wint =Wel =-kx

=-k and Uelastic =-Wint =
k
22 2

To define a potential energy U(x); define the system; chose a coordinative frame; note the location where
U = 0; -calculate the work (Wi) by the internal force from there till a location “x” and use relation (5).

NOTES: A potential energy is due to the interaction between system’s constituents.

-So, it does not make sense to talk about potential energy of a single object or particle alone. When one
says that the potential energy of an object with mass ‘m’ at height ‘h‘ is U = mgh , actually, this means the
energy UG of the ‘system object – earth due to their gravitational interaction.’
-The mechanical potential energy is due to gravitational and restoring (elastic) forces.
- From a general point of view, one may define a potential energy only if the internal force at origin of this
energy is conservative. In the following we explain the meaning of a conservative force.

b) One can define a potential U(x,y,z) (or potential energy) which is a function of space coordinates.
The physics history showed that this function is very useful for solving difficult problems.

-Remember that a mechanic force is produced by an object “source” and is applied on another object. So,
we may talk always about the system of two objects. Meanwhile, only if this force is conservative, one can
use efficiently the concept of system and define the potential energy of the system (due to this force).

Related Notes:

a)
Any two or more objects undergo their gravitational attraction forces. As gravitational forces are conservative,
one may always define a system and a gravitational potential energy for this system. If an elastic (restoring) force
acts over one of these objects, this is an additional conservative force. So, one includes the source of this force
into the system and adds the corresponding PE term to potential energy of system.

b) If one of contributing forces is much bigger than others, at a first approximation, one may consider only the major
term of the potential and neglect the smaller ones.
c) As non-conservative forces cannot provide a potential function their source is not considered as part of a
system and non conservative forces are considered as external forces to the system.

3]
INTERNAL FORCES. TOTAL MECHANIAL ENERGY OF AN ISOLATED SYSTEM

-If one has defined a simple system (one object & one source) in the upper terms, the source of the considered
conservative force is part of the system and the force is an internal force of this system . Then, one may fix
the frame origin at “source”, select Ox axe along the direction “source - object” and find out U-expression.
This potential energy function U(x) depends only on x-coordinate and, one may show that, the internal
force acting on the “object”, can be derived from the potential function U(x) as

dU

f =-
(6)intdx
The U-function (potential energy) is a physical parameter of the system; very often it is noted PE.
(If the “system” contains several “sources” and “objects”, the potential function will depend on a set of coordinates and there will
be several internal forces.)

Exemple: An aeroplane with mass m is flying at height y from earth surface. The gravitational force acting
on the plane (weight) is a conservative force. The source of the weight is the earth. We define the system
earth-plane. We select Oy-axe along vertical and the origin O earth surface (or its center).
The potential energy of this system (we are used to call it aeroplane potential energy) is U =
E =
m * g * y

p

dU

Then, the plane weight (internal force of system) can be calculated by use of formula f =-=-mg

dy

The “–“ shows the weight direction opposite to selected Oy positive direction.

-Assume that only a single force is applied on an object and it is conservative. So, we can define a system,
tie a reference frame Ox to the source of force and calculate its potential energy U(x). Next, we calculate the
kinetic energy of the object K (with respect to this frame) and apply the work-energy theorem over the object.
Wnet = Kfin - Kin = .K and referring to a small shift “dx” we get the differential form dWnet = dK (7)

As dWnet = dWint + dWext = dWint (because dWext = 0) and dWint = fint*dx we get

dU

te (8)

dWnet =
dWint =
fint * dx =-
* dx =-dU =
dK .
dK +
dU =
d(K +
U ) =
0 .
K +U =
c

dx

So “the sum of kinetic energy K and potential energy U of the “object” remains constant in time”
The potential energy concerns the object and the “source” of force. As the kinetic energy of the “source” is
zero (reference frame is tied to it), it comes out that the kinetic energy of the system is equal to that of the
object under study. So, the sum K + U presents actually the total mechanical energy Emech (or ME) of the
system and the expression (8) tells that

ME=KE+PE= const or Emech = K + U = const (9)

Remember that we assumed no action from outside, i.e. an isolated system. So, the expression (9) shows
that the total mechanical energy of an isolated system is conserved.

4] CONSERVATION OF MECANICAL ENERGY FOR NON ISOLATED SYSTEMS

How to deal with situations where any kind of forces (conservative and non-conservative) act on the same
object while it is shifted from location “1” to location “2” of space ?

Step_1 Identify the conservative forces(weight, restoring) acting on the objects under study.

Step_2 Define a potential function for each conservative force. Chose the reference frame and define
clearly(write) where each potential is zero. Take their sum as common U potential for system.

Step_4 Divide the set of all acting forces on the object into internal (that do contribute to total U) and
external (do not contribute to U).

Step_5 Note that if there are no external forces the total mechanical energy of the system
remains constant; Emech-fin = Emech-in.

So, if the external forces are doing a positive work Wext_net on the system (actually on “the objects”
which are part of the system) the total energy of the system will be increased by this amount and

Emech-fin = Emech-in + Wext_net or .Emech = Emech_fin – Emech_in = Wext_net (10)

Notes: - The relation (10) is valid even for the case of a negative work by external forces. In this case
it shows a decrease of total mechanical energy of the system.

-Expression (10) presents the general form of the principle of mechanical energy conservation.
It is valid for any kind of system (isolated Wext_net = 0 or not isolated Wext_net . 0).
REMEMBER: The kinetic, potential and total mechanical energies of a system are mathematical
functions defined at a reference frame.

-While the definition of kinetic energy does not need the concept of system, the potential and total
energy functions cannot be defined without going through system concept. The values of KE, PE,
ME functions do not have any precise physical meaning by themselves because they depend on the
choice of reference frame (the change of the reference frame changes the value of those functions). But, the change
of those functions has a very precise physical meaning. It is related to the external work achieved
on the system and this quantity does not depend on the selected frame for calculations. This is the
basic issue one must not forget when dealing with energy related problems.

-The zero value of U-function (PE) is chosen in such a way that makes easier the solution of problem.

Electromotive Force and Kirchoff's Rules

2014-10-18T17:15:38Z

Kevin: /* Examples */

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricCurrent/ElectricCurrentWorksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricCurrent/ElectricCurrentWorksheet.pdf Worksheet.pdf] 

 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/6.pdf Additional Problems] 
 
Other Resources:
*Haliday & Resnick, Fundamentals of Physics 27.1-9
*[http://cnx.org/content/m42354/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_6.pdf?FCItemID=S0027D196 Printable version]
 
==ELECTROMOTIVE FORCE==
The terminal of a charged battery contains stored electric charges that build an electric field in
the space around it. Once we fix the end of a conducting it wire to (+) terminal, this electric field
attracts the free electrons from the wire into the battery and a current is produced inside the wire. But
this is an instantaneous current because the wire gets quickly the potential of terminal and the charge
movement stops. If we fix the other end of wire to the (-) negative terminal, we close the circuit and after this the
current through the wire becomes steady. The steady current is due to the fact that there is a constant
potential difference between the two wire ends, all the time. 
Consider an electric circuit (Fig. 1) where the current I is steady and flows only in one direction;
we call this a DC current. The positive charges leave the higher potential (terminal +) and move
[[File:EMF1.png|left|250x175px|alt=Figure 2]]
through the resistor R toward the lower potential(terminal -). Once inside
the battery, the positive charges have to move from the lower to the higher
potential so that the dc current in the circuit be steady.
Note that the electrostatic field inside the battery has opposite direction to
their motion and its action would block the current. We say that the positive
charges overcome an electrostatic potential barrier (hill) inside the battery.
How does this happen? There is only one answer; the necessary energy is
provided by external (non-electrostatic) sources of energy and for a general
purpose battery the external source is a source of chemical energy.

==How a Battery Works==
Actually, this energy is related to a set of chemical reactions that happen inside a battery. In a lead–
acid cell, a PbO2 plate and a Pb plate are immersed into an aqueous solution of H2SO4. One part of
H2SO4 molecules dissociate under the action of water molecules into positive ions (H+) and negative
ions (SO4-). Due to respective chemical affinity the following reactions happen at interface of plates: 
At the PbO2 plate: <math>PbO_2 +4H^+ +SO_4^{2-}+2e^- \rightarrow PbSO_4 +H_2O</math>. 
2 electrons removed from the PbO2 plate; it becomes Terminal + 
At the Pb plate: <math>Pb +SO_4^{2-}\rightarrow PbSO_4 +2e^-</math>. 
2 electrons stored at the Pb plate; it becomes Terminal - 
[[File:EMF2.png|left|250x175px|alt=Figure 2]]
So, the first reaction “stores q= +2e” on plate “+”while the second
reaction “removes q = +2e” from plate “-“. Then, those charged plates
build (inside the battery) an electric field that pushes ions away from the
“chemically affine plate”. In an open circuit situation, the chemical
reactions at plates stop when the charge of plates gets such that the
related electric field blocks further movement of ions versus plates. 
For a lead battery, this situation corresponds to the potential difference 2.05V between the two plates
in open circuit. When the battery is switched into a closed circuit, the positive charges go via the
circuit into the terminal “-“ ( or negative charges into terminal “+”). This decreases the net charge of each
terminal and the related internal electric field. As consequence, inside the battery, other ions move
toward the plates and the rate of chemical reactions increases instantaneously. The moving ions
transport “+” charges toward “+” plate and “-“ charges toward “-“ plate. This is equivalent with the
current (direction of + charges) from the lower potential to the higher potential inside the battery (Fig.2). 
A dynamic equilibrium between the field and chemical reactions rate is built inside the battery due to
the principle of charge conservation condition: The magnitude of current must be the same outside in
circuit) and inside the battery. This way, the number of e- leaving Pb plate versus circuit each second
is equal to the number of e- stored into it chemically.
+
-
I
R through the resistor R toward the lower potential(terminal -). Once inside
the battery, the positive charges have to move from the lower to the higher
potential so that the dc current in the circuit be steady.
Note that the electrostatic field inside the battery has opposite direction to
their motion and its action would block the current. We say that the positive
charges overcome an electrostatic potential barrier (hill) inside the battery.
How does this happen? There is only one answer; the necessary energy is
provided by external (non-electrostatic) sources of energy and for a general
purpose battery the external source is a source of chemical energy.

The chemical reactions decrease the ions concentrations around plates and produce a
variation of ions’ concentration in the solution. The diffusion processes which tent to uniform the
concentrations bring other ions close to plates moving them “uphill the potential” and so on. 
Simply put, from current modelling point of view, the chemical source provides an external work
which moves the positive charges uphill the electric potential inside the battery.
The work provided by a non-electrostatic source to move the charge +1C from the lower potential to
the higher potential terminal is known as emf1 and presented by the symbol Ɛ.
When moving the charge q uphill the potential difference, the emf source achieves the work 
<math>W_{emf} = q \times \mathcal{E} </math>(3) 
==Definition of EMF==
Like potential difference, the emf is defined as the work done during the displacement of +1C
charge but one must not forget their different physical origin. The source of potential difference
is the electrostatic field while the source of emf is a non-electric phenomenon.
In an electric circuit, there is one emf, at least. It provides the charge distribution at the origin of the
electrostatic field which drives the current. Note that the source of emf is not always chemical. It may
be magnetic (electric generator), mechanical (Van de Graff generator), etc.. 
When a charged battery is part of an open circuit, there is a potential difference <math>V_{open} = (V_+ - V_- )</math>
between its terminals. In this case, if a charge +q is moved from “-“ plate to “+” plate, the non-electric
source emf provides the amount of work <math>W_{emf} = W_{ext} = q \times \mathcal{E}</math> which goes to increase the
electrostatic energy of battery by <math>\Delta U = q\times( V_+ - V_- )</math> 

<math>W_{ext} = \Delta U \rightarrow q \times \mathcal{E} = q (V_+ - V_- )\rightarrow \mathcal{E} = V_+ - V_- \rightarrow \mathcal{E} = V_{open} </math> (4) 

In a closed circuit like that in Fig. 1, the potential V applied between the ends of the resistor R is
expressed (Ohms’law) as <math>V = I\ R</math> and it is equal to that on battery terminals . So, considering an
ideal source (without internal resistance), one would find 
<math>\mathcal{E}=V = I\ R</math> (5) 
Equation (5) tells that, in a closed circuit containing an ideal source emf and a resistor, the potential
rise (Ɛ ) inside due to emf source is equal to potential drop (I*R ) through the resistor in the circuit.
A real battery in a closed circuit gets heated. This shows that it has a resistance-like behaviour, or in
other words it presents an internal resistance to the current. So, it comes out that a real source in a
circuit is equivalent to an ideal source plus a resistance (r) in series (Fig.3). In this case, one
[[File:EMF3.png|center|250x175px|alt=Figure 2]]
part of work Wext provided by emf goes to compensate for
potential drop through the resistance I*r and the remaining part
to pull “+” charges uphill the potential barrier inside the battery.
 <math>\mathcal{E} = V_{closed} + I*r\ \ \ \ and\ \ \ \ V_{closed} = \mathcal{E} - I*r </math> (6) 
 
So, the terminal potential difference (Vclosed) of a battery in a closed circuit is smaller than its emf Ɛ. 
Equation (4) shows that one may measure Ɛ by Vopen measurements. For a real battery in a circuit
with a resistance R, the equation (5) transforms to 
<math> \mathcal{E} - I*r = V_{closed} \equiv V = I*R </math>(7) 
and <math>\mathcal{E} = I*r + I*R =I* (r +R) \rightarrow I = \mathcal{E}/(r +R) </math>(8)
1 “Electromotive force”. This wrong nomination (because Ɛ is energy and not force) remained for historical purposes.
2 In an electrical circuit, one uses the notation Vab = Va - Vb , where Va > Vb

==KIRCHHOFF RULES==
There are two circuit-analysis laws that are so simple that you may consider them “statements of the obvious” and yet so powerful as to facilitate the analysis of circuits of great complexity. The laws are known as Kirchhoff’s Laws. The first one, known both as “Kirchhoff’s Voltage Law” and “The Loop Rule” states that, starting on a conductor , if you drag the tip of your finger around any loop in the circuit back to the original conductor, the sum of the voltage changes experienced by your fingertip will be zero. (To avoid electrocution, please think of the finger dragging in an actual circuit as a thought experiment.) 
 
 
==Kirchhoff’s Voltage Law (a.k.a. the Loop Rule)==
To convey the idea behind Kirchhoff’s Voltage Law, I provide an analogy. Imagine that you are exploring a six-story mansion that has 20 staircases. Suppose that you start out on the first floor. As you wander around the mansion, you sometimes go up stairs and sometimes go down stairs. Each time you go up stairs, you experience a positive change in your elevation. Each time you go down stairs, you experience a negative change in your elevation. No matter how convoluted the path of your explorations might be, if you again find yourself on the first floor of the mansion, you can rest assured that the algebraic sum of all your elevation changes is zero. 
 
To relate the analogy to a circuit, it is best to view the circuit as a bunch of conductors connected by circuit elements (rather than the other way around as we usually view a circuit). Each conductor in the circuit is at a different value of electric potential (just as each floor in the mansion is at a different value of elevation). You start with your fingertip on a particular conductor in the circuit, analogous to starting on a particular floor of the mansion. The conductor is at a particular potential. You probably don’t know the value of that potential any more than you know the elevation that the first floor of the mansion is above sea level. You don’t need that information. Now, as you drag your finger around the loop, as long as you stay on the same conductor, your fingertip will stay at the same potential. But, as you drag your fingertip from that conductor, through a circuit element, to the next conductor on your path, the potential of your fingertip will change by an amount equal to the voltage across the circuit element (the potential difference between the two conductors). This is analogous to climbing or descending a flight of stairs and experiencing a change in elevation equal to the elevation difference between the two floors. 
 
If you drag your fingertip around the circuit in a loop, back to the original conductor, your finger is again at the potential of that conductor. As such, the sum of the changes in electric potential experienced by your finger on its traversal of the loop must be zero. This is analogous to stating that if you start on one floor of the mansion, and, after wandering through the mansion, up and down staircases, you end up on the same floor of the mansion, your total elevation change is zero. This is another way of applying the conservation of energy: The change in gravitational potential energy is zero, if there is no change in vertical height. 
 
In dragging your finger around a closed loop of a circuit (in any direction you want, regardless of the current direction) and adding each of the voltage changes to a running total, the critical issue is the algebraic sign of each voltage change. In the following example we show the steps that you need to take to get those signs right, and to prove to the reader of your solution that they are correct. 
 
 
===Example===
 
Find the current through each of the resistors in the following circuit. 
 [[File:KirchoffsLawsFig1.png|center]] 
Before we get started, let’s define some names for the given quantities: 
 [[File:KirchoffsLawsFig2.png|center]] 
Each two-terminal circuit element has one terminal that is at a higher potential than the other terminal. The next thing we want to do is to label each higher potential terminal with a “+” and each lower-potential terminal with a “-”. We start with the seats of EMF. They are trivial. By definition, the longer parallel line segment, in the symbol used to depict a seat of EMF, is at the higher potential. 
 [[File:KirchoffsLawsFig3.png|center]] 
Next we define a current variable for each “leg” of the circuit. A “leg” of the circuit extends from a point in the circuit where three or more wires are joined (called a junction) to the next junction. All the circuit elements in any one leg of the circuit are in series with each other, so, they all have the same current through them. 
 [[File:KirchoffsLawsFig4.png|center]] 
Note: In defining your current variables, the direction in which you draw the arrow in a particular leg of the circuit, is just a guess. Don’t spend a lot of time on your guess. It doesn’t matter. If the current is actually in the direction opposite that in which your arrow points, you will simply get a negative value for the current variable. The reader of your solution is responsible for looking at your diagram to see how you have defined the current direction and for interpreting the algebraic sign of the current value accordingly. 
 
Now, by definition, the current is the direction in which positive charge carriers are flowing. The charge carriers lose electric potential energy when they go through a resistor, so, they go from a higher-potential conductor, to a lower-potential conductor when they go through a resistor. That means that the end of the resistor at which the current enters the resistor is the higher potential terminal (+), and, the end at which the current exits the resistor is the lower-potential terminal (-) of the resistor. [[File:KirchoffsLawsFig5.png|center]] 
Now let’s define some variable names for the resistor voltages: 
 [[File:KirchoffsLawsFig6.png|center]] 
Note that the + and – signs on the resistors are important parts of our definitions of <math>V_{R1}</math> and <math>V_{R2}</math>. If, for instance, we calculate <math>V_{R1}</math> to have a positive value, then, that means that the left (as we view it) end of <math>V_{R1}</math> is at a higher potential than the right end (as indicated in our diagram). If <math>V_{R1}</math> turns out to be negative, that means that the left end of R1 is actually at a lower potential than the right end. We do not have to do any more work if <math>V_{R1}</math> turns out to be negative. It is incumbent upon the reader of our solution to look at our circuit diagram to see what the algebraic sign of our value for <math>V_{R1}</math> means. 
 
With all the circuit-element terminals labeled “+” for “higher potential” or “–” for “lower potential,” we are now ready to apply the Loop Rule. I’m going to draw two loops with arrowheads. The loop that one draws is not supposed to be a vague indicator of direction but a specific statement that says, “Start at this point in the circuit. Go around this loop in this direction, and, end at this point in the circuit.” Also, the starting point and the ending point should be the same. In particular, they must be on the same conductor. (Never start the loop on a circuit element.) In the following diagram are the two loops, one labeled (1) and the other labeled (2) . 
 [[File:KirchoffsLawsFig7.png|center]] 
Now we write KVL (1) to tell the reader that we are applying the Loop Rule (Kirchhoff’s Voltage Law) using loop (1) , and transcribe the loop equation from the circuit diagram: 
 
KVL (1) 
<math>+ V_1 - V_{R1} + V_2 = 0</math>
 
The equation is obtained by dragging your fingertip around the exact loop indicated and recording the voltage changes experienced by your fingertip, and then, remembering to write “= 0.” Starting at the point on the circuit closest to the tail of the loop 1 arrow, as we drag our finger around the loop, we first traverse the seat of EMF, <math>V_{1}</math>. In traversing <math>V_{1}</math> we go from lower potential (-) to higher potential (+). That means that the finger experiences a positive change in potential, hence, <math>V_{1}</math> enters the equation with a positive sign. Next we come to resistor R1. In traversing R1 we go from higher potential (+) to lower potential (-). That’s a negative change in potential. Hence, VR1 enters our loop equation with a negative sign. As we continue our way about the loop we come to the seat of EMF <math>V_{2}</math> and go from lower potential (-) to higher potential (+) as we traverse it. Thus, <math>V_{2}</math> enters the loop equation with a positive sign. Finally, we arrive back at the starting point. That means that it is time to write “ = 0.” 
 
We transcribe the second loop equation in the same fashion: 
 
KVL (2) 
<math>- V_2 + V_{R2} - V_3 = 0</math>
 
 
With these two equations in hand, and knowing that <math>V_{R1} = I_1R_1</math> and <math>V_{R2} = I_2R_2</math>, the solution to the example problem is straightforward. (We leave it as an exercise for the reader.) It is now time to move on to Kirchhoff’s other law. 
 
 
 
 

==Kirchhoff’s Current Law (a.k.a. the Junction Rule)==

Kirchhoff’s junction rule is a simple statement of the fact that charge does not pile up at a junction. (Recall that a junction is a point in a circuit where three or more wires are joined together.) I’m going to state it two ways and ask you to pick the one you prefer and use that one. One way of stating it is to say that the net current into a junction is zero. Check out the circuit from the example problem:

 [[File:KirchoffsLawsFig8.png|center]] 
In this copy of the diagram of that circuit, I put a dot at the junction at which I wish to apply Kirchhoff’s Current Law, and, I labeled that junction “A.”

Note that there are three legs of the circuit attached to junction A. In one of them, current I1 flows toward the junction. In another, current I2 flows toward the junction. In the third leg, current I3 flows away from the junction. A current away from the junction counts as the negative of that value of current, toward the junction. So, applying Kirchhoff’s Current Law in the form, “The net current into any junction is zero,” to junction A yields: 

Kirchoff Current Law applied to point A 
<math>I_1 + I_2 - I_3 = 0</math>

Note the negative sign in front of I3. A current of – I3 into junction A is the same thing as a current of I3 out of that junction, which is exactly what we have.

The other way of stating Kirchhoff’s Current Law is, “The current into a junction is equal to the current out of that junction.” In this form, in applying Kirchhoff’s Current Law to junction A in the circuit above, one would write:

Kirchoff Current Law applied to point A 
<math>I_1 + I_2 = I_3 </math>

Obviously, the two results are the same.

These two laws give you a set of equations that you can use to solve for unknowns.

==COMBINATIONS OF RESISTORS==
A common question is to find “the equivalent resistor to a set of resistors into a circuit”. To find
the equivalent resistor group them into sets of resistors in series and resistors in parallel.
===Resistors in Series ===
When two resistors are connected to
the battery, the terminal difference of
potential (V) applies to the end points
(1,4) and the same current passes
through each resistor. Ohm’s law tells
that, following the current sense, the
potential drops by I*R1 between points
(1-2), remains constant in wire section
(2-3), drops by I*R2 in section (3-4). 
As the total potential drop (-V) between points (1-4) is the sum of two subsequent potential drops: 
<math>-V =-V_1+ ( -V_2 )\rightarrow V= V_1+ V_2= I \times R_1+ I \times R_2 =I * (R_1+ R_2 )= I \times R_{eqivalent}</math> (9) 
So, 
<math>R_{equiv}= R_1+ R_2 </math> . 
For N resistors in series one gets 
<math>R_{equiv}=\sum_{i=1}^N R_i=R_1+ R_2 + R_3...+ R_i</math> 

Remember that if <math>R_{largest}</math> is largest resistance in the set, <math>R_{equiv}>R_{largest}</math>. 

===Resistors in Parallel ===
Fig.9
In this case one applies the first Kirchhoff’s rule :
<math>I = I_1 +I_2 </math>(11)
Next, one expresses the current at each resistor through the potential drop at its ends (the same V) 
By substituting at (11) one gets 
<math>I = \frac{V}{R_{equiv}}= \frac{V}{R_{1}}+\frac{V}{R_{2}}=V\times (\frac{1}{R_{1}}+\frac{1}{R_{2}})\Rightarrow </math> 
<math>\frac{1}{R_{equiv}} = \frac{1}{R_{1}}+\frac{1}{R_{2}}</math> 
(13)
For a set of N resistors in parallel one gets the equivalent resistor value as 
<math>\frac{1}{R_{equiv}} = \sum_{i=1}^N \frac{1}{R_{i}}</math> (14) 
Remember that if Rsmall is smallest resistance of the set, Req < Rsmall.

==DIRECT CURRENT INSTRUMENTS==
===Current Measurement: The Ammeter===
The Ammeter is a device that measures the current passing by a given point of circuit. This
instrument has a very low resistance. One can measure the current by introducing the ammeter in
series with the circuit (Fig.10) at the point of interest. As its resistance is very low, its introduction in
circuit does not alter significantly the current in circuit. The found value of current is practically the
same as in the closed circuit without ammeter. 
[[File:EMF10.png|center|250x175px|alt=Figure 10]] 
'''Take Home Message: Ammeters have very low resistance. They must be put in series with the circuit because they do not affect the current. If they were put in parallel they would short out the circuit.'''
===Voltage Measurement: The Voltmeter===
The voltmeter is used to measure the potential difference between two points in a circuit. This
instrument has a very large resistance. You can measure the difference of potential between two points
by connecting the voltmeter in parallel to the circuit between these two points (Fig.11). 
[[File:EMF11.png|center|250x175px|alt=Figure 11]]
As the resistance of voltmeter is large, its introduction does not affect
significantly the current in circuit. Thus, the potential drop to the resistor R
(V=I*R) is essentially the same as if the voltmeter were not in circuit. 
'''Take Home Message: Voltmeters have a very high resistance, so practically no current flows through them. They must be put in parallel with the circuit.'''
====Example====
For an ideal source Ɛ =12V and R = 20Ω in circuit I = 12/20 =
0.6A. When using a voltmeter with R=10KΩ, 1/Req = (1/20) + (1/104) = 0.0501
Req = 1/0.0501 = 19.96Ω and I1=12/19.96 = 0.601A. As the current passing
through voltmeter is 12/10000 = 0.012A, it turns out that the current
through the resistance R is practically unchanged (0.601-0.012~0.6A).

== RC CIRCUITS==
RC circuits are so named because they have a resistor and and a capacitor in series. 
When a capacitor is connected to a battery in a circuit, the electric charge will be stored into its plates
until it gets to a certain maximum value <math>Q_{max}</math>. During this time a current will pass through the circuit and
a potential difference <math>V_C(t)= \frac{q(t)}{C} </math> exist between capacitor plates. 
Once the capacitor gets the charge
<math>Q_{max}</math> no more current pass through the circuit; the potential difference between its plates becomes equal
to ε and <math>Q_{max}=C\ \mathcal{E}</math>. 
Let’s see what way happens this process in the case of a circuit that contains an
emf source ε, a resistor R and a capacitor C (fig.14).
[[File:EMF14.png|center|250x175px|alt=Figure 14]]
In the following proofs for discharging and charging a capacitor in an RC circuit, the quantities that change with time are noted by lowercase letters and the ones that do not change with time are noted by uppercase letters.
===CHARGING A CAPACITOR===
By using the 2nd Kirchoff’s rule we get <math>\mathcal{E} -V_{C} -V_{R} =0</math> (21) 
where <math>V_C = q / C</math> is the potential drop through the capacitor and
<math>V_R = i\ R</math> is the potential drop through the resistor. 

We can rewrite eq.21 in the form 
<math>\mathcal{E} -\frac{q}{C} -iR =0</math> 
<math>C\mathcal{E} -q= RCI = 0</math> (22) 
Now<math> i = dq / dt </math>, so eq.22 takes the form 

<math>C \ \mathcal{E} - q = (RC)\frac{dq }{ dt }</math>. 
We transform it to 
<math>\frac{dq }{(C \ \mathcal{E} - q)} = \frac{dt }{(RC)}</math> 
 
Note that <math>C \ \mathcal{E}= Q_{max}</math> and also <math>dq = -d(Q_{max}-q)</math>, because <math>Q_{max}</math> is a constant.
 
Therefore: 
<math> - \frac{d (Q_{max} - q)}{( Q_{max} - q )}= \frac{dt}{ RC}</math> 

After substituting <math>Z = Q_{max} - q </math> and taking the integral of both sides 
<math>\int_{Z=Q_{max}}^{Z=Q_{max}-q} \frac{dZ}{Z} = -\frac{1}{RC}\int _0^t dt</math> 
Therefore: 

<math>ln (Q_{max} - q)-ln(Q_{max}) = -\frac{t}{RC}</math> 
<math>ln \frac{(Q_{max} - q)}{(Q_{max})} = -\frac{t}{RC}</math> 
<math> \frac{(Q_{max} - q)}{(Q_{max})} = e^{-\frac{t}{RC}}</math> 
<math> q = Q_{max}(1-e^{-\frac{t}{RC}} ) </math> 
Introducing the notation <math>\tau = RC</math> which is known as the “time constant” of the RC circuit, and substituting: 
<math> q = Q_{max}(1-e^{-\frac{t}{\tau}} ) </math> 
This charge function is shown in the Figure 15a. 
[[File:EMF15.png|center|1000x700px|alt=Figure 15]]
====Meaning of the time constant <math>\tau</math>====
At <math>t = \tau</math> the charge in capacitor plates gets to 
<math> q = Q_{max} (1-1/ e) = Q_{max} (1-1/ 2.73) = 0.63Q_{max}</math> 
Thus, just looking at the time constant can give an idea of how fast or slow the circuit will charge the capacitor. The longer the time constant, the longer the circuit takes to charge the capacitor.

====Current in a charging circuit====
From the expression <math> q = Q_{max}(1-e^{-\frac{t}{\tau}} ) </math> you can calculate the evolution of the current in the circuit. 
As i = dq/dt you can just take the derivative of the charging function to find the current: 
<math>i = \frac{dq}{dt} =\frac{d}{dt}Q_{max}(1-e^{-\frac{t}{\tau}} ) = \frac{Q_{max}}{\tau}e^{-\frac{t}{\tau}} = \frac{C\ \mathcal{E}}{R\ C}e^{-\frac{t}{\tau}} = \frac{\mathcal{E}}{R}e^{-\frac{t}{\tau}}</math> 
 
Initially, the charge rushes onto the capacitor, but as more charge accumulates, the current is reduced. The current will be maximum at t=0, and have a value of <math>I_{max} = \mathcal{E} / R</math> 
Therefore: 
<math>i = I_{max}e^{-\frac{t}{\tau}}</math> 
This function is shown in figure 15.b.

===DISCHARGING THE CAPACITOR===
You discharge a capacitor by removing the battery and shorting out the two ends of the circuit. This is shown in the Figure 16: 
[[File:EMF16.png|center|250x175px|alt=Figure 16]]

By using Kirchoff rule #2 for this circuit you get:
<math>V_C - V_R =0</math> (25) 
where <math>V_C =q /C</math> is the potential across the capacitor terminals and
<math>V_R =i\ R</math> is the potential across the resistor. 
You can rewrite eq.25 in the form <math>q/C - i\ R =0</math> 
Substituting i = −dq / dt into this equation:
<math>\frac{q}{RC} = -\frac{dq}{dt}</math> 
which can be transformed to 
<math>\frac{dq}{q} = -\frac{dt}{RC}</math> 
We take the integral of both sides 
<math>\int_{Q_{max}}^q \frac{dq}{q} = -\int_0^t\frac{dt}{RC}</math> 
and get
<math>ln(q)-ln(Q_{max}= -\frac{t}{RC}</math> 
<math>q = Q_{max}e^{-\frac{t}{RC}}</math> 
<math>q = Q_{max}e^{-\frac{t}{\tau}}</math> 

For t = τ we get q = Qmax / e = 0.37Qmax. The “ half-time” noted as <math>T_{1/2}</math> is the time it takes for the charge of the capacitor to get to half of its initial charge(q = 0.5Qmax); 
so, <math>\frac{Q_{max}}{2} = Q_{max}e^{( -T_{1/2} / \tau )}</math> 
<math>\frac{1}{2} = e^{( -T_{1/2} / \tau )}</math> 
<math>2 = e^{( +T_{1/2} / \tau )}</math> 
<math>T_{1/2} / \tau = ln2 = 0.693</math> 
We finally get <math>T_{1/2} = \tau \ 0.693</math>(28) 
====Current in a Discharging RC Circuit====
Taking into account that during the discharge of a capacitor the current in the circuit is I = - dq/dt 
<math>i = -\frac{dq}{dt}= -\frac{d}{dt}q = Q_{max}e^{-\frac{t}{\tau}}= \frac{Q_{max}}{\tau}e^{-\frac{t}{\tau}}</math> 
Using the same notation as for the charging situation, that is <math>I_{max} = \mathcal{E}/R</math>, 
<math>i = \frac{Q_{max}}{\tau}e^{-\frac{t}{\tau}}= \frac{C\mathcal{E}}{RC}e^{-\frac{t}{\tau}}</math> 
<math>i = I_{max}e^{-\frac{t}{\tau}}</math> 
Note that this equation is the same as that for charging the capacitor! 
Of course the current is opposite to the charging case, but the current function is the same. 

Note that for <math>t = \tau </math> we get <math>i(\tau) = I_{max} / e = 0.37I_{max} </math> and for <math>t = T 1/2</math>: 

<math>i(T_{1/2}) = I_{max}e^{-0.693}=0.5I_{max}</math>
 
The graph of the current in the circuit during
the discharge of capacitor is shown in Figure 17. [[File:EMF17.png|center|1000x700px|alt=Figure 17]]

===Examples===
====Kirchoff's Laws====
:::::<youtube>0vwe1XLZsCo</youtube>
:::::<youtube>YWAuoy_qleU</youtube>
====Combining Resistors====
:::::<youtube>YsvAtGHHHWI</youtube>
====RC Circuits====
:::::<youtube>9fRIW6ZN6n0</youtube>
:::::<youtube>C1zmDdPvzcM</youtube>

Kirchoff's Laws

2014-10-18T16:52:57Z

Kevin: /* Kirchhoff’s Current Law (a.k.a. the Junction Rule) */

''Calculus-Based Physics by Jeffrey Schnick as modified by Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricCurrent/ElectricCurrentWorksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricCurrent/ElectricCurrentWorksheet.pdf Worksheet.pdf] 

 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/6.pdf Additional Problems] 
 
Other Resources:
*Haliday & Resnick, Fundamentals of Physics 27.1-9
*[http://cnx.org/content/m42354/latest/-collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_6.pdf-FCItemID=S0027D196 Printable version]
 
There are two circuit-analysis laws that are so simple that you may consider them “statements of the obvious” and yet so powerful as to facilitate the analysis of circuits of great complexity. The laws are known as Kirchhoff’s Laws. The first one, known both as “Kirchhoff’s Voltage Law” and “The Loop Rule” states that, starting on a conductor , if you drag the tip of your finger around any loop in the circuit back to the original conductor, the sum of the voltage changes experienced by your fingertip will be zero. (To avoid electrocution, please think of the finger dragging in an actual circuit as a thought experiment.) 
 
 
==Kirchhoff’s Voltage Law (a.k.a. the Loop Rule)==
To convey the idea behind Kirchhoff’s Voltage Law, I provide an analogy. Imagine that you are exploring a six-story mansion that has 20 staircases. Suppose that you start out on the first floor. As you wander around the mansion, you sometimes go up stairs and sometimes go down stairs. Each time you go up stairs, you experience a positive change in your elevation. Each time you go down stairs, you experience a negative change in your elevation. No matter how convoluted the path of your explorations might be, if you again find yourself on the first floor of the mansion, you can rest assured that the algebraic sum of all your elevation changes is zero. 
 
To relate the analogy to a circuit, it is best to view the circuit as a bunch of conductors connected by circuit elements (rather than the other way around as we usually view a circuit). Each conductor in the circuit is at a different value of electric potential (just as each floor in the mansion is at a different value of elevation). You start with your fingertip on a particular conductor in the circuit, analogous to starting on a particular floor of the mansion. The conductor is at a particular potential. You probably don’t know the value of that potential any more than you know the elevation that the first floor of the mansion is above sea level. You don’t need that information. Now, as you drag your finger around the loop, as long as you stay on the same conductor, your fingertip will stay at the same potential. But, as you drag your fingertip from that conductor, through a circuit element, to the next conductor on your path, the potential of your fingertip will change by an amount equal to the voltage across the circuit element (the potential difference between the two conductors). This is analogous to climbing or descending a flight of stairs and experiencing a change in elevation equal to the elevation difference between the two floors. 
 
If you drag your fingertip around the circuit in a loop, back to the original conductor, your finger is again at the potential of that conductor. As such, the sum of the changes in electric potential experienced by your finger on its traversal of the loop must be zero. This is analogous to stating that if you start on one floor of the mansion, and, after wandering through the mansion, up and down staircases, you end up on the same floor of the mansion, your total elevation change is zero. This is another way of applying the conservation of energy: The change in gravitational potential energy is zero, if there is no change in vertical height. 
 
In dragging your finger around a closed loop of a circuit (in any direction you want, regardless of the current direction) and adding each of the voltage changes to a running total, the critical issue is the algebraic sign of each voltage change. In the following example we show the steps that you need to take to get those signs right, and to prove to the reader of your solution that they are correct. 
 
 
===Example===
 
Find the current through each of the resistors in the following circuit. 
 [[File:KirchoffsLawsFig1.png|center]] 
Before we get started, let’s define some names for the given quantities: 
 [[File:KirchoffsLawsFig2.png|center]] 
Each two-terminal circuit element has one terminal that is at a higher potential than the other terminal. The next thing we want to do is to label each higher potential terminal with a “+” and each lower-potential terminal with a “-”. We start with the seats of EMF. They are trivial. By definition, the longer parallel line segment, in the symbol used to depict a seat of EMF, is at the higher potential. 
 [[File:KirchoffsLawsFig3.png|center]] 
Next we define a current variable for each “leg” of the circuit. A “leg” of the circuit extends from a point in the circuit where three or more wires are joined (called a junction) to the next junction. All the circuit elements in any one leg of the circuit are in series with each other, so, they all have the same current through them. 
 [[File:KirchoffsLawsFig4.png|center]] 
Note: In defining your current variables, the direction in which you draw the arrow in a particular leg of the circuit, is just a guess. Don’t spend a lot of time on your guess. It doesn’t matter. If the current is actually in the direction opposite that in which your arrow points, you will simply get a negative value for the current variable. The reader of your solution is responsible for looking at your diagram to see how you have defined the current direction and for interpreting the algebraic sign of the current value accordingly. 
 
Now, by definition, the current is the direction in which positive charge carriers are flowing. The charge carriers lose electric potential energy when they go through a resistor, so, they go from a higher-potential conductor, to a lower-potential conductor when they go through a resistor. That means that the end of the resistor at which the current enters the resistor is the higher potential terminal (+), and, the end at which the current exits the resistor is the lower-potential terminal (-) of the resistor. [[File:KirchoffsLawsFig5.png|center]] 
Now let’s define some variable names for the resistor voltages: 
 [[File:KirchoffsLawsFig6.png|center]] 
Note that the + and – signs on the resistors are important parts of our definitions of <math>V_{R1}</math> and <math>V_{R2}</math>. If, for instance, we calculate <math>V_{R1}</math> to have a positive value, then, that means that the left (as we view it) end of <math>V_{R1}</math> is at a higher potential than the right end (as indicated in our diagram). If <math>V_{R1}</math> turns out to be negative, that means that the left end of R1 is actually at a lower potential than the right end. We do not have to do any more work if <math>V_{R1}</math> turns out to be negative. It is incumbent upon the reader of our solution to look at our circuit diagram to see what the algebraic sign of our value for <math>V_{R1}</math> means. 
 
With all the circuit-element terminals labeled “+” for “higher potential” or “–” for “lower potential,” we are now ready to apply the Loop Rule. I’m going to draw two loops with arrowheads. The loop that one draws is not supposed to be a vague indicator of direction but a specific statement that says, “Start at this point in the circuit. Go around this loop in this direction, and, end at this point in the circuit.” Also, the starting point and the ending point should be the same. In particular, they must be on the same conductor. (Never start the loop on a circuit element.) In the following diagram are the two loops, one labeled (1) and the other labeled (2) . 
 [[File:KirchoffsLawsFig7.png|center]] 
Now we write KVL (1) to tell the reader that we are applying the Loop Rule (Kirchhoff’s Voltage Law) using loop (1) , and transcribe the loop equation from the circuit diagram: 
 
KVL (1) 
<math>+ V_1 - V_{R1} + V_2 = 0</math>
 
The equation is obtained by dragging your fingertip around the exact loop indicated and recording the voltage changes experienced by your fingertip, and then, remembering to write “= 0.” Starting at the point on the circuit closest to the tail of the loop 1 arrow, as we drag our finger around the loop, we first traverse the seat of EMF, <math>V_{1}</math>. In traversing <math>V_{1}</math> we go from lower potential (-) to higher potential (+). That means that the finger experiences a positive change in potential, hence, <math>V_{1}</math> enters the equation with a positive sign. Next we come to resistor R1. In traversing R1 we go from higher potential (+) to lower potential (-). That’s a negative change in potential. Hence, VR1 enters our loop equation with a negative sign. As we continue our way about the loop we come to the seat of EMF <math>V_{2}</math> and go from lower potential (-) to higher potential (+) as we traverse it. Thus, <math>V_{2}</math> enters the loop equation with a positive sign. Finally, we arrive back at the starting point. That means that it is time to write “ = 0.” 
 
We transcribe the second loop equation in the same fashion: 
 
KVL (2) 
<math>- V_2 + V_{R2} - V_3 = 0</math>
 
 
With these two equations in hand, and knowing that <math>V_{R1} = I_1R_1</math> and <math>V_{R2} = I_2R_2</math>, the solution to the example problem is straightforward. (We leave it as an exercise for the reader.) It is now time to move on to Kirchhoff’s other law. 
 
 
 
 

==Kirchhoff’s Current Law (a.k.a. the Junction Rule)==

Kirchhoff’s junction rule is a simple statement of the fact that charge does not pile up at a junction. (Recall that a junction is a point in a circuit where three or more wires are joined together.) I’m going to state it two ways and ask you to pick the one you prefer and use that one. One way of stating it is to say that the net current into a junction is zero. Check out the circuit from the example problem:

 [[File:KirchoffsLawsFig8.png|center]] 
In this copy of the diagram of that circuit, I put a dot at the junction at which I wish to apply Kirchhoff’s Current Law, and, I labeled that junction “A.”

Note that there are three legs of the circuit attached to junction A. In one of them, current I1 flows toward the junction. In another, current I2 flows toward the junction. In the third leg, current I3 flows away from the junction. A current away from the junction counts as the negative of that value of current, toward the junction. So, applying Kirchhoff’s Current Law in the form, “The net current into any junction is zero,” to junction A yields: 

Kirchoff Current Law applied to point A 
<math>I_1 + I_2 - I_3 = 0</math>

Note the negative sign in front of I3. A current of – I3 into junction A is the same thing as a current of I3 out of that junction, which is exactly what we have.

The other way of stating Kirchhoff’s Current Law is, “The current into a junction is equal to the current out of that junction.” In this form, in applying Kirchhoff’s Current Law to junction A in the circuit above, one would write:

Kirchoff Current Law applied to point A 
<math>I_1 + I_2 = I_3 </math>

Obviously, the two results are the same.

These two laws give you a set of equations that you can use to solve for unknowns.

File:KirchoffsLawsFig8.png

2014-10-18T16:36:07Z

Kevin:

File:KirchoffsLawsFig7.png

2014-10-18T16:32:52Z

Kevin:

File:KirchoffsLawsFig6.png

2014-10-18T16:32:29Z

Kevin:

File:KirchoffsLawsFig5.png

2014-10-18T16:32:04Z

Kevin:

File:KirchoffsLawsFig4.png

2014-10-18T16:31:11Z

Kevin:

File:KirchoffsLawsFig3.png

2014-10-18T16:30:54Z

Kevin:

File:KirchoffsLawsFig2.png

2014-10-18T16:30:34Z

Kevin:

File:KirchoffsLawsFig1.png

2014-10-18T16:30:15Z

Kevin:

ClassNotes and Compiled Class Problems: 203NYA A14

2014-10-10T15:39:12Z

Kevin:

Electric Potential Energy and Electric Potential

2014-10-09T21:03:11Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricPotential/ElectricPotential%20Worksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricPotential/ElectricPotential%20Worksheet.pdf Worksheet.pdf] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/3.pdf Additional Problems] 
 
 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 21.1-5
*[http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.8:145 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_3.pdf?FCItemID=S00286123 Printable version]
 

Electric Potential and Electric Potential Energy are important extensions from the Electric Field, because they are scalar quantities, rather than vectors. For this reason they are very practical in examining the effects of charges on other charge. 
==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* define potential energy and electric potential for a test charge in a uniform electric field.
* calculate the potential energy and electric potential for a small set of point charges.
* calculate the force as the derivative of a potential energy function in one dimension.
* calculate the field as the derivative of a potential function in one dimension.
* recognize conductors as regions of uniform potential.
 

==A Short Review of Potential Energy (U)==
 
The potential energy can be defined for a system only if conservative forces act between its constituents. 
 
Conservative forces depend only on the position (not on speed or acceleration). 
 
Only the difference between two values of potential energy (not the values by themselves) has a physical meaning. i.e. it is the relative value, not the absolute value that is important. This allows us to choose the location where potential energy is zero such that it simplifies the calculations. (Example; you might choose U = 0 for an object on the earth’s surface but if the object can move only inside the lab you should choose U=0 at floor level and you could even choose U=0 at the centre of the earth or at infinity if this choice simplifies the solution of considered problem. You are free to choose this reference because it is the value of potential energy relative to this reference position that is important. 
 
The principle of energy conservation; 
<math>\Delta U+\Delta K = W_{ext}</math> tells that, provided that the change in kinetic energy <math> \Delta K=0</math> , the change of potential energy <math> \Delta U</math> is equal to the work done by external forces. 
 
Consider an object (m) at rest on the surface of the earth; U(0) = 0. If we move it up “ slowly “and leave it at rest on a table of height y (fig.1), the object receives potential energy U(y) = mgy. The external force1 (the force you apply on it to make it move) would do a positive work Wext which increases the potential energy of the object (more correctly of the earth-object system). 
<math> \qquad\qquad\qquad(1)</math> 
<math>W_{ext} =\Delta U=U(y)-U(0) \qquad\qquad\qquad(1)</math> 

Meanwhile, the internal conservative force <math>F_{int}</math> (internal to the object-earth system that is, i.e. the weight), does the same amount of work but this is negative work. So, 
<math> W_{int}=- W_{ext} =-\Delta U \qquad\qquad\qquad(2)</math> 
What this means is the change in potential energy is equal to the work you do on the mass (the external work), but opposite in sign to the work that the force of gravity (the internal force) does in the same movement.

 [[File:ElectricPotentialFig1.png|center|500x372px|alt=Figure 1]] 
 
For an infinitesimal displacement dx, the internal force <math> F_{int}</math>, does infinitesimal work equal to 

<math> dW_{int}= F_{int} * dy = -dU \qquad\qquad\qquad(3)</math> 
 
The sign (-) in (3) shows that when a conservative force does positive work the potential energy decreases. 

From (3) we get the relation between a conservative force and the potential energy
<math> F_{int}=-\frac{dU}{dy} dU \qquad\qquad\qquad(4)</math> 

Ex: the Earth-object system; If the y direction is directed vertically up U(y) = mgy and <math>F_{int}=-\frac{dU}{dy}=-mg\ (directed\ down)</math>

 
The potential energy of the object is actually the energy of the system earth-object and it is due to the gravitational field of the earth. You can define a function called the gravitational potential as the potential energy normalized for mass as described below. 
 
===Gravitational Potential VG is Gravitational Potential Energy Normalized for Mass===
The gravitational potential <math>V_G(y)</math> at a location “h” of space is equal to the potential energy of a 1 Kg mass located at this point of the space. i.e. it is the potential energy normalized for mass. 
Remember the gravitational potential energy is mgh.
Therefore <math>V_G(y)=\frac{mgh}{m}=g\cdot h</math> 
In this way you: 
a) Define a function <math>V_G(y)=g\cdot h</math> which is a characteristic of the earth's gravitational field. 
b) For a mass m ≠1Kg, you get the equations 
<math> V_G=\frac{U}{m}\ \ \ and\ \ \ \Delta V_G=\frac{\Delta U}{m}\ \qquad\qquad\qquad(5)</math> 

c) Can calculate the internal work done by gravitational field during the shift of a mass “m” inside it as 
 
<math> W_{int} =-\Delta U =- m \cdot \Delta V_G \qquad\qquad\qquad(6)</math> 

==Definition of Electric Potential Energy for Point Charges==
Consider one charge <math>q_1</math> being brought towards another charge <math>q_2</math> from infinity to a distance r. 
The work done by you pulling the charges together will be equal to the increase in electric potential energy of the system. 
<math> W_{ext} =\Delta U = -W_{int}=-\int^r_{\infty} F_{electric}= -\int^r_{\infty} \frac{kq_1q_2}{r^2} =+\frac{kq_1q_2}{r}\qquad\qquad\qquad(6a)</math> 
Any other charges brought into this system will have to do work against the other charges, and so on as the charge distribution is assembled. 
Note that the potential energy of the system increases if you do work bringing two positive charges together. However, the potential energy of the system decreases if you bring two opposite sign charges together: the charges want to rush together, and the potential energy is released as kinetic energy! 
i.e. for electric potential energy, the signs of the charges should be included in the calculation.

==Where to define the reference position U=0 ?==
Remember that when dealing with gravitational potential energy, what is important is the change in potential energy, not the energy itself. This means that you can define where the potential energy is zero. You are free to define any position, but usually some points make better reference points than others. For instance, for gravitational potential energy, choosing the lowest point in the problem as the reference (U=0) is recommended. 
For electric potential energy of point charges, it is recommended to choose infinity as the reference position. The electric force is zero at infinity, anyway, so that makes sense. 
For something like a capacitor, which has zero field outside the plates, the reference position could be on the negatively charged plate. In this way the positive plate will have a positive potential.

==Electric Potential (V) is Electric Potential Energy (U) Normalized for Charge==
At this point remember that the Coulomb force is a conservative force, too. So, you can just adopt all the previous steps to the electric field and get a similar definition of electric potential V. 
 
The electric potential is equal to the potential energy of a 1C charge at location r of an electrostatic field. 
Similarly to Equations (1-6) you can calculate it from the work done by the electric field (Win) (7, 8) for a charge q=+1C.

<math> V = \frac{U}{q}=-\frac{W_{int_{relative\ to\ reference}}}{q} \qquad\qquad\qquad(7)</math> 

<math> \Delta V = \frac{\Delta U}{q}=-\frac{W_{int }}{q} \qquad\qquad\qquad(8)</math> 
Therefore the change in potential between infinity and point r is (substituting (6a) into 8): 

<math> \Delta V = \frac{\Delta U}{q_2} = \frac{\frac{kq_1q_2}{r^2}}{q_2} = \frac{kq_1}{r} \qquad\qquad\qquad(8a)</math> 

 
These mean: 
(7) the electric potential is equal to the “-“ internal work spent by field when “+1C” charge moves from the point with 0-potential energy to the considered space location. 
 
and (8) the change of electric potential between any two space locations is equal to “-“ amount of work done by the field when a charge +1C is shifted between these two locations. 
Note: Potential has units of Volts (V), yes the same volts as your car battery!. Using the definition of potential, 1V=1J/C.
 
When the electric field moves a positive charge along its lines, it does positive internal work. The equation (8) shows that the charge is moved from higher to lower potential because ∆V = V2 -V1< 0. So, if a positive charge is placed inside an electric field, the electric field tends to move it towards locations with lower V-values(2.b). A similar situation is met inside gravitational fields; objects in free fall move towards the earth, decreasing their potential energy. If a negative charge is placed in an electric field the electric force will move it against the field lines. As the electric field does positive internal work
(Wint>0), the equation (8) shows that in this case ∆V = V2-V1 > 0 and V2 > V1. This means that the electric field pushes the negative charges from lower towards higher V values(2.c). 
 [[File:ElectricPotentialFig2.png|center|1000x744px|alt=Figure 1]] 
The electric field will push a “+” charge towards lower V-values and a “-“ towards higher V-values. 
 
 
When a test charge (+1C) is moved by a distance s inside an electric field, the field does work (see 8) 
<math> W_{int}=W_{electric}= \vec{E}\cdot \vec{\Delta s} =-\Delta U=-\Delta V \qquad\qquad\qquad(9)</math> 
For an infinitesimal displacement ds you get 
<math> \vec{E}\cdot \vec{d s} =-dV \qquad\qquad\qquad(10)</math> 

So, dV E* dsEs * ds(11) and Es dV <math> dV=- \vec{E}\cdot \vec{\Delta s} =-E_s\ ds \qquad\qquad\qquad(11)</math> 
And 
<math>E_s=-\frac{dV}{ds}\qquad\qquad\qquad(12)</math> 

 
The equations (11-12) are very important because they relate the two main parameters of the same electric field and show how to calculate one of them if the other one is known. 
 
As electric potential is potential energy normalized for charge, its value depends only on the charge's location. So, its change depends only on the initial and final positions and not on the path followed for a shift from “A” to “B”. 
This shows how you could calculate the work done by the electric field when a charge is moved inside a non uniform field (fig.3), avoiding complicated integral calculations. 
So, for a test charge “+1C” moving inside a non uniform field the work done by the field is calculated as follows: 
<math> W^{A\rightarrow B}_{int}= \int ^B_A \vec{E}\cdot\vec{ds} = \int ^B_A Eds = \int ^B_A -dV =-(V_B-V_A)\qquad\qquad\qquad(1)</math> 

Better, keep in mind: 

<math> V_B-V_A=-W^{AtoB}_{int}=-\int^B_A \vec{E}\vec{ds} \qquad\qquad\qquad(13)</math> 
 [[File:ElectricPotentialFig3.png|center|500x372px|alt=Figure 1]] 

==Potential and Potential Energy inside a Uniform Field (e.g. inside a Capacitor) ==
Consider that an external force shifts the positive unit charge (+1C) inside the uniform electric field E from A to B following an irregular path (Fig.4). As we know, the work by electric force ( Fe E ) is 

<math> W_{int}= \int_{path}\vec{F}\vec{ ds} =\int_{path}\vec{E}\vec{ ds}= \vec{E}\int_{path}\vec{ ds} =\vec{E}\cdot\vec{\Delta s} \qquad\qquad\qquad(1)</math> 

because the field is constant (take out integral) and the
vector sum of infinitesimal shifts is equal to <math>\vec{\Delta s}=\vec{AB}</math> .
As <math>\Delta V = - W_{int} </math> we get 
<math> V_B-V_A=\Delta V=-\vec{E}\cdot \vec{\Delta s} =-E\cdot\Delta s_E = - E\cdot d \qquad\qquad\qquad(15)</math> 
Note that the same result would be for all positions of point B along the direction CB and this means that for all these positions VB = constant. In other words: 
The line CB is an equipotential line.
 
 [[File:ElectricPotentialFig4.png|center|500x372px|alt=Figure 1]] 

We fix the origin of an axe Ox at point A and select its direction the same as the field. If V=V(0) at origin, then at point B with coordinate x on Ox axis, the eq. (15) is written: 
<math> V (x) - V (0)=E \cdot \ \ and \ \ V (x) = V (0) - E \cdot x \qquad\qquad\qquad(16)</math> 
Equation (16) shows us that the potential decreases (Fig.4) while x increases ( E along field sense). 
 
We can conclude that in a uniform field: 
a) The potential value is the same on each plane perpendicular to electric field (equipotential planes). 
b) The potential decreases while moving along the direction of field; it increases in opposite sense. 
c) Dimensional rule applied at (16) gives [V]=[E]*[x] and [V]≡ volt = (N/C)*m= N*m/C=J/C 
The equipotential plane is a special case of an equipotential surface “the geometrical place of points with the same potential”. This concept comes from gravitational equipotentials which are depicted as lines in a 2-dimentional plane (see Fig.5). In the case of uniform field we found that: 

THE ELECTRIC FIELD LINES ARE PERPENDICULAR TO 
THE EQUIPOTENTIALS AND POINT “DOWNHILL” FROM 
THE HIGHER TO THE LOWER VALUES OF POTENTIAL. 
 
This is a general rule that applies for all electric fields. 
 
In fact, for a shift ds of test charge along an equipotential line (ΔV = 0), 

the work by electric field is zero becauseE* ds Wint V 0
 [[File:ElectricPotentialFig5.png|center|500x372px|alt=Figure 1]] 

As E* ds 0 means E ds it comes out that the electric field is perpendicular to equipotential line.
 
Consider a moving charged particle that enters in an electric field. If other forces are negligible, one may apply the energy conservation law for the system field-charged particle. So, if the particle has the kinetic energy K(1) and the potential energy due to the field U(1) at the moment t1, as long as it is inside
the field its total energy will be constant i.e. E = E(1) = K(1) + U(1). This means that ΔE = ΔK+ ΔU = 0
and ΔK = - ΔU . Then, as ΔU = q* ΔV one gets
<math> \Delta K= -q \cdot \Delta V \qquad\qquad\qquad(17)</math> 

 
It turns out that, the kinetic energy of particle may increase or decrease depending on sign of its charge. If q > 0, and the charged particle enters the field moving 
 
a) along the direction of potential decrease (“downhill”), ΔV < 0 and the field will increase its K 
b) along the direction potential increase (“uphill”) , ΔV > 0 and the field will decrease its K. 
If q < 0, and the charged particle enters the field moving : 
a) along the direction of potential decrease (“downhill”), ΔV < 0 and the field will decrease its K 
c) along the direction potential increase (“uphill”), ΔV > 0 and the field will increase its K. 
 
These rules are important when dealing with electrons or protons accelerated by use of electrostatic fields. One has introduced a new energy unit “electronvolt (eV) “ which is widely used in chemistry, atomic and nuclear physics. It is the kinetic energy gained by a particle with charge +e when moving downhill through a potential difference of 1V. 

<math>1eV = (1.602\times 10^{- 19} C) \cdot 1V = 1.602\times 10^{- 19} J \qquad\qquad\qquad(18)</math> 

 

==The Electric Potential of Point Charges ==
 
a) One point charge 
 
The electric field created by a point charge +Q in the space around it is a radial field 
<math> \vec{E}=k\frac{Q}{r^2} \hat{r}= E_r \hat{r} \qquad\qquad\qquad(1)</math> 

We may find the expression for the potential through the work done by
electric force on a charge (+1C) when it moves on a path AB. So, 
<math> V_B-V_A=-\int^B_A \vec{E}\vec{ds}=-\int^B_AE(r)de=-\int^B_Ak\frac{Q}{r^2}dr = -kQ\int^B_A\frac{dr}{r^2}=-kQ(-\frac{1}{r})^B_A=kQ( \frac{1}{r_B} -\frac{1}{r_A} ) \qquad\qquad\qquad(20)</math> 
Note that <math> \vec{E}\vec{ds}=E(r)\hat{r}d\vec{s}=E(r)[\hat{r}d\vec{s}]=E(r)dr </math>
 [[File:ElectricPotentialFig6.png|center|500x372px|alt=Figure 1]] 

By convention, you fix the zero value of potential at <math>r_A = \infty</math>,V(∞) = 0. 
Then one gets the potential at point rB ≡ r as 

<math> V (r) - V ( \infty )= V (r) = kQ\frac{1}{r} \qquad\qquad\qquad(21)</math> 

 
Figure 7 presents graphically this function in 3D space. The dashed circles present the equipotential lines. They are closer to each other near the charge because the potential changes rapidly in this region. The field lines are normal to equipotential lines. Note that the magnitude of field is bigger at locations where the equipotential lines are denser (near the charge which is the source of field) and it decreases at locations where they are rare (big r-values). 
 
 [[File:ElectricPotentialFig7.png|center|500x372px|alt=Figure 1]] 
 
b) Two point charges 
 
Consider that two point charges Q1 and Q2 are present at the same space. Let E1_ and E2 be the electric fields corresponding to each of them at a given point of this space. We can find the resultant electric field at this point by applying the superposition principle 
 

<math> \vec{E} = \vec{E} _1+ \vec{E}_ 2\qquad\qquad\qquad(1)</math> 

 
As in the case of one particle, we chose the starting point A of a path at infinity (VA= V(∞) = 0) and express the difference of potential through the work done by net electric field during the shift A to B. 
<math> V_B - V_A = V_B = k\frac{Q_1}{r_{1-B}} + k\frac{Q_2}{r_{2-B}} \qquad\qquad\qquad(1)</math> 
In the general case, when the net field is due to several charges Q1,Q2,..Qi the net potential is calculated through the expression 
<math> V_B=\sum_i k\frac{Q_i}{r_{i-B}} </math> 

or more generally
 
<math> V(r)=\sum_i k\frac{Q_i}{r_{i}} \qquad\qquad\qquad(1)</math> 

Notes: a) ri is the distance from the “ith-charge” to the location “r” where we calculate the potential b) you have to include the charge signs inside the sum (23). 
 
Figure 8.a shows the potential evolution (up) in the electric field built by a dipole and the equipotential lines and the fields’ lines (down). Figure 8.b shows the potential evolution (up) in the electric field due to two equal positive charges, the equipotential lines and the field lines in the plan that contains charges. 

 [[File:ElectricPotentialFig8.png|center|500x372px|alt=Figure 1]] 
 

==Conductors ==
 
As explained in section 2.3 the electrostatic field is zero inside a conductor and perpendicular to its surface in the surroundings. As <math> V_B- V_A=-\int^B_A\vec{E}\cdot\vec{ds}=0 </math> for a path connecting two points located inside
it or on its surface, it turns out that the electric potential of a conductor is the same everywhere on its surface and inside it. Note that this is valid for any charged conductor and any neutral conductor placed inside an electric field. 
 
It is important to know that the presence of a conductor in an external field modifies the field pattern in the close region around its surface so that the field lines become perpendicular to its surface.
 [[File:ElectricPotentialFig11.png|center|500x372px|alt=Figure 1]] 

Fig.11 shows the deformation of equipotential surfaces and field lines of a uniform external field around a spherical uncharged conductor. Near the sphere, the equipotential planes become spherical and the field lines become radial. 
 
 
As the charges are distributed on the conductor surface and the field inside it is zero, one may open a cavity inside the conductor and use the conductor surface as a shield from external fields. This is what happens in a Faraday cage, a device widely used in as a shield against electric fields. 
If a conducting wire connects two charged conductor objects, their charge redistributes so that the potential be the same and the field be zero within them and parallel to their surface. 
 
Assume that we place the charge +Q on a metallic sphere. It will be distributed uniformly on the spherical surface and this will bring a constant charge density <math> \sigma = \frac{+Q}{4\pi R^2}</math> on the sphere surface. As you may see, the charge density on this surface is bigger for small values of radius R. 
This has important implications for all conductors because even when the surface of the metallic object is not spherical, the charge density is always higher at the sharp points (smaller R) and the electric field ( <math>|\vec{E}| \propto \sigma </math> ) is higher at those locations as a consequence. 
For this reason, lightning conductors usually end in sharp points to attract the lightning. 
This effect can also be used to produce electric fields with very high magnitude. On the other hand, this effect may become problematic because very high E fields can ionize air and may be the origin of corona discharges (=sparking!). 

==Examples==
Example 1: Potential energy of point charges 
A positive charge 2.00µC is held on a horizontal plane and a charge of -1µC is held 5.00 cm North of it, and a charge of 3.00µC is held 5.00 cm West of it. 
a) What is the total potential energy of the system? 
b) What is the potential energy of the 3.00 µC charge? 
c) If the 2.00 µC charge is released (let's say the object has a mass equal to 1 gram) what is its speed when it is far far away from the other 2 charges? 
:::::<youtube>jnzbv0oSY78</youtube> 
Example 2: Potential energy of point charges 
A small sphere carrying an unknown charge q1 is held fixed in space. A second small sphere of mass 1 mg carrying charge q2=5.00 µC is projected toward q1. 
When q2 is 1.00 m from q1, its kinetic energy is 0.500J. 
When q2 is 0.500 m from q1, its kinetic energy is 0.250J. 
Does q2 stop and turn around, or does it hit q1? 
:::::<youtube>bEnmo3qPdnI</youtube> 
Example 3: Potential energy of point charges 
Charge q1 (-10.0µC) is placed at the origin and charge q2 (1.00µC) is at (3.00,-1.00)cm. 
a) What is the electric potential at point P (3.00,2.00)cm? 
b) Compare the potential energy of a 10.0 µC charge at point P with that of a -10.0µC charge at point P. 
:::::<youtube>TXzS1NksFhQ</youtube> 
Example 4: Potential energy of point charges 
A small object with mass 1.00mg carries <math>3.125\times 10^{12}</math> extra electrons. When it passes point A, it is moving at 14.15m/s and when it passes point B it is moving at 10.0m/s. Electric forces are the only ones acting on it. 
a) Sketch electric field vectors at point A and B, and sketch an electric field line through point C. 
b) What is the potential difference between points B and A (=<math>V_{AB}</math> or <math>V_{B}-V_{A}</math>)?
 
c) If A is at 25.0V what is the electric potential at B? 
:::::<youtube>Af7ZCXqtqzo</youtube> 
Example 5: Drawing equipotential surfaces 
A positive charge 2Q is held on a horizontal plane and a negative charge -Q is held left of it. 
Sketch the equipotential surfaces around these 2 charges. 
:::::<youtube>nQEgkZ8WyFs</youtube>

Electric Current

2014-10-06T15:54:28Z

Kevin:

Capacitors

2014-10-06T15:54:04Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/Capacitors/Capacitors%20Worksheet.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/Capacitors/Capacitors%20Worksheet.pdf PDF Worksheet] 
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/4.pdf Additional Problems] 
 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 21.1-5
*[http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@8.9:150 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_4.pdf?FCItemID=S002862FE Printable version]
 
==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* understand the relation between charge and voltage in a pair of parallel conducting plates.
* define capacitance.
* define and calculate equivalent capacitance for series and parallel connections.
* understand the mechanism and calculate the amount of energy storage in a capacitor.

 

 
Capacitors are an extremely common electrical component, because they can store charge and energy. In addition they charge and discharge over time intervals determined by their value and therefore they are essential for any timing circuit (see AC circuits).
 
===Introduction to Capacitors===
As explained previously, the presence of electric charges in a space creates an electric field. 

If another charged particle (for instance an electron e-, or a proton p+ or an ion) appears somewhere in this space, the electric field will exert a force on it and when this particle moves inside the electric field, the electric force or its “source”, i.e. the electric field, does work. This means that the electric field possesses energy and it can be thought of as an “energy container” or “energy source”. 
 
In principle, you can “store energy” in space by using any possible configuration of electric charges. One common configuration, called a capacitor, corresponds to two charged conductors (presented as parallel plates ┤├) that are separated by an insulator. The most common insulator in our examples will be air. The plates may be in a plane, cylindrical or spherical. 
 
Consider that the battery B (shown in fig.1) is a “box” that keeps constant the difference of potential V1 between its terminals, no matter what circuit is connected to them. Once the plates of capacitor C are connected (by metallic wires) to the battery terminals, they start getting charge from the terminal and this builds the difference in potential V between the capacitor plates. The electromotor force “emf”
of battery removes some free electrons from one plate of capacitor , and in this way charges it positively “ +” and transfers the same amount of electrons to the other plate of the capacitor (charging it negatively “-“ ). At the end of transfer, each plate gains the same magnitude of charge, but of different signs (+Q,-Q). For a perfect capacitor, these charges are distributed uniformly on the surface of the conducting plates. As the area of both plates is equal, the same magnitude of charge density  (but opposite sign) is present on each of them. It turns out that, for a given potential difference between them, the plates of capacitor can only hold a limited amount of charge “Q”. 
Once this amount Q is stored on the plates, the battery is not able to remove any more charges from the plates of capacitor and the charge transfer is stopped. 
Assume that, after this, that someone disconnects the battery. The charges will remain on plates due to the mutual attraction with the other sign charge on the other side (Fig.2) and also because the charges have nowhere to go! 
So, the capacitor keeps the charges +Q, -Q and the difference of potential V between its plates. 
 [[File:CapacitorsFig1_2.png|center]] 

 

==Capacitance: The Relationship between the Charge Q and the Potential V==
The strength of the electric field E close to the surface of a conductor is proportional to the charge density (<math>E \propto \sigma </math>) on its surface. As, for a given area and uniform distribution
<math>\sigma = Q/A </math>, we get (<math>E \propto Q </math>) where Q is the total charge on surface. 
The difference of potential ∆V between two locations inside a field is proportional to the field strength (<math>\Delta V= -E \cdot \Delta x</math>). Therefore the (<math>\Delta V </math>) between the two capacitor plates (the two locations in the field) is proportional to the charge Q on each of them. Then if we take the potential on the negative plate V(-) = 0, then V(+) known as the electric potential of the capacitor is proportional to Q and vice-versa. So 
<math>Q=C\ V \quad\quad\quad(1)</math> 
Or 
<math>C=\frac{Q}{V} \quad\quad\quad(2)</math> 

where C is a constant of proportionality and is called the capacitance of the capacitor. Note that C is constant for any particular capacitor. Its unit in the SI system is the Farad (1F); 1F = 1Coulomb/Volt 
 

For a given difference of potential V between the terminals, the greater the capacitance C, the greater the quantity of stored charge Q and also the greater the amount of electric energy stored inside the capacitor. 
 
The capacitance is the main parameter of a capacitor. It measures its efficiency in storing energy. For general purpose capacitors, the capacitance values are in pF(picofarad), nF(nanofarad), <math>\mu F</math>(microfarad). 
 

==The Capacitance only depends on the Capacitor Dimensions, The E Field only depends on the Charge Density==
Fig.2 presents the electric field in the empty space between two plates of a capacitor. If the plate separation d is small, you may ignore the edges and consider only the field in central region. The more you ignore edge effects, the more ideal the capacitor becomes. In this region you may use the model of the field from a charge Q distributed uniformly on a large surface. 
We have
found that this field is uniform, directed vertically to the surface and its magnitude is 
<math>|E|=\frac{\sigma}{2\varepsilon _0}</math> 

Since there are two plates with the same charge density <math>\sigma</math> producing fields with the same magnitude and direction (from the “+” to the “-“ plate), It turns out that, for an area plate “A” the magnitude of net electric field in a capacitor is
 
<math>|E|=2\frac{\sigma}{2\varepsilon _0} = \frac{\sigma}{\varepsilon _0} = \frac{Q}{\varepsilon _0 A} \quad\quad\quad(3)</math> 
This field is uniform; so the potential difference between the two plates is
<math>V=E\cdot d</math> 
Then 
<math>V=\frac{Q\cdot d}{\varepsilon _0 \cdot A} \quad\quad\quad(5)</math> 
And 
<math>C=\frac{Q}{V }= \varepsilon _0 \frac{A}{d} \quad\quad\quad(6)</math> 

The expression (6) shows that, in a vacuum, the capacitance of a given capacitor depends only on its geometry and dimensions. If the capacitor shape is not plate, the expression for its capacitance is more complex than (6) but in all cases it depends only on its dimensions. 
 
 

==CAPACITORS IN SERIES AND PARALLEL==
 
For certain situations you may want to combine capacitors, either because you have a limited selection of capacitors, or there are other constraints e.g. exceeding the breakdown voltage, when a spark goes from one plate to the other. There are two main ways to connect several capacitors; in series and in parallel. 
 
===Capacitors in Series===
Figure 3 presents the scheme of two capacitors connected in series. Consider if you connect the points a and b to

the battery terminals for some time; then disconnect them. 
 [[File:CapacitorsFig3.png|center]] 

During the contact with the battery, the capacitors C1 and C2 acquire the same charge amount Q on their plates because the exterior charges +/- Q shifted by battery induce the same magnitude of charge -/+Q to the interconnected plates. 
Tip 1: Capacitors in series always carry same charge on each plate. 

Using this tip, and knowing the value of capacitances C1 and C2, you may calculate the potential difference across each capacitor, V1 and V2 respectively, as follows: 
<math>V_1 = \frac{Q}{C_1}=V_{ac} \ \ \ and\ \ \ V_2 = \frac{Q}{C_2}=V_{cb} \quad\quad\quad(7)</math> 
Then 
<math>V=V_{ac}+V_{cb} = \frac{Q}{C_1}+ \frac{Q}{C_2} = Q(\frac{1}{C_1}+ \frac{1}{C_2} ) =Q\frac{1}{C_{eq}} \quad\quad\quad(8)</math> 

This expression shows that two capacitors connected in series function the same way as an equivalent capacitor with capacitance Ceq such that 
<math> \frac{1}{C_{eq}} = \frac{1}{C_{1}} +\frac{1}{C_{2}} \quad\quad\quad(9) </math> 
In the general case of N capacitors connected in series, the equivalent capacitance is 
<math> \frac{1}{C_{eq}} = \sum^N_{i=1}\frac{1}{C_{i}} \quad\quad\quad(10) </math> 
 
Note that, for the same charge amount “Q” stored on the plates, this “equivalent capacitor” supports higher voltage on its terminals” V=V1+V2+V3+…”. 
On the other hand, the capacitance of the combined capacitors is always ''less'' than the value of the capacitors themselves. One way of looking at it is that the voltage across each capacitor is always less than the total voltage and therefore the charge each capacitor carries is less than if that capacitor was attached directly to the battery.
 

===Capacitors in Parallel===
 [[File:CapacitorsFig4.png|center]] 
Figure 4 presents the scheme of two capacitors connected in parallel. In this case, the same potential difference is applied to the two capacitors and the battery emf has moved the total charge amount Q1 + Q2. The charges Q1 and Q2 depend on the capacitance of capacitors as follows 
<math>C_1=\frac{Q_1}{V}</math> and <math>C_2=\frac{Q_2}{V}\quad\quad\quad(11)</math>
So, the equivalent capacitor would get the total charge 
<math>Q=Q_1+Q_2\quad\quad\quad(12)</math>
and its capacitance would be 
<math>C_{eq}=\frac{Q}{V}= \frac{Q_1+Q_2}{V} =\frac{Q_1}{V} +\frac{Q_2}{V}=C_1+C_2\quad\quad\quad(13) </math> 
In the most general case of connecting N capacitors in parallel, the equivalent capacitance is equal to 
<math>C_{eq}=\sum^N_{i=1}C_i\quad\quad\quad(14)</math>

 
The “equivalent capacitor” offers higher storage capacity of charge ”Q1+Q2+..” for the same voltage. 
Another way of looking at it is that the area on which charge can be stored is increased, therefore the charge storage must increase as well.
 
 
 
 

===ENERGY STORED IN A CAPACITOR===
 
The stored energy in a capacitor comes from the work done in accumulating the charges onto the plates. Importantly, another way of looking at the stored energy is that is contained in the electric field created in the space between the plates by the charge on the plates. This stored energy is due to the work done by the emf of the battery as it moves “free electrons” from one plate (that becomes charged “+”) to the other plate (that becomes charged “-”). Note that the electron path passes through the wires and battery and not through the space between the plates. 
Let’s consider a circuit containing only a battery connected to a capacitor at an intermediary point in time when the plate’s charge is q (not the final charge Q) and the potential difference between them is v (not the final volage V). The work dW done by the battery, for the transport of the next elemental charge “-dq” from the “+” plate to the “-“ plate is external work done on the system electric field- charge. That is to say the battery loses work and the electric field (arising from the charge arrangement), gains work. 
The work done will be equal to the potential difference times the charge transferred, as shown here: 
<math>dW_{ext}=\Delta U = (-dq)\times\Delta V=-dq\times (0-v)=dq\ v\quad\quad\quad(15)</math>

 
As <math>v=\frac{q}{C}</math> then from (15) 
<math>dW_{ext}=\frac{q}{C}dq\quad\quad\quad(16)</math> 
The total work for the battery to move the charge is: 
<math>W_{total}= \int ^W _0 dW_{ext} = \int ^Q _0 \frac{q}{C}dq=\frac{1}{2}\frac{Q^2}{C}\quad\quad\quad(17)</math> 
You can rewrite this expression, if you assume the initial potential was zero, then: 
<math>W_{total}= \frac{1}{2}\frac{Q^2}{C} =\frac{1}{2}\frac{(CV)^2}{C}=\frac{1}{2}CV^2 \quad\quad\quad(18)</math> 
Which is the formula most quoted.
 
The principle of energy conservation tells that this external work goes towards increasing the potential energy stored inside the capacitor. 
 

 

===Real Capacitors and Dielectrics===
 
Capacitors come in all sorts of different shapes and sizes. The capacitors you often use look small, but in reality contain large sheets of conductor folded neatly into these small packages. 
Most real capacitors contain an insulating material between the plates. This material is called a dielectric (insulator) material. 
The dielectric is very practical for several reasons: 
 
a) It stops the plates from shorting out; 
b) It increases the maximum potential difference supported by the capacitor, by reducing the electric field between the plates; 
c) It increases the capacitance of the capacitor compared to the capacitor without dielectric; 
These factors make your capacitor with dielectric more efficient: your capacitor can store more charge, for a given potential.
 

 
==Examples==
How to combine multiple capacitors in parallel and in series 
<youtube>CfloCOQ2_1U</youtube>
 

What happens when you charge two capacitors and then connect them together? 
<youtube>3iH5vs-2rN0</youtube>

The Electric Field

2014-10-06T15:50:05Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricField/ElectricField%20Worksheet.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricField/ElectricField%20Worksheet.pdf Worksheet.pdf] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/2.pdf Additional Problems] 
 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 21.1-5
*[http://cnx.org/contents/031da8d3-b525-429c-80cf-6c8ed997733a@7.61:140 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_2.pdf?FCItemID=S00285D80 Printable version]
 
 
The Electric Field is an important extension of the Electric Force. The Electric Field allows you to describe the effect of charges on a point in space, without having to know what those source charges are, or what the distribution of charge is. In addition, if you know the Electric Field at any point, you can immediately calculate the force on any charge at that point.
===Comparison with other Scalar and Vector Fields===
If you use a very sensitive thermometer to measure the temperature at different locations in the city you will have a set of different values as a function of position. This distribution of temperature values in space is known as a temperature field. Similarly, a pressure field consists of a distribution of pressure values in different locations of atmosphere . In these two examples we deal with scalar fields because of the scalar nature of field parameter. The basic advantage of such kind of fields is the possibility to compare the values for a physical parameter at different space locations. 
Similarly you can have a vector field, in which you can define the value and direction of the parameter at each point in space. Consider this diagram showing the wind vector field for the earth. The wind velocity is shown at each point with an arrow showing the direction and its length showing the value. 

https://www.ncl.ucar.edu/Applications/Images/wind_3_lg.png
 
The electric field is a vector field. But how is the vector of the electric field defined ? 
 

===Definition of the Electric Field Vector E===
Initially, the electric field definition appeared as an intermediary concept to answer questions: What way interact two electric charges? How do they know the presence of the other charge? 
 
But when the electromagnetic theory was completed, the scientists understood that the electric field is not just an intermediary useful parameter; it is one of the basic constituents (matter and field) of the nature. 
 
Nowadays, the following model of interaction between two electric charges is accepted: Once an electric charge Q is placed at a given point of space it creates its electric field at all space locations around it. Then, if another electric charge q appears at a point P of this space, it interacts straight away with the local electric field (created by the charge Q) at point P. 
 
Definition: The electric field vector at a given point P of space is equal to the Coulomb force exerted
on the unit positive charge (+1C) placed at point P. In practice, one measures the force F exerted on a positive test charge qt (the test charge is very small, much less than 1C) and then finds out the electric field from the relation: 
<math>\vec{E}=\frac{\vec{F}}{q_t} \qquad \qquad\qquad(1)</math>

===Electric Field of a Point Charge===
The electric field created by a point charge Q at the location r is defined by the expression 
<math>\vec{E}=\frac{\vec{F}}{q_t}=k\frac{q_1q_t}{r^2}\hat{r}\cdot\frac{1}{q_t}=k\frac{q}{r^2}\hat{r}\qquad \qquad\qquad(2)</math>
 where
<math> \vec{r}=r \hat{r} </math>
 
Note: <math>\hat{r}</math> is a unit vector in the direction of r. It simply means: in the direction of r

From Equation (1), it turns out that the SI unit of electric field is N/C.
 [[File:ElectricFieldFig1.png|left|250x188px|alt=Figure 1]] 
Fig 1. Shows how the Electric field can be found by considering the force on a test point charge at different locations of the space surrounding charge. 

Equation (2) tells us that the electric field at a given point P of space depends on the location of P and the location of field source (that is to say the position of the charge producing the field). But, you can figure out that, for an electric charge q placed at P point, the electric field is defined at this location, no matter what charges produce it. 
The force exerted on a charge q placed at point P can be found straight away from the local electric field EP as 
 
<math>\vec{E}=q\ \vec{\vec{E_P}} \qquad \qquad\qquad(3)</math>
 
If q > 0 (the charge is positive) the force has the same direction as the field and if q < 0 (the charge is positive) it has the opposite direction. 
Since the principle of superposition is valid for Coulomb forces, it is valid for electric fields, too. 
<math> \vec{E_{net}} = \frac{\vec{F_{net}}}{q_t} =\frac{\vec{F_{1}}}{q_t} +\frac{\vec{F_{2}}}{q_t} + ...\frac{\vec{F_{N}}}{q_t} =\vec{E_{1}} +\vec{E_{2}} +...\vec{E_{N}} \qquad \qquad\qquad(4)</math> 
That is to say you can add electric field vectors just like you can add electric force vectors.

 

===ELECTRIC FIELD LINES===
Field Lines are a way of visualizing the electric field produced by charge in a particular space. 
Electric Field lines originate with positive charge and end with negative charge, as shown in the figure. 
[[File:electricfieldlines.jpg|center]]
 
Let’s consider a positive charge +Q. The map of its electrical field in the surrounding space is shown in fig.2a. The electric field for a set of two equal charges +Q and +Q can be found as shown at fig.2b. If we draw the electric field at any point of space the picture becomes very complicated. So, you can use a system of field lines which helps to visualize the basic information on the field pattern.
 
[[File:ElectricFieldFig2.png|center|500x376px|alt=Figure 1]]
 

Fig. 2a Electric field map (+Q) 
Fig.2b This shows how you could build up the Electric field map (+Q and -Q) by constructing the field contributions from each of the charges at any point. Field lines are a convenient way of plotting the results of this process. 

Here are the rules of interpreting field lines and the resulting electric field vectors: 

a) The electric field lines emerge from (+) charges and go towards (-) charges. 

b) At any point, the direction of the electric field vector is tangential to the field line passing through the point in question.
 
c) The field lines never cross. Otherwise, it would mean there would be more than one direction for the vector of electric field at a given point!! (=not allowed)
 
d) The number of field lines around a space point (more precisely the field line density) is proportional to the magnitude of electric field at this point. Thus, the magnitude of electric field is large if the field lines are dense and small if the field lines are far apart.
 
e) While the electric field is a physical reality, the field lines are not. They are just an intermediary tool used to give quick visual and general information about the field. 

 [[File:ElectricFieldFig3.png|center]] 

Fig.3a The field lines for (+Q and –Q) Fig.3b The field lines for (+Q and +Q) 
Electric dipole 

[http://www.youtube.com/watch?v=LBr63hA4mww Video Example 1 Courtesy of John Abbott College]
:::::<youtube>LBr63hA4mww</youtube>
[http://www.youtube.com/watch?v=mjQ6MJss_oI Video Example 2 Courtesy of John Abbott College]
:::::<youtube>mjQ6MJss_oI</youtube>
 

===CONTINUOUS CHARGE DISTRIBUTIONS===
We have already seen how to calculate the electric field for a point charge. But most practical applications do not involved point charges, but continuous charge distributions. For instance, charge distributed on a line (=charge on a wire), and charge on a surface (e.g. = charge on a capacitor plate). 
Remember that Coulomb’s law is valid for charged point particles. In order to find the electric field due to a continuous charge distribution, you must: 
 
a) Divide the total charge into infinitesimal elements each of charge dq. i.e. consider the charge distribution as a sum of charges small enough that they look like point charges. 

b) Apply the Coulomb’s law to each of these point-like charges and find the field due to each of them as 
<math>d\vec{E}=k\frac{dq}{r^2}\hat{r} \qquad\qquad\qquad(8)</math>
c) Apply the principle of linear superposition and get the sum of all infinitesimal fields as 

<math>\vec{E}=\int_{chargedobject}d\vec{E}=k\ \int_{chargedobject}\frac{dq}{r^2}\hat{r} \qquad\qquad\qquad(9)</math>
 
Here it is important to remember that you have to always express dq as a function of location coordinates dq(x, y, z) so that you can calculate the value of the integral (9) practically.

==Electric Field from an Infinite Line of Charge of Line Charge Density <math> \lambda </math> ==
Previous examples have involved point charges. In this example the charge distribution is continuous. 
Recall that for a system of charges you calculate the net electric field by taking the vector sum of all the contributions (principle of superposition). For a continuous charge distribution the sum becomes an integral over all the contributions. 
You can describe the charge distribution using the line charge density <math> \lambda </math>. Where <math> Q_{total}=\lambda_{total} \times Length</math>. 
[[File:Infinitelinecharge.png|center|500x376px|alt=Figure 1]]
 
First consider the electric field due to a small segment A of the line of charge to the left of P. The magnitude of this field is given by Coulomb’s law, and the direction of the field points directly away from the segment. Next, consider the electric field due to a small segment B, of the line of charge to the right of P. The magnitude of the electric field from segment B is the same as the magnitude of the electric field from A since A and B are equidistant from point P. However, the electric field from B has a negative x component that exactly cancels the positive x component of the electric field from A. 
As we integrate from <math> - \infty </math> to <math> + \infty </math> you can see that for every segment of the line to the left of P there is another segment of the line to the right of P whose x component of the field exactly cancels the x-component from the initial segment, just as we saw for A and B. Consequently the total field at point P must point in the y direction. 
To calculate the magnitude of the net field at P, we begin by writing an expression for the y component of the electric field due to a small segment of the charges line that we can treat as a point source. The magnitude dE of the electric field due to this segment is proportional to the amount of charge in the segment dq and is inversely proportional to s^2, the square of the distance from P. 
<math> dE = k\frac{dq}{s^2}</math> 
To find the total field, we will only need to consider the y component:
<math> dE = k\frac{dq}{s^2}cos\theta</math> 
Because both the distance s and the angle change with x, you should rewrite the integral in terms of one variable. Let’s choose <math> \theta</math>! 
Use: 
<math> s = \frac{r}{cos\theta}</math> and <math> x = rtan\theta</math> 
Take the derivative of this last equation to yield <math> dx = \frac{r}{cos^2\theta}d\theta</math> 
The equation for the net electric field then becomes: 
<math> E = \int dE_y = \int ^{+\pi/2}_{-\pi/2} k \frac{\lambda}{r}cos\theta d\theta = 2k\frac{\lambda}{r}</math> 
<math> E = 2k\frac{\lambda}{r}</math> 
Phew! 
The good news is you do not need to know this proof, although you need to know that the electric field from an infinite line charge is equal to <math> E_{infinite line charge} = 2k\frac{\lambda}{r}</math> and is always perpendicular from the wire. 
However the proof shows some principles that you will use in future: 
1. The physical meaning of integration is the summation of elements (in this case all the dE s). You are using the principle of superposition here. 
2. You can use symmetry as a way of simplifying the problem In this case realizing that the x components cancel out. This only happens when you have an infinite line charge, non-infinite makes it a bit more complicated. Note that any line charge looks infinite if you are close enough to it. 

 
Want more? 
Consult these video explanations: 
[http://www.youtube.com/watch?v=WmZ3G2DWHlg Electric Field of a finite line charge Video 1];
[http://www.youtube.com/watch?v=1p1nCs1775E&list=PL694FA187D8C2028A&index=7 Video 2]
 
[http://www.youtube.com/watch?v=80mM3kSTZcE&list=PL694FA187D8C2028A&index=8 Electric Field of a ring of charge]

===The Electric Field of an Infinite Plane of Charge===
http://physics-help.info/physicsguide/electricity/gauss_law_images/image030.gif
 
This Figure shows part of an infinitely large, and very thin, conducting plane with a uniform distribution of charge. The electric field vector at any point of space in either side of this sheet is perpendicular to the sheet plane because the parallel components wipe out each other. A closer consideration shows that the
magnitude of its electric field is the same in the surrounding space. If the vector E is the same at every point of space, the field is called a uniform electric field. 
The electric field of an infinite plane of charge is defined by: 
<math>\vec{E}_{infinite plane}=\frac{\sigma}{2\varepsilon_0}\qquad\qquad\qquad(5)</math> 

Where 
<math>\sigma</math> is the area charge density. <math>\sigma=\frac{Total\ Charge}{Total\ Area}</math> Units: C/m^2 
<math>\varepsilon_0 </math> is the permittivity of free space <math>\varepsilon_0 = 8.854 \times 10^{-12} [F/m]</math>.
Note that <math>k=\frac{1}{4\pi\varepsilon_0}</math> 
Note: 
The field is: 
*Constant to infinity
*Perpendicular to both surfaces
*Proportional to the charge density
*For a negative charge density, the E field goes into the plane, not away from.

=== The E Field inside a Conductor is Zero===
When a neutral homogeneous conductor is placed inside an external electric field <math>\vec{E}_{ext}</math> , its free electrons move
quickly in the opposite direction of <math>\vec{E}_{ext}</math> (Fig.5a) [because the electrons are negative] and leave an unbalanced positive charge on the other side. This charge redistribution inside the conductor creates an internal field <math>\vec{E}_{int}</math> that is directed in the opposite direction to <math>\vec{E}_{ext}</math> . As long as <math>\vec{E}_{total}</math> is not equal to zero, the electrons continue to move in a way that increases the magnitude of <math>\vec{E}_{int}</math> and consequently decreases the magnitude of the total E filed inside; when <math>\vec{E}_{total}=0</math> they stop moving. So, under static conditions, the net macroscopic field inside a homogeneous conductor is zero. 
This is to say inside a conductor, the static E field is zero because if it were not zero, charge would move, until it is zero, and the situation would not be static. 
Another consequence of this is that the free charge always accumulates on the outside of a conductor. The charge moves until it cannot move anymore, i.e. the outside edge of the object. 
Take note that this is not true for insulators, because the charge is not free to move.
 

[[File:ElectricFieldFig5a.png|center|250x188px|alt=Figure 1]] 
Figure 5a: Showing how charge moves in a conductor such that the electric field inside is always equal to zero.

==Electric Field Close to the Surface of a Conductor==
The easiest way to look at the field close to the surface of a conductor is to get infinitely close to that surface. If you get infinitely close then what you see is actually a flat infinite surface of charge. Therefore the electric field will always be perpendicular to the surface! And the magnitude very close to the surface is given by equation (5) 
Note that this approximation is true even for curved surface, because even a curved surface is flat if you are infinitely close to it. We are in the realm of classical physics, where you can get infinitely close, not considering atoms and all that modern physics malarky. 

Another way of looking at it is that like charge pushes all other like charge away from it. Therefore the charges will distribute themselves on the surface such that they cannot move sideways anymore, that is to say the net force (and therefore net E field) parallel to the surface is zero. 
 
Remember the field inside a conductor is zero, and the field just outside a conductor is always perpendicular and proportional to the charge density at that point.
 
 [[File:ElectricFieldFig5b.png|center|250x188px|alt=Figure 1]] 
Figure 5b: Showing how the electric field just outide a conductor is always perpendicular to the surface.

==MOTION OF CHARGES IN UNIFORM FIELDS==

If a free charged particle (such as an electron or a free proton) is moving inside a static uniform electric field it is subjected to the electric force with magnitude 
<math>|\vec{F}_{electric}|=|q\vec{E}| \qquad\qquad\qquad(6) </math>

Due to their small mass, in general, the gravitation force on elementary
particle is much smaller than electric force, and is usually ignored. 
So, it turns out that the net force exerted on particle is equal
to the electric force. 
In these circumstances, the Newton’s second law gives 
 
<math>\sum\vec{F}=\vec{F}_{electric}=m\vec{a}\qquad\qquad\qquad(7) </math> 
and the particle moves with a constant acceleration (E is constant if the field is uniform). 
 
<math> \vec{a}=\frac{q\vec{E}}{m}\qquad\qquad\qquad(8) </math> 

 
Note that the electric field gives the direction of the force, and therefore the acceleration only. If the particle has a velocity, it will not follow the electric field lines. Rather, the acceleration vector will be in the same direction as the field lines.
 
In general you only have to consider the motion of charge in uniform electric fields (e.g. produced by an infinite plane of charge). Therefore you have a force and acceleration in only one direction = parabolic motion!!. 
Review your kinematic formulas. 
:::::<youtube>y-3Ow_-JpbI</youtube>

==Examples==
1. Electric Field of Point Charges 
Two charges are held fixed on a horizontal plane: q1(1.00μC) is at (-1,3)cm and q2 (-2μC) is at the origin. 
Find the net electric field at point P3 (2,1)cm. 
:::::<youtube>4xuM8HN6Vms</youtube>
 
2. Motion of a charged particle in an electric field 
In a poorly aligned cathode ray tube, electrons enter a vertical uniform electric field with a velocity of 5x10^7 m/s at 20 deg above the horizontal. 
The electrons hit the top plate, 1.00 cm up and 2.00cm to the right from the place where they entered the electric field. 
What is the magnitude of the electric field, and in which direction does it point?
 
:::::<youtube>4aRUcVEgIAM</youtube>

Charge and Electric Force

2014-10-06T15:47:20Z

Kevin:

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricForceandCharge/ElectricChargesandForces%20Worksheet.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/ElectricForceandCharge/ElectricChargesandForces%20Worksheet.pdf PDF Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Problems/1.pdf Additional Problems] 
==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand the nature of charge.
* Know Coulomb's Law: the magnitude and direction of the Electric Force.
* Apply the concepts to some problems.
 

 

==What leads us to think that there are such things as ''charges'', anyway?==
Think about what you have learned about “forces.” A force can be thought of as a push or a pull that may act on an object. Usually, the force that acts on some object “A” is caused by some other object which is in direct contact with object “A.” For instance, a hammer might strike a nail; then the hammer is exerting a force on the nail. According to Newton’s third law, the nail also exerts a force on the hammer (equal in magnitude, opposite in direction). This type of force is called a “contact” force. 
There is another type of force, in which direct contact between objects is not required. For example the force of gravity or magnets interacting while not in direct contact. These forces are sometimes called “action-at-a-distance” forces. 
Now consider some other phenomena you have seen. When you rub a comb through your hair, you may have noticed the comb pick up small bits of paper. You may have seen balloons appearing to “stick” to walls, without any glue. You have probably seen various types of plastic materials (such as thin plastic wrap) adhere to other objects, as if they were “attracted” to each other across space. 
 
In experiments using certain simple materials such as rubber and glass rods, when they are rubbed with fur or silk, certain phenomena consistently recur: (1) when two of these objects made from identical materials are prepared the same way (e.g., rubber rubbed with fur), the objects appear to exert small repulsive forces on each other (2) when two of these objects made from ''different'' materials are held near each other (e.g., a rubber rod held near a glass rod), the objects appear to exert attractive forces on each other. This is due to the “electrical” force. It can be attractive, or repulsive. 

==Two Types of Charge==
The experiments described above, and many other similar ones, lead us to conclude that there are two different properties of matter that, in some sense, ''cause'' this force. When two objects with property “A” are near each other, they push each other apart (i.e., exert repulsive forces on each other). The same thing happens when two objects with property “B” are near each other. But when an object with property “A” is near an object with property “B,” they exert attractive forces on each other. These properties of matter have been called “charge,” and instead of “A” charge and “B” charge, the terms “positive” [symbol: +] and “negative” [symbol: –] charge are used. The symbols usually used for charge are q or Q, and it is measured in units called “coulombs” [symbol: C]. 

==Two Types of Material==
Very broadly, materials can be described as either conductors or insulators. Conductors allow charge (usually free electrons) to move easily. Conductors are often metals. This is why wires are made of metal. 
Insulators can be charged, but the charge is trapped in or on the material and cannot move easily. Plastics are often good insulators. This is why metal wires are covered in plastic to isolate them. 
Semi-conductors fall somewhere in between.

==Nature of the Electric Force: Coulomb's Law==
From the experiments with “charged” objects (i.e., objects with the charge property), we are led to conclude that the magnitude of the electrical force depends strongly on the distance between the objects. The effects of the repulsive and attractive forces are much more noticeable when the charged objects are close together, than when they are far apart. Many careful experiments have led to the following relationship for the magnitude of the electrical force between two objects separated by a distance r, when one object has charge q1 and the other object has charge q2: <math>|F_{electric}|=k \frac{|q_1||q_2|}{r^2}</math>
 
This relationship is called “Coulomb’s law.” Here, the letter k represents a proportionality constant, which has the value <math>9 \times 10^9 N m^2/C^2</math>. 
You may see k written in terms of another constant, the permittivity of free space <math>\varepsilon_0</math>. In fact <math>k = \frac{1}{4\pi\varepsilon_0}</math>. 
Note how the function depends on the variables. The bigger the charges, the bigger the electric force. However, the function is an inverse r squared function. Therefore the bigger the r the smaller the force, the smaller the r the much bigger the force.
 The absolute value signs on the charge symbols are there because the magnitude of the force does not depend on whether the charges are positive or negative – it only depends on the amount of charge present on the objects. The direction of the force does depend on the charge sign. Note that the electrical force does not depend on the mass of the objects, or any other property – only on the charge. 
'''Note: When doing problems it is recommended to calculate the magnitude, and then think about the direction of the force separately'''

==Direction of the Electric Force==
The direction of the force depends on the relative types of the charges. If q1 and q2 both have the same type of charge – positive or negative – then the force between them is repulsive, and is directed along a straight line connecting the two charges (see figures below, left and center). If one charge is positive is the other is negative, then the force is attractive, but still directed along the line connecting the charges (see right figure below). 
 
 [[File:ElectricChargeFigure1.gif|center]] 
 
In the SI system of units, distance is measured in meters (m), charge is measured in coulombs (C), and force is measured in newtons (N). In these units, the constant k has the value <math>9 \times 10^9 N m^2/C^2</math>. So, for instance, the magnitude of the electrical force between an object with a charge of 3 C and one with a charge of 6 C, separated by a distance of 4 m, is: <math>F_{electric} =k\frac{q_1q_2}{r^2} =9\times 10^9 Nm^2/C^2\frac{3C 6C}{(4m)^2} = 1.01\times 10^{10}N</math>
 

 
This is a huge force, but a charge of 3 C is far larger than would be found on any ordinary object.
==Charge is quantized==
Although typical amounts of charge are much smaller than that, it turns out that charge never appears in quantities below a certain minimum value. This value, symbolized by the letter e, is called the “elementary charge.” It is equal to <math> 1.6 \times 10^{-19} C</math>. This can be thought of as the “minimum package size” for charge. Charge always appears in quantities that are some integer multiple of e, e.g. 35e, 17984e,<math> 3 \times 10^{6} e</math> , etc. That is to say charge is quantized, the charge always comes as a multiple of a base unit. 
So, can you obtain a quantity of charge equal to 3.5 e? [Answer: No]. 
Atoms are composed of smaller, “sub-atomic” particles, such as the proton (p), neutron (n), and electron (e). It turns out that while the neutron has no charge (and so does not experience electrical forces), both the proton and the electron have an amount of charge whose magnitude is e. However, the proton has +e (a positive charge) and the electron has –e (a negative charge). The fact that the charges on these two particles are the same magnitude is curious because their masses are so different (the proton has a mass nearly 2000 times larger than that of the electron). 
==Charge is Conserved==
An important principle about charge that that has been discovered through many experiments is called “conservation” of charge. This principle states that the total amount of charge in any closed system never changes. A “closed” system is one in which charges can neither enter nor leave. By “total amount of charge,” we mean the algebraic sum of all positive and negative charge quantities. 
 
Example: If a sealed box initially contains positive charges equal to +6 e, and negative charges equal to –11 e, the total amount of charge in that box must always remain at –5 e. It is possible that some of the positively charged particles and some of the negatively charged particles may actually disappear – such phenomena do occur (and energy is then given off in some form) – but nonetheless, the charge on the particles that would remain must always sum up to –5 e in this case. 

==The Electric Force obeys the Superposition Principle==
An important property of electrical forces is known as the “superposition” principle. This states that the net electrical force acting on any charged object is equal to the vector sum of all the individual forces on that object, where each individual force results from the interaction of the object and one other charged object. Each individual interaction is unaffected by any of the other interactions. 
That is to say: each interaction is independent. 
Algebraically, this can be expressed as: 
<math> \vec{F}_{net}=\Sigma\vec{F} = \vec{F}_{1} +\vec{F}_{2}+\vec{F}_{3}...+\vec{F}_{n}</math>
 
Here, the net force on the object is found from the '''vector sum''' of the forces from n other charged objects. This equation implies that the net x component of the force equals the sum of the individual x components, and that the net y component equals the sum of the individual y components. 
 
[http://www.independent.co.uk/news/science/power-of-the-web-the-secret-of-how-spiders-catch-their-prey-8990850.html Flying insects pick up static charge (also spiders on LSD...)] 

''Need to review Vector Algebra? See [http://gauss.vaniercollege.qc.ca/gwikis/mechanicsModules/index.php/Module_2:_Intro_to_Vectors here]''

==Short Example Questions==
Example 1: Suppose each grid square below is one meter long, and suppose that all three charges are identical, that is: Q1 = Q2 = Q3 = <math> 3\mu C</math>. What is the net force acting on charge Q2? 
 [[File:ElectricChargeFigure2.gif|center]]
 
 
Answer: Net force equals the vector sum of [force from Q3] and [force from Q1]. These forces are represented by left and right arrows, respectively, in the diagram below: 
 [[File:ElectricChargeFigure3.gif|center]]
 
Since all three charges are identical, we can use the same symbol Q to represent all of them; that is: Q1 = Q2 = Q3 = Q. 
You have to consider both the magnitude and direction of each force. The arrow pointing to the right is the force due to charge Q1; its magnitude is 
<math>|F_{1\ on\ 2}|=\frac{kQ1Q2}{(12)^2}= \frac{kQ^2}{144}</math>;
 the arrow pointing to the left is the force due to the charge Q3; its magnitude is 
<math>|F_{3\ on\ 2}|=\frac{kQ2Q3}{(6)^2}= \frac{kQ^2}{36}\ or\ 4\frac{kQ^2}{144}</math>; so it is four times larger than the magnitude of the rightward-pointing force: because it is twice as close. 
 
''Note that we are only looking at the forces acting on Q2. Q1 experience forces from Q2 and Q3. Q3 experiences forces from Q1 and Q2. Newton's 3rd Law!'' 
These forces are directed along the line connecting the charges, and so we can see that each force in this case only has an x component – the y components are zero. The rightward pointing force has a positive x component, and the leftward pointing force has a negative x component. Then the net x component is equal to
<math>F_{x_{net}}=\frac{kQ^2}{144}-4\frac{kQ^2}{144}= -3\frac{kQ^2}{144} =-1.68mN</math>
. This would be represented by an arrow pointing to the left (negative x direction) with a length of three grid squares (three times the length of the rightward pointing force). That arrow would then represent the net electrical force acting on Q2, the middle charge. 
 
----
Example 2: Suppose an object with zero net charge (i.e., exactly equal quantities of positive and negative charges) is located in the neighborhood of another object with zero net charge. What will be the net electrical force experienced by either object? 
 
Answer: There will be virtually no net electrical force on either object, because all of the repulsive and attractive forces will cancel each other out (i.e., their vector sum will be nearly zero). Show this by considering two objects, each containing two protons and two electrons, and drawing all of the force vectors on all of the charged particles. You should be able to see that the net force on each object is nearly zero, as long as they are not located too close together. 

==Long Answer Example Problem==
Three charges <math>q1 = 1.0 \mu C</math>, <math>q2 = -2.0 \mu C</math> and <math>q3 = 3.0 \mu C</math> are held fixed on a horizontal plane at positions (x=-1,y=3)cm for q1, (0,0)cm for q2 and (2,1)cm for q3. 
Find the net electric force that q2 and q3 exert on q1. [Ans: 12.8 N@205deg] 
[http://www.youtube.com/watch?v=iLd38Yu8hqA Video Solution Courtesy of John Abbott College]
:::::<youtube>iLd38Yu8hqA</youtube>

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Kevin:

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Kevin:

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Kevin:

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Kevin:

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Kevin:

ClassNotes and Compiled Class Problems: 203NYA H14

2014-05-26T12:33:35Z

Kevin:

These are files used in class containing some notes, problems worked on in class. 
The original Smart Notebook files are included, as well as a pdf export file. 
Smart Notebook files can be viewed and modified using [http://express.smarttech.com/# SmartNotebook Express] 

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_vector_WebWork.pdf VectorWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_1DMotionWebwork.pdf 1DMotionWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_1DKinematicsWebwork.pdf 1DKinematicsWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_2DKinematics.pdf 2DKinematicsWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_RotationalKinematics.pdf RotationalKinematicsWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/Test1ProblemSolvingStrategies.pdf Test1ProblemSolvingStrategies] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_Newtons_Laws_Webwork.pdf NewtonsLawsWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_CircularDynamics.pdf CircularDynamicsWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_StaticEquilibriumWebwork.pdf StaticEquilibriumWebwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/Conservation_of_Energy_Webwork.pdf Conservation of Energy Webwork] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/Strategies_for_Test2.pdf Strategies for Test2] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/203NYAH14/203NYAH14_MomentumWebwork.pdf MomentumWebwork]

Module : Rotational Dynamics

2014-05-08T00:48:25Z

Kevin:

Module : Torque

2014-04-15T17:33:20Z

Kevin: /* Magnitude of the Torque */

''Kreshnik Angoni and Kevin Lenton''

Torque ( given symbol <math>\vec{\tau}</math>) is a measure of how much a force acting on an object causes that object to ''rotate'', or more precisely to impart an angular acceleration. 
It is super important because we use rotation frequently in our everyday lives. 
For instance, the motor of your car ''rotates'' the crankshaft, which rotates the wheels. This is why we talk about the [http://www.autosnout.com/Car-Torque-List.php torque of car engines]. Each time you take a step your arms and legs are ''rotating'', your muscles are applying forces to make your arms and legs turn. 

[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Torque_StaticEquilibrium/ModuleTorque.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/Torque_StaticEquilibrium/ModuleTorque.pdf PDF Worksheet] 

==Instructions on How '''Not''' to Open a Door==
Think of how many doors you have opened in your lifetime...or more accurately how many doors you have ''rotated'', because when you open or close a door you make it rotate. 
Not surprisingly, doors are designed to be easy to open: they are designed to maximize the torque you apply on the door from the force you apply on the door handle. The bigger the torque applied, the easier the door is to open (the bigger the angular acceleration) 

Imagine pushing a door to open it. What are the parameters that influence how easy or hard it is to open the door? 
*Pivot Point
First, to rotate any object you need a pivot point. The pivot point of the door is the hinge axis: the door rotates about this point. Sometimes the pivot point is labelled O, like the origin. 
*Force
Second, you need a force! The force of your push (F) causes the door to rotate about its hinges (the pivot point, O). The bigger the force, the easier it is to close the door, and the bigger the associated torque. 
*[http://www.youtube.com/watch?v=uN14tefYzGA Distance r from pivot to where force is applied]
:::::<youtube>uN14tefYzGA</youtube> 
Next think about the position of the door handle, where you are applying the force. The door handle is almost always opposite the hinge, as far away from the hinge as possible. Let's call the distance from the hinge (pivot) to where the force is applied displacement r. The closer the force is to the hinge (i.e. the smaller r is), the harder it is to close the door. The torque you created on the door is smaller than it would have been had you correctly pushed the handle (the handle is far away from the hinge). The bigger the distance to the pivot point, the easier it is to rotate an object; the bigger the torque. 
Have you ever opened a door by pushing on the hinge? 
Well, actually, if you push hard enough on the hinge the frame will collapse and the door will open. But this would be ''translational'' motion. The door will not ''rotate'' open. 
Any force pushing on or through the hinge will never make any object rotate. The torque is always zero in this case (r=0). Is a force still being applied? Yes. 
*Direction of the Force
When you apply a force on the handle, the force is almost always perpendicular to the door itself. Only forces (or components of forces) which are '''perpendicular to the door <math>\vec{F_{\perp}}</math>can make it open'''. Forces (or components of forces) parallel to the plane of the door <math>\vec{F_{\parallel}}</math> can never make the door open: these forces are effectively pushing on the hinge (torque = zero, see above).
 
In summary, to maximize the ease of opening doors you need to: Have a big force acting perpendicular to the door plane and as far a distance as possible from the pivot point. 
[http://www.youtube.com/watch?v=BJkGIe271B4 Video showing a Force opening door: Note how the torque value at the bottom changes]
:::::<youtube>BJkGIe271B4</youtube> 
Now can you answer, how ''not'' to open a door while still applying a force? 

==Torque: Putting it together==
==Magnitude of the Torque==
The magnitude of the torque is given by: 
[[File:Torque.png|right]]
<center><math>|\vec{\tau}| = |\vec{r}||\vec{F}|\sin{\theta}</math></center> 
<math>\vec{r}</math>: the displacement vector from the pivot point to the point where force is applied, 
<math>\vec{F}</math>: The force vector applied. 
<math>\theta</math>: the angle between <math>\vec{r}</math> and <math>\vec{F}</math>. 
Note that this equation contains all the elements discussed above: 
<math>|\vec{\tau}| \propto |\vec{r}|</math>. 
<math>|\vec{\tau}| \propto |\vec{F}|</math>. 
<math>|\vec{F}|\sin{\theta}</math> is the component of the force <math>\vec{F}</math> perpendicular to <math>\vec{r}</math>, written as <math>\vec{F_{\perp}}</math> 
 
Following from this last statement, torque can also be written as: 
<center><math>|\vec{\tau}| = |\vec{r}||\vec{F_{\perp}}|</math></center> 
Where <math>|\vec{F_{\perp}}|=|\vec{F}|\sin{\theta}</math> 
===The Lever or Moment Arm===
Also following from the above equation you can write the torque equation as 
<math>|\vec{\tau}| = |\vec{r}||\vec{F}|\sin{\theta}=|\vec{r}||\vec{F_{\perp}}|\ but\ also\ as\ =\ |\vec{r_{\perp}}||\vec{F}|</math> where <math>|\vec{r_{\perp}}|=|\vec{r}|\sin{\theta}</math> 
<math>\vec{r_{\perp}}</math> is called '''lever or moment arm''' and is the perpendicular distance between the line of action of the force and the pivot. 
The line of action of a force F is the line through the point at which F is applied and along the direction in which F is applied. That is to say it is the continuation of the line the force acts through.
 [[File:MomentArm1.jpg|center]] 

==Cross Product Definition==
All the above equations can be summed up as a cross (or vector) product. Technically, torque is the cross (or vector) product between the distance vector <math>\vec{r}</math>(the distance from the pivot point to the point where force is applied) and the force vector <math>\vec{F}</math>.
This is written as: 
<center><math>\vec{\tau} = \vec{r} \times \vec{F}</math></center>

==Units of Torque==
Note that in terms of units, torque is the product of force and distance, therefore the SI unit of torque is the Newton-meter <math>N\centerdot m</math> (American units are lb-ft (pound-feet)). The Work Done (or energy) also has units of Newton-meters, but is given the special unit of Joule (the unit for energy). 
However, torque is not energy. So, to avoid confusion, make sure you use the units N.m, and not J, when referring to torque. The distinction arises because energy is a scalar quantity, arising from the scalar product, whereas torque is a vector, arising from the vector product of r and F. 

==Direction of Torque==
Torque is a vector and therefore has a direction. 

The direction of torque is always perpendicular to <math>\vec{r} </math>and to <math>\vec{F}</math>. Using the right hand rule (wait for your Physics NYB course if you don't know this), you can find the direction of the torque vector. If we put our right hand fingers in the direction of r, and curl them to the direction of F, then the thumb points in the direction of the torque vector. 
However this course will be using special cases where the axis of rotation is given.
You will be using the following convention: 
The torque is '''positive''' when the torque causes the object to rotate in the '''counter-clockwise''' direction and 
The torque is '''negative''' when the torque causes the object to rotate in the '''clockwise''' direction. 

[http://www.physics.uoguelph.ca/tutorials/torque/Q.torque.example.torque.html SIMPLE EXAMPLE PROBLEM ON TORQUE: The Swinging Door]
 
[[http://gauss.vaniercollege.qc.ca/pwiki/index.php/Exercises_on_Torque OTHER EXERCISES ON TORQUE]] 

==Misconception: Torque is Force ==
'''False''' 
Torque is a rotational equivalent of force, but it is not force itself. 
As you have seen, <math>|\vec{\tau}| = |\vec{r}||\vec{F}|\sin{\theta}</math>. 
Note that the equation contains a force term, so you can use the concept of torque to find forces in your system.

==Static Equilibrium ==

First, what does it mean for an object to be in equilibrium? 
A body is at translational equilibrium if the linear acceleration of its Centre of Mass is zero ( <math>\vec{a_{CM}} =0</math>). That is to say <math>\Sigma\vec{F}_{net} = 0</math> 
A body is at rotational equilibrium if the angular acceleration is zero ( <math>\vec{\alpha} =0</math>). That is to say <math>\Sigma\vec{\tau}_{net} = 0</math> 

*An object is in static equilibrium '''if and only if''' <math>\vec{\tau}_{net} = 0</math> and <math>\vec{F}_{net} = 0</math>, where the net torque is the sum of all torques and the net force is the sum of all forces acting on an object. 

For Static Equilibirum the object has to have zero speed (linear & angular velocities are zero)it is in static equilibrium. If it does have speed, but not acceleration (i.e. it is moving but not accelerating), then it is in dynamic equilibrium. 

==A note about the Centre of Mass of an Object==
The weight of a body is a force that participates in all static’s problems. It appears in the conditions for
translational equilibrium and its torque appears in the condition for rotational equilibrium. Actually, the
weight of an object is the sum of all gravitation forces exerted on all the particles distributed over the body
extension. We locate this sum at the center of gravity of object. The center of gravity (CG) is the point about
which the net gravitational force exerted on body has a zero torque. It is the point at which the mass is said to act. 
The position of CG is very critical for the rotational
equilibrium. How can you find the CG position? 
The easiest way is to balance the object on a pivot. When the pivot is located at the centre of mass, the object will balance on the pivot. 
For a uniform object, the centre of mass is right in the middle of the object.

===Example Problem: The Hanging Traffic Light===
 [[File:TrafficLight.jpg|center]] 
A traffic light hangs from a pole as shown in the figure. The uniform aluminum pole AB is 8 m long and has a mass of 15 kg. The mass of the traffic light is 21 kg. 
a) Determine the tension in the horizontal massless cable CD. 
b) Determine the horizontal and vertical components of the force exerted by the pivot A on the aluminum pole.
 
* To tackle the problem you have to realize that there are torques present in the situation. The forces are acting at distances from each other. 
Think about what will happen if the cord is cut: the pole and traffic light will ''rotate'' about the attached point at A. 
* The system is in static equilibrium: everything is stationary, so you can solve the problem by applying sum of the forces is zero, sum of the torques is zero 
As previously, you will get a system of equations to solve. 
* Next you should draw an ''Extended Body Diagram''. An extended body diagram is like a free body diagram in that it shows all the forces, but at the correct distances. To find torque you need three parameters: F, r and <math> \theta </math>. Remember, each force will have its own torque about the pivot point. 
All these parameters should be shown on the extended body diagram for each force (and therefore for each torque). You will have to get all these parameters from information in the problem. For instance, you are told the pole is uniform, therefore its weight, which acts at its centre of mass, acts right in the middle of the pole.
 [[File:TrafficLightStaticEquilibriumExtendedBodyDiagram.jpg|center]] 
* Note the coordinate system: counterclockwise torques are positive (torques only!), positive x and y are also indicated (forces only!).
* Pay attention to the forces <math> F_x</math> and <math> F_y</math>. These are the horizontal and vertical forces acting at the attachment point. Really there is only one force <math> F_{total}</math>, you are splitting this up into components to fit your coordinate system. How do you know these forces are there? Imagine if there were no forces at A - that is to say the pole is not attached. What would happen? The whole system would collapse.
* Choose a pivot point for the problem. In static equilibrium you can choose any pivot point you want and solve the problem. However, some pivot points are more judicious than others. Choose pivot points where you have unknown forces. In this case you have unknown forces at A. It makes sense to choose this as your pivot point because the torque of these forces about this pivot are equal to zero. They will not appear in your torque equation. There is an unknown force of Tension T in the cable, but that is what you are looking for, so you want to keep that term in your equation. 
* Just to make the point, here is a table of all the values to calculate the torques for each of the forces.
{| class="wikitable"
|-
! Force !! F(N) !! r (m) !! <math> \theta </math>!! Torque Direction!! <math> \tau =rFsin\theta</math> (Nm)
|-
| <math> F_x</math>|| <math> F_x</math>|| 0|| -|| -|| 0
|-
| <math> F_y</math>|| <math> F_y</math>|| 0|| -|| -|| 0
|-
| <math> W_{pole}</math>|| =mg=15kg*9.8 =147|| 4 m|| 53|| Clockwise|| -469.6
|-
| <math>Tension T</math>|| T|| <math> =\frac{3.8m}{sin37}=6.31m</math>|| 37|| CounterClockwise|| +3.8T
|-
| <math> W_{light}</math>|| =mg=21kg*9.8 =205.8N|| 8 m|| 53|| Clockwise|| -1314.9
|}
Note that for the Tension, you could use the concept of line of action to get the same result. It doesn't matter. 
Also note that according to the right hand rule, you should take the external angle between r and F. However, sin(37)=sin(180-37) (Try it if you don't believe it). Practically it does not matter which angle you choose, in either case you are finding the component of force perpendicular to r. 
Now: 
<math>\sum \tau =0</math> 
<math>\therefore \sum \tau = \tau_{F_x}+ \tau_{F_y}+\tau_{W_{pole}}+\tau_{Tension}+\tau_{W_{light}}=0</math> 
<math> 0+ 0-469.6+3.8T+-1314.9=0</math> 
<math> T=469 N</math> 
You can find the forces <math> F_x</math> and <math> F_y</math> by using the fact that for static equilibrium <math> \sum F=0</math> and therefore <math> \sum F_x=0</math> , <math> \sum F_y=0</math>. 
{| class="wikitable"
|-
! Axis !! Sum of forces
|-
| Horizontal x|| <math> \sum F_x= F_x - T = 0</math> <math>\therefore F_x = T = 469N</math>
|-
| Vertical y|| <math> \sum F_y= F_y - W_{pole} - W_{light} = 0</math> <math> F_y = W_{pole} + W_{light} = 352N</math>
|}

===General Static equilibrium '''problem solving strategy''':===
#Draw an extended body diagram
#Select the coordinate system
#Determine all forces acting on an object and draw an arrow for each force indicating:
#*A point where this force is applied
#*In which direction this force acts
#Define a pivot point: Any pivot point will work. But to make your life easier, choose a pivot point through which unknown forces act. The torque of these forces about this pivot point will be zero.
#For each force:
#*Identify the magnitude and the direction of lever arm
#*Draw a diagram consisting of the force vectors, r vectors and determine the angle between these two vectors
#*Determine the direction of the torque and then compute its magnitude.
#Substitute into the equation <math>\Sigma \tau = 0</math>
#Compute the components of the forces
#Substitute into the equations <math>\Sigma F_x = 0</math> and <math>\Sigma F_y = 0</math>
#Solve the system of three equations using the method of substitution
 
[http://www.youtube.com/watch?v=oB6cC6LPZ_0 Video Explanation+ExamplePartI]

:::::<youtube>oB6cC6LPZ_0</youtube> 

[http://www.youtube.com/watch?v=dbjNX91EQh0 Example Part II] 
:::::<youtube>dbjNX91EQh0</youtube> 
[http://gauss.vaniercollege.qc.ca/pwiki/index.php/Exercises_on_Static_Equilibrium EXERCISES ON STATIC EQUILIBRIUM]

Module : Work and Kinetic Energy

2014-04-13T00:31:19Z

Kevin:

''Kreshnik Angoni and Kevin Lenton'' 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.docx Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/NYB/MagneticInduction/MagneticInduction.pdf Worksheet.pdf] 

 
Other Resources: 
*Haliday & Resnick, Fundamentals of Physics 30.1-5
*[http://cnx.org/content/m42146/latest/?collection=col11406/1.7 Openstax]
* [http://fclass.vaniercollege.qc.ca/~angonik/FOV1-00075435/FOV1-0007AA75/NYB_Lect_10.pdf?FCItemID=S0027DBAB Printable version]
 

Two concepts, work and energy, are widely used when studying mechanics because they are scalar quantities and therefore much easier to work with (excuse the pun) than force (a vector). Especially, if the force changes (either magnitude, direction or both) during the object's motion (for example; the force of a spring , the normal or friction force along a curved surface), it is easier to use work and energy instead of Newton’s laws. 
Consider the following scenario: 
When lifting a body up with zero acceleration you have to exert a force with magnitude equal to that of the body's weight. Everyday experience shows that: 
• For the same vertical distance moved, the bigger the weight is, the more muscular exertion is needed; 
• For the same weight, the bigger the vertical distance moved is, the more muscular exertion is needed; 
This shows that the quantity (force times distance) can be a meaningful measure to characterize the activity exerted by a “source” that moves up an object from one level to higher one. This essential information, intuitively confirmed, opens the way for the precise definition of mechanical work. 
This is linked to the concept of energy, a first step approach would be to figure it out as the “capacity of a source to provide mechanical work”. This way of looking at energy means that “the work provided by a source should be equal to the change of energy in the source that provides it “ and this definition is good enough as long as we deal only with mechanical phenomena without heat production. 
Work is energy transferred by a force.
== THE WORK DONE BY A CONSTANT FORCE ==

A constant force F.
has done work only if the point where the force is applied moves. 
If the object displacement is
s , the force must also have moved by this amount and the work done is equal to: 
<math>W=\vec{F} \cdot \vec{s}=F\ s\ cos\theta \qquad \qquad\qquad(1)</math> 

<math>\theta</math> (constant) is the angle between the direction of force
and the direction of displacement vectors. Note that: 
•The work is done by the source that exerts the force. 
•If the displacement is zero, there is zero work even if
the source makes an effort (applies the force). 
•Only the component of the force along the displacement
direction achieves work. The force component
perpendicular to displacement achieves zero work
because in this case <math>\theta = 90^\circ </math>.

 [[File:FigureWork1.png|left|250x188px|alt=Figure 1]] 
As it is defined by a scalar product, the work is a scalar physical quantity. 
Its unit is a derived unit; in the SI system its unit is the Joule (J);
1J = 1N*1m (force applied in the displacement direction) (2)

 
You can calculate the work done by a constant force F during a displacement s by using the
components of those vectors in any Cartesian frame through the formula : 
<math>
W =
\vec{F} \cdot \vec{s}=
F_x \times s_x +
F_y \times s_y +
F_z \times s_z =
F_x \Delta x +
F_y \Delta y +
F_z \Delta z
\qquad\qquad\qquad(3) </math> 

Remember a) <math>\vec{F} =
F_x \mathbf{\hat{ \imath } } +
F_y \mathbf{\hat{ \jmath } }+
F_z \mathbf{\hat{ k } } </math> and <math>\vec{s} =
x \mathbf{\hat{ \imath } } +
y \mathbf{\hat{ \jmath } }+
z \mathbf{\hat{ k } } </math> and 
b) Rules of scalar product
 

Meanwhile, the numerical value of work does ''not'' depend on the frame of reference, because the magnitudes of the vectors F , s and the angle between them (see expression 1) do not depend on any frame. 

The work done by <math>\vec{F_G}</math> or <math>\vec{N}</math> when a block slides over a horizontal plane (fig.2) is zero because both these forces are
both perpendicular to the displacement vector. For the same reason '''the work done by centripetal forces is
always zero'''.
When considering the work done by friction forces you have to verify carefully the directions of the displacement and the friction force vectors before starting any numerical calculations. In the case shown in fig 3.a the friction force on the block is ''opposite'' in direction compared todisplacement vector and
<math> W_f=\vec{f_k}\cdot \vec{s}=f_k\ s\ cos 180^\circ =-f_k.s \qquad\qquad\qquad(4)</math> 
So, the friction does ''negative work'' on the block.
Note that the source of the friction force is the floor.
In the case of figure 3.b, the force <math>F_0</math> pushes the block B forward. Due to its inertia, the block A tends

 [[File:FigureWork3.png|left|250x188px|alt=Figure 1]] 
to keep its motion status and this way it would tend to slide left over block B. Meanwhile, block B pulls A to the right via the friction force. In this case the (static or kinetic) friction force exerted on
block A has the same direction as the displacement of A and its work is positive. 
<math>W_f = + fk\ s </math>
 
When a set of forces <math>F_1, F_2 ,..., F_n</math> is acting on the same body that is displaced by <math>\vec{s}</math>, each force
does work independently of the forces. 
<math>W_1=
\vec{F_1}\cdot \vec{ s};W_2=
\vec{F_2}\cdot \vec{ s}_............W_n=
\vec{F_n}\cdot \vec{ s}</math> (5) 
The total work done on the object is <math>W_{net} =
W_1 +W_2 +
..W_n =
(\vec{F_1} +
\vec{F_2} +
...\vec{F_n} )\cdot \vec{s} =
\vec{F_{net}} \cdot \vec{s} \qquad\qquad\qquad(6)</math>

 

<math>\vec{F_{net}} = \sum ^n_i \vec{F_i}\qquad </math> is the vector sum of all forces exerted on the body. 
From (1) you can see that the work done by a force can be positive (if <math>\theta < 90 ^\circ </math> ) or negative (if <math>\theta > 90 ^\circ </math>).<be>
In figure 2, the hand “object A” pushes the block “object B” and moves
it through the displacement s. This way, via the force exerted on B,
“object A” does positive work 
<math>W_{onBbyA}=\vec{F_{BA}}\cdot \vec{s} = F_{BA}\ s\ cos 0 ^\circ \qquad\qquad\qquad(7)</math>

 

 [[File:FigureWork2.png|left|250x188px|alt=Figure 1]] 
The Newton's third law states that the force that block “object B “exerts on the hand ''“object A”'' is equal in
magnitude but opposite in direction to <math>\vec{F_{ BA}} </math>. 
<math>\vec{F_{AB}}=-\vec{F_{BA}}\qquad\qquad\qquad(8)</math> 

The work done by the force of the block on the hand ( <math>W_{\vec{F_{AB}}}</math> work) after the hand is moved by <math>\vec{s}</math> is 
<math>W_{onAbyB}=\vec{F_{AB}}\cdot \vec{s}=-\vec{F_{ABA}}\cdot \vec{s}=-W_{onBbyA}\qquad\qquad\qquad(9)</math> 

So, when object A (the hand) does positive work to move object B (the block), then object B (block)
does the same amount of work (but negative) on object A, which is the “source of force that moves it”.

The work done by the force of gravity is an important first step in introducing the concept of energy . Consider a block that is moved up an inclined plane under the effect of a pulling force (fig.4). Let’s ignore the action of other forces and concentrate on the work done by the weight. We define the displacement
vector and select a frame of axes as shown. The selected system for the axes is not appropriate to study the motion using Newton's laws but it allows us to infer some important results and simplifies the calculation of work by gravity. The components of the force of gravity along these x,y coordinates are <math>F_{G_x}=0; F_{G_y}=-mg</math> and those of the displacement are <math>s_{x}=\Delta x; s_{y}=\Delta y</math>. 
So, equation (3) gives: 
<math>W_{onBlockBbyGravity}=(0 \mathbf{\hat{ \imath } }-mg \mathbf{\hat{ \jmath } })\cdot (\Delta x \mathbf{\hat{ \imath } } + \Delta y \mathbf{\hat{ \jmath } })=-mg\Delta y</math> 

<math>W_{onBlockBbyGravity}=-mg(y_f-y_i \qquad\qquad\qquad(10)</math> 

When the block is moved up, yf > yi and the work
done by gravity on the block is negative.
When the block ''slides down'', yf < yi the work done by gravity on the block is positive.
 [[File:FigureWork4.png|left|250x188px|alt=Figure 1]] 
Note: If, initially, the block is moved up from height yi to yf and then left to slide down
to yi , the total work done by gravity force on the block is zero. 

<math>W^{up}_G +W^{down}_G =[-mg(y_f-y_i)]+[-mg(y_i-y_f)]</math> 

<math>W_{G-total}=-mg\ [(y_f -y_i )+(y_i -y_f )]=0</math> 
 

So: The work done by gravity depends only on the initial and final heights and not on the
path followed during the displacement. This means that, the work done by gravity is zero
for any closed path (final location is the same as the initial location).

==THE AREA TECHNIQUE FOR WORK CALCULATIONS IN ONE DIMENSION SPACE ==

For a constant force in ''one dimension space'': 
<math> \vec{F} =
F_x \hat{\imath}\ ;\ \vec{s} =
x \hat{\imath}\qquad and \qquad W =
F_x \Delta x =
F (x_f -x_i )\qquad\qquad\qquad(11) </math> 

This work is equal to the shaded area under the graph f = f(x) shown in figure 5.a. If you set the
origin at xi=0 then <math>W =F_xx_f</math> corresponds to the area under the force graph from xi= 0 to x = xf.

 [[File:FigureWork5.png|center|500x376px|alt=Figure 1]] 
Note that when this area is located in first or third quadrant it is positive i.e. force and displacement are in the same direction, (both either positive or negative)work is positive ; when it is located in the second or fourth quadrant it does negative work ( fig.5.c, d)[the force is opposite to the displacement].

==Work done by the Force of a Spring==
The area technique offers an easy way for calculating work in the case of variable forces. In this case, the force has a different value for
each location of the object under study. The elastic force of a spring is a
good example of such type of force; when compressed or extended, an ideal (massless) spring, exerts an “elastic or restoring force” given by Hooke’s law:
<math> F_{spring} =-k x \qquad\qquad\qquad(12) </math> 
k [N/m] is the spring constant; x is the displacement from equilibrium; +x=extended; -x=compressed. 

This expression assumes that xi = 0 (at the equilibrium position, the spring is unextended); in this way the displacement
becomes equal to the algebraic value of x. 
When x > 0 (spring extended) the force Fsp < 0 and when x < 0 (spring
compressed) the force Fsp > 0. So, in any case, the spring’s force tends to
restore the equilibrium position of spring. The forces that tend to bring the system to its equilibrium configuration are known as restoring forces. 
 [[File:FigureWork6.png|center|500x376px|alt=Figure 1]] 
Figure 7 shows the graph of equation (12). The work done by elastic
force when its application point (spring end) is shifted from xi to xf is given
by the shaded area under the graph. By calculating this as the difference of
the two triangle areas one finds out that 
<math>
W_{spring} =(-kx_f \frac{x_f}{2}) -
(-kx_i \frac{x_i}{2}) = \frac{k}{2}
(x_i^2 -
x_f^2 )\qquad\qquad (13)
</math> 
In general, the work done by a spring is often written as
<math>
W_{spring} = \frac{1}{2}kx^2
</math>, just remember it is always the change that is important. 
Expression (13) shows that the work done by a spring force (an elastic force)
depends only on the initial and final positions of its moving end.
The work done by a spring is positive when the direction of its force is the same as that of
displacement. This happens when the block is moving towards the equilibrium
point, i.e. when <math>|x_i|>|x_f|</math> 

The spring’s work is negative when the direction of the restoring force is opposite to that of displacement. This happens when the block is moving away from the equilibrium point, i.e. when <math>|x_i|<|x_f|</math> 
 [[File:FigureWork7.png|center|500x376px|alt=Figure 1]] 
Also, for a full oscillation, the spring’s work is zero because xf = xi (the spring comes back to its initial position).
 
Note: The area technique may be used for any variable force but you should keep in mind that the sign
of the area (i.e. work sign) under the graph depends on the location of starting point versus the location of
final point. The sign of “areas” in figures 5.a,b,c,d is related to the location of starting point at x = 0.

==THE WORK – ENERGY THEOREM ==

Consider a particle (an object) with mass m in translational motion with acceleration a along the x-direction
under the effect of a net constant force <math>\vec{F}_{NET}</math> directed along the same direction. The work done
by the force <math>\vec{F}_{NET}</math> for the displacement of the particle by <math>\Delta x = x_f - x_i </math>, is
 
<math>
W_{NET} =
F_{NET} \cdot \Delta x =
(ma)* \Delta x =
m *(a\Delta x)\qquad\qquad\qquad (14)
</math>
 [[File:FigureWork8.png|center|500x376px|alt=Figure 1]] 
By using the kinematic relationships between initial and final velocities you get: 
<math>
v_f^2-v_i^2=
2a \Delta x
\rightarrow
a\Delta x = \frac{1}{2}v_f^2-\frac{1}{2}v_i^2
\qquad\qquad\qquad
(15)
</math>
 
and by substituting in eq. (14) 
<math>W_{NET} =\frac{1}{2}mv_f^2-\frac{1}{2}mv_i^2 \qquad\qquad\qquad
(16)
</math>

==Definition of Kinetic Energy==
The quantity 
<math>K =\frac{1}{2}mv^2 \qquad\qquad\qquad
(17)
</math> 
is called the KINETIC ENERGY of the particle. 
This means that you can express the work WNET done by the net force FNET through a difference of kinetic energy. 
<math>
W_{NET} =\Delta K =
K_f -
K_i \qquad\qquad\qquad(18)
</math> 

==The Work-Kinetic Energy Theorem==
Equation (18) is the mathematical expression of the Work-Kinetic Energy theorem (or simply the Work-Energy Theorem) which states that: 

'''The net work done on a particle is equal to the change of its kinetic energy. ''' 
So, the energy is defined as a mathematical function whose change is equal to physics concept of
work. Equation (17) shows that the numerical value of the kinetic energy does depend on the
frame of reference. In another frame of reference (in uniform motion versus a first one), the velocity of the particle is different and therefore
so is its kinetic energy. This helps to keep in mind that the numerical value of kinetic energy is not very meaningful. However, the ''change'' of kinetic energy is independent on the reference
frame and does have a precise physical meaning; it is equal to the net work done on the particle. 
For problems related to our experience here on earth, you could use an inertial frame tied to the earth. In this case, the kinetic energy of the terrestrial object is the work needed to increase its speed from zero to the given value
( put Ki = 0 in eq. (18) and get <math> W_{NET} = K_f = \frac{1}{2} m v_i^2 </math>
). 
In the same way, consider the situation when the object velocity is decreased from v to zero due to the stopping force external to the object.
In this case, Kf = 0 and the net work done on the object is negative 
<math> W_{NET} = -K_f = -\frac{1}{2}mv_i^2 </math>
 
 

Remembering the third law, one infers that in this case, the object under consideration does positive work on
the surrounding medium and in doing so it loses some of its kinetic energy. Note that this assertion is consistent with the
idea about the energy as ''the capacity of this particle to provide work''. 

Important note: We derived the work-energy theorem for the case of constant force acting in a
one-dimensional space. This theorem remains true in all cases, that is: 
-variable forces 

-acting in a two or three-dimensional space.

:::::<youtube>9dtOMK6cnMY</youtube>

==POWER==

Suppose that one needs to shift an object from point A to point B. This requires providing a certain
amount of work and one has to choose between different sources that can provide this amount of work.
Which one to chose? If one decides to use “the source that achieves faster displacement” one has to
consider another physical quantity; the mechanical power. 
 
[[File:FigureWork9.png|center|500x372px|alt=Figure 1]] 

The mechanical power is the rate at which a given work is done. If the “source” delivers a portion of
work <math>\Delta W</math> in the time interval <math>\Delta t</math>, you can say that it is delivering an average power equal to: 
<math>
P_{average}=\frac{\Delta W}{\Delta t}
\qquad
\qquad
\qquad
(19)
</math>

===Units of Power===

The unit of power in SI system is the “Watt “ 1W = 1J /1s (20)

Note: In the car industry anoth widely used unit of power is ”horsepower”; 1hp = 745.669W 

The instantaneous mechanical power is defined as 
<math>
P = lim_{|Delta t \rightarrow 0}\frac{\Delta W}{\Delta t}\equiv \frac{d W}{d t}
\qquad
\qquad
\qquad
(21)
</math> 
We may assume that the “source” is acting on the object with a constant force F during an infinitesimal displacement ds. This means that it has provided the work 
<math>dW =
\vec{F}\cdot d\vec{s}
\qquad
\qquad
\qquad(22)
</math> 
If this happens during the infinitesimal time interval dt, it means that the object was moving with an instantaneous velocity 
<math>\vec{v}=\frac{d\vec{s}}{dt}(23)</math> from which we derive <math>d\vec{s} =\vec{v}dt</math> 

After substitution in (22) we get 
<math>dW =
\vec{F}\vec{v}dt
\qquad\qquad\qquad(24) </math>

 
and finally, based on (21) 
<math>P=\frac{dW}{dt}=\vec{f}\cdot \vec{v}
\qquad
\qquad
\qquad
(25)
</math>
 
The expression (25) is based on instantaneous vectors (force and velocity). Note that the power is a
physical parameter that may change in time. But, if the force and velocity are constant vectors, the
power is constant, too. In this case, the source provides a constant power in time. 
From a formal point of view, the work done by the source of the force on an object can be seen as the portion of
mechanical energy transferred from the “source to the adjacent regions of space”. This way, the delivered power [watt]
from a “source” would be expressed as 
<math>

P =-\frac{dE}{dt}
\qquad
\qquad
\qquad (26)
</math>
 

where E [Joules] stands for the total mechanical energy of the “source” that provides the mechanical power . The sign “ - “
in expression (26) is related to the fact that the energy of the source is decreased when providing energy. The power received by the object is
positive in the sense that positive work was done on the object.

Module : Applications of Newton' Laws : Circular Motion

2014-03-18T00:57:29Z

Kevin: /* Example 1: The Conical Pendulum */

''Kreshnik Angoni, Karen Tennenhouse and Kevin Lenton''
 [http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/CircularMotionDynamics/ModuleCircularMotionDynamicsWorksheet.docx Word Worksheet] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/CircularMotionDynamics/ModuleCircularMotionDynamicsWorksheet.pdf pdfWorksheet] 
==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand why uniform circular motion requires a radial force.
* Know some of the common pitfalls of circular motion dynamics.
* Apply the concepts of circular motion to some circular motion dynamics problems
 

===THE DYNAMICS OF CIRCULAR MOTION===
 
[http://gauss.vaniercollege.qc.ca/gwikis/mechanicsModules/index.php/Module_6:_Circular_Motion In kinematics] we found that a particle moving at constant speed v around a circle with radius r has acceleration. This acceleration is directed towards the circle center at all times, and has constant magnitude: 

<math>|\vec{a_c}| = \frac{v^2}{r}</math> 

[[image: Kreshnik_Circ Dyn_Fig1.png]]

 The vector <math>\vec{a_c}</math> is called '''''centripetal acceleration''''' because it is directed all times towards the circle center. The 2nd law tells that a particle with mass <math>m</math> which is moving with centripetal acceleration is under the action of a net force directed along <math>\vec{a_c}</math> (fig.1), that is to say towards the centre of the circle (or the negative radial axis). 

<math>\sum \vec{F}_{radial} =\sum \vec{F_c} = m \vec{a_c} </math> ........ (2) 
That is to say the direction of the net radial force must be in the same direction as the centripetal acceleration, towards the centre of the circle. 
Or: 

<math>|\sum \vec{F}_{radial} |= m\frac{ v^2}{r}</math> ........ (3) 
This force, with magnitude <math>m\frac{ v^2}{r}</math>, is called centripetal force because it is directed at all times towards the circle center. Note that it is named only from its ''direction'' and not from its physical origin.
 
For example, an object of mass 123 kg going around a circle of radius 10m with a velocity equal to 25m/s requires a net radial force equal to: 
<math>|\sum \vec{F}_{radial} |= m\frac{ v^2}{r}= 123\frac{ 25^2}{10}= 7687 N</math> 
If the force is greater than this, the object will be pulled in closer to the centre (no longer a uniform circle). If the force is smaller than this, the object will fall out of the circle.
 
This force could be a tension, a normal force, a weight, or a combination of different forces. The point is that the net force must be exactly that to pull the object into a circle. Also, the forces must be real physical forces in the system.

===Period of the Circular Motion===
The period of the circular motion is the time for the object to complete one revolution, one circle. 
The distance travelled in this time is the circumference of the circle <math> 2\pi r</math>. 
Therefore <math> v =\frac{2\pi r}{T_{period}}</math>. 
Do not confuse <math>T_{period}</math> with <math>T_{Tension}</math>
 

==Misconception #1: Magical Forces==
Some students think that whenever an object moves in a circle there is a magical centripetal force which should appear on their free body diagrams. 
'''FALSE''' 
The radial forces come from the radial forces already present in the system you are studying e.g. A force of gravity, friction, …or their sum can contribute to a centripetal force. See a fuller list [http://gauss.vaniercollege.qc.ca/gwikis/mechanicsModules/index.php/Module_8:_Newton%27s_Laws_of_Motion#An_overview_of_the_forces_acting_on_a_body_in_a_mechanical_system here] 

'''''Remember: You must not add a magical centripetal force to a free body diagram. The required radial forces come from physical forces already present.'''''. 
 

==Misconception #2: Centripetal or Centrifugal?==
When you go around a circle at speed, you seem to be thrown out of the circle. Some people interpret this as a force pushing them out, a centrifugal force. 
'''FALSE''' 
If you think about circular motion, the forces are always pushing into the circle, that is to say pulling the object into the centre of the circle. 
The reason you may think otherwise is due to the ''inertia'' of the object. Newton 1 tells us that objects want to continue at constant velocity, in a straight line. This is the tangent to the circle, in the case of circular motion. It requires a force to pull the object into the circle, so it doesn't move in a straight line. 
The action of centripetal force consists of ''not'' letting the object follow its path along the tangent “as required by its inertia”. 
Centrifugal effects occur as a mathematical consequence of changing your frame of reference to a non-inertial frame. This is not covered in this course.
 
'''''Remember: For this course, there is no such thing as a Centrifugal force: the net radial force always pulls to the centre of the circle.'''''. 

=='''Example 1: The Conical Pendulum'''==
A conical pendulum is a pendulum made to go around in a horizontal circle, rather than back and forth. 

A mass m is swung by a string length l at an angle <math>\theta</math> as shown. Find expressions for the speed v and the period of rotation. 
[[image: conicalpendulum.png]]
 

===Choose your axes===
First note the chosen axes. There is a vertical y axis, and a radial r axis. Positive y is up, positive r is towards the centre of the circle. Don’t forget to always define this. The radial axis in fact goes around the circle with the object, always pointing towards the centre. Therefore, the axes shown are at this point in time. 
Why have you chosen these axes? You know there must be an acceleration towards the centre of the circle. Choose your axes to correspond. 
There are only two forces acting on the mass: the tension, and the weight of the object. The weight is straight down: y only. Therefore it must be a component of the tension which pulls into the circle. 
 

===Equations in the y and radial directions===
As before you can write equations of motion from Newton’s first and second laws in the radial and y direction. You usually solve these problems by getting one equation from axis, one equation from the other, and solving simultaneously 
{| class="wikitable"
|-
! Equation in y direction !! Equation in radial direction
|-
|
<math> \sum F_{y}= T\cos\theta -mg =ma_y=0</math> 
Note that the acceleration is zero because the mass is not moving up or down along the y axis. 
This is therefore an example of Newton's 1st Law. 
Also note that the tension is ''not'' equal to the weight of the mass. Rather <math> T = \frac{mg}{\cos\theta } </math> 
 
||
In the radial direction, there is circular motion. 
There must be a force pulling the mass into the circle. 
This force comes from the component of the tension. 
<math> \sum F_{radial}= T\sin\theta </math> 
Note that there is a net force, therefore the system is not in equilibrium. =Newton's 2 Law 
Now this radial force gives rise to radial acceleration. Therefore: 
<math> \sum F_{radial}= T\sin\theta =ma_{centripetal}=m\frac{v^{2}}{r} </math> 
Rearranging: 
<math> v^2=\frac{rT\sin\theta}{m}</math>

|}

 

===Resolve the Equations===
 
As in many Newton2 problems, you take an equation from one axis and substitute it into the other axis and solve for what you are looking for. In this case, let's find an expression for the speed of the mass going around, and the period of rotation.

 
Substitute in the expression for T from the y axis, into the radial equation. 
<math> v=\sqrt{\frac{r}{m}(\frac{mg}{\cos\theta})\sin\theta}=\sqrt{\frac{rg\sin\theta}{\cos\theta}}</math> 
Rather than r we would like an expression in <math> l</math> so we sub in <math> r=l\sin\theta</math> 

<math> v=\sqrt{\frac{lg\sin^{2}\theta}{\cos\theta}}</math> 
The period of rotation, also sometimes labelled T is given by: 

<math> T_{period}= \frac{2\pi r}{v}= 2\pi l \sin\theta
\sqrt{\frac{\cos\theta}{lg\sin^{2}\theta}}=2\pi\sqrt{\frac{l\cos\theta}{g}}</math>
 
!!!!Note: This equation only applies to the specific case of the conical pendulum. It is not a general equation for all pendula. What you need to remember is the '''strategy''' for solving circular motion problems!!!!
 

=='''Example 2: Car going around a bend in the road'''==
A car travels at high speed around a horizontal curved path of radius r. 
The coefficient of static friction between the tires and the road is <math> \mu_s</math> 
The coefficient of kinetic friction between the tires and the road is <math> \mu_k</math> 
Find the maximum speed that the car can go around the circle. 

First visualize the problem: 
[[image: cargoingaroundbend.jpeg]] 
The car is going around at constant speed. However, the velocity vector is tangential to the circle at any particular point. 
How does friction affect the car? 
You can test how friction affects the system by thinking about how the car will react if the road is covered in ice (i.e. no friction) The car will go straight (Newton 1)! 
Therefore it is the force of friction which keeps the car in the circle. 
But what sort of friction is it? The tires are not spinning on the road. Therefore the rubber tire surface is static relative to the road surface. It is the force of static friction acting here. 
The net force must be radial pulling the car into the circle. 
[[image: cargoingaroundbendFBD.jpeg]] 

===Choose your axes===
First note the chosen axes. There is a vertical y axis, and a radial r axis. Positive y is up, positive r is towards the centre of the circle. Don’t forget to always define this. The radial axis in fact goes around the circle with the car, always pointing towards the centre. Therefore, the axes shown are at this one point in time. 
Why have you chosen these axes? You know there must be an acceleration towards the centre of the circle. Choose your axes to correspond. 
There are only three forces acting on the mass: the force of friction, the normal, and the weight of the object. The weight and normal force are straight down: y only. Therefore it must be the force of friction which pulls into the circle. 
 

===Equations in the y and radial directions===
As before you can write equations of motion from Newton’s first and second laws in the radial and y direction. You usually solve these problems by getting one equation from axis, one equation from the other, and solving simultaneously 
{| class="wikitable"
|-
! Equation in y direction !! Equation in radial direction
|-
|
<math> \sum F_{y}= n -mg =ma_y=0</math> 
Note that the acceleration is zero because the car is not moving up or down along the y axis. 
This is therefore an example of Newton's 1st Law. 
<math> \therefore n = mg</math> 
 
||
In the radial direction, there is circular motion. 
There must be a force pulling the mass into the circle. 
This force comes from the force of friction. 
<math> \sum F_{radial}= F_{friction} </math> 
We are asked for the ''max'' speed, therefore we want the ''max'' force of static friction. 
<math> F_{friction_{max}} = \mu_s n</math> 
Note that there is a net force, therefore the system is not in equilibrium. =Newton's 2 Law 
Now this radial force gives rise to radial acceleration. Therefore: 
<math> \sum F_{radial}= \mu_s n =ma_{centripetal}=m\frac{v^{2}}{r} </math> 
Rearranging: 
<math> v^2=\frac{r\mu_s n}{m}</math>

|}

 

===Resolve the Equations===
 
As in many Newton2 problems, you take an equation from one axis and substitute it into the other axis and solve for what you are looking for. In this case, let's find an expression for the speed of the car going around, and the period of rotation.

 
Substitute in the expression for n from the y axis, into the radial equation. 
<math> v_{max} =\sqrt{\frac{r}{m}(\mu_s mg)}=\sqrt{r\mu_s g)}</math> 

Say r = 157 m, <math>\mu_s =0.5</math>, then the max velocity is 28 m/s. Note that it is independent of mass.
 

What happens if you go faster then this? 
The force of friction cannot provide the necessary centripetal force. The car will slip off the road. 
What happens if you go slower than this? 
The force of static friction will exactly provide the force required to get the car around the bend. 
[http://www.youtube.com/watch?v=aFEnKRQMjgQ&feature=related Video Explanation] 
:::::<youtube>aFEnKRQMjgQ</youtube>

=='''Example 3: Vertical Circle'''==
The classic case of a vertical circle is a car going over a hill and through a valley. We are most interested in the special cases of just at the bottom of the valley and just at the top of the hill. Other examples of a vertical circle are a ball on a string being swung in a vertical circle and a person going around on a Ferris wheel (la grande roue). 
For the car, the only forces acting on it are the normal force from the road and the weight. Only these forces can contribute to pulling the car into a circle. 

First visualize the problem: 
[[image: verticalcircle.jpeg]] 
The car is going around at constant speed. However, the velocity vector is tangential to the circle at any particular point. 
We can start by looking at the special case of the car down in the valley. 
The net force must be radial pulling the car into the circle. The centre of the circle is up, so the acceleration is up and the net force is up. Therefore the normal force must be greater than the weight. 
[[image: verticalcirclebottomFBD.jpeg]] 

===Choose your axes===
First note the chosen axes. There is a horizontal x axis, and a radial r axis. Positive r is up towards the centre of the circle, positive x is right. Don’t forget to always define this. The radial axis in fact goes around the circle with the car, always pointing towards the centre. Therefore, the axes shown are at this point in time. 
Why have you chosen these axes? You know there must be an acceleration towards the centre of the circle. Choose your axes to correspond. 
There are only two forces acting on the mass: the normal, and the weight of the object. The weight and normal force are straight up and down: r (or y if you insist) only. 
 

===Equations in the y and radial directions===
As before you can write equations of motion from Newton’s first and second laws in the radial and y direction. You usually solve these problems by getting one equation from axis, one equation from the other, and solving simultaneously 
{| class="wikitable"
|-
! Equation in x direction !! Equation in radial (or y) direction
|-
|
<math> \sum F_{x}= 0</math> 
Note that the acceleration is zero because the car is not accelerating along the x axis. 
This is therefore an example of Newton's 1st Law. 

 
||
In the radial y direction, there is circular motion. 
There must be a force pulling the mass into the circle. 
This force comes from the combination of the normal and weight. 
<math> \sum F_{radial}= n - w </math> 

Note that there is a net force, therefore the system is not in equilibrium. =Newton's 2 Law 
Now this radial force gives rise to radial acceleration. Therefore: 
<math> \sum F_{radial}= n- mg =ma_{centripetal}=m\frac{v^{2}}{r} </math> 
Rearranging: 
<math> v^2=\frac{r(n-mg)}{m}</math>

|}

 

===Resolve the Equations===
 
As in many Newton2 problems, you take an equation from one axis and substitute it into the other axis and solve for what you are looking for. In this case, you only have one equation. A common question is to find the normal force.

 
Rearrange the equation for n. 
<math> n=\frac{v^2m}{r}+mg</math>

 

Note that the normal force is not equal to mg in this case. It is always greater than mg because there must be a net force up. 

'''Can you find an expression for the normal force at the top of the hill?''' 
[http://www.youtube.com/watch?v=AtJBl9N9o6E&feature=related Video Solution of this question] 
:::::<youtube>AtJBl9N9o6E</youtube>

Module 5: 2D Projectile Motion

2014-03-04T19:24:51Z

Kevin: /* Derivation of the Equation for Horizontal Range */

''Helena Dedic, Kreshnik Angoni and Kevin Lenton''
 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/2DProjectileMotion/Module%202D%20Motion%20Worksheet.docx Module 2D Motion Worksheet.docx] 
[http://gauss.vaniercollege.qc.ca/~physics/MechanicsModules/Module%20Worksheets/2DProjectileMotion/Module%202D%20Motion%20Worksheet.pdf Module 2D Motion Worksheet.pdf] 

==Learning Objectives==
After reading this page, watching the videos and reviewing the exercises, you will be able to: 
* Understand 3D (2D) motion as the vector combination of motion along 3(2) 1D-Independent axes. What links the axes is ''time''.
* Understand Projectile motion as a special case of 2D motion with a constant acceleration in one axis, and no acceleration in the other.
*Describe the independence of horizontal and vertical components of motion.
*Predict the landing point of a projectile, given the initial conditions
 

==Extending the 1D case to 2 and 3 Dimensions==
===1D Motion: Choose an Axis which corresponds to the Straight Line Motion===
When a path of a particle is a straight line we say that its motion is one dimensional. In such cases we orient one of the axis along the path traveled by the particle and we can describe its position or velocity by one of the components:

[[image: 2D intro.png|TOP]] 

===2D Motion: Vector Sum of 1D Coordinates===
Now consider when the path of a particle is a curve in a plane (let's say either the horizontal or vertical plane) then we say that its motion is a '''two dimensional 2D''' motion. The Figure shows the y vs x graph (not y vs t!).

[[image: 2D introB.png|TOP]] 

The position vector at any instant t has two components each of which is a function of time.

<math>\vec r(t) = x(t)\hat i + y(t)\hat j</math>

The velocity vector points in the direction of the tangent to the path. We cannot deduce its magnitude because we do not have any time information at the moment, but we know that it has two components neither of which is zero at t: 

<math>\vec v(t) = v_x(t)\hat i + v_y(t)\hat j</math>

''If you knew the components of the velocity vector <math>\vec v(t) </math>, how could you calculate the magnitude of the velocity (otherwise known as the speed)?'' 
<math>\Rightarrow\ \ |\vec v(t)| = \sqrt{v_x(t)^2 i + v_y(t)^2}</math>

 
The acceleration vector also has two components. We cannot deduce either the direction or its magnitude from the diagram above. But if we knew then, we could write: 

<math>\vec a(t) = a_x(t)\hat i + a_y(t)\hat j</math>

In this course you will study two cases of two-dimensional motion: projectile motion and motion with constant acceleration. This module looks exclusively at Projectile Motion.

===Equations of Motion in 3D===

The <math>x, y, z</math> axes are independent of each other because they are orthogonal (at 90 degrees to each other). Therefore what is important to remember is that the motion along the horizontal (x) axis does not affect the motion along the vertical axis and vice-versa. Horizontal motion and vertical motion are totally independent of each other. 
This means that you can rewrite the equations of motion independently for each axis. What links the axes at a particular point in the motion is ''time''. 

{| class="wikitable"
|-
! !! X Axis !! Y Axis!! Z Axis
|-
| Equation 1 || <math>v_{f_{x}} = v_{i_x} + a_xt</math> || <math>v_{f_{y}} = v_{i_y} + a_yt</math>|| <math>v_{f_{z}} = v_{i_z} + a_zt</math>
|-
| Equation 2|| <math>x_f-x_i = v_{i_x}t + \frac{1}{2}a_xt^2</math>|| <math>y_f-y_i = v_{i_y}t + \frac{1}{2}a_yt^2</math>|| <math>z_f-z_i = v_{i_z}t + \frac{1}{2}a_zt^2</math>
|-
| Equation 3|| <math> v_{f_{x}}^2 = v_{i_x}^2 + 2a(x_f-x_i)</math> || <math> v_{f_{y}}^2 = v_{i_y}^2 + 2a(y_f-y_i)</math>|| <math> v_{f_{z}}^2 = v_{i_z}^2 + 2a(z_f-z_i)</math>
|-
| Equation 4|| <math> (x_f-x_i)= + \frac{1}{2}(v_{f_{x}} +v_{i_x})t</math>|| <math> (y_f-y_i)= + \frac{1}{2}(v_{f_{y}} +v_{i_y})t</math>|| <math> (z_f-z_i)= + \frac{1}{2}(v_{f_{z}} +v_{i_z})t</math>
|}
 

==Projectile Motion: A special case==
Note: Using axes where up is positive y, and to the right positive x. 
A particle is in projectile motion when: 
1. It has a horizontal component of velocity (i.e. <math>v_x \neq 0</math> ; <math>v_x </math> is different from zero) and 
2. The force of gravity is the only force acting on it. This means you are ignoring other forces such as air resistance (you suck all the air out of the problem), and you therefore only have one acceleration, g in the negative y direction (<math>-\hat j</math>). 
You are saying the acceleration in the x direction is zero <math>a_x = 0</math> and the acceleration in the y direction is -9.8 m/s/s <math>a_y = -9.8 ms^{-2}</math>. 
The implication of <math>a_x = 0</math> is that, in the x direction, the velocity does not change and the only equation you need to describe the motion is essentially <math>speed=\frac{distance}{time}</math> 
More correctly, the kinematic equations can be rewritten for this case of <math>a_x = 0</math> and <math>a_y = -g</math>: 

{| class="wikitable"
|-
! !! X Axis !! Y Axis
|-
| Equation 1 || <math>v_{f_{x}} = v_{i_x} </math> || <math>v_{f_{y}} = v_{i_y} + (-g)t</math>
|-
| Equation 2|| <math>x_f-x_i = v_{i_x}t </math>|| <math>y_f-y_i = v_{i_y}t + \frac{1}{2}(-g)t^2</math>
|-
| Equation 3|| <math> v_{f_{x}}^2 = v_{i_x}^2 </math> || <math> v_{f_{y}}^2 = v_{i_y}^2 + 2(-g)(y_f-y_i)</math>
|-
| Equation 4|| <math> (x_f-x_i)= + \frac{1}{2}(v_{f_{x}} +v_{i_x})t</math>|| <math> (y_f-y_i)= + \frac{1}{2}(v_{f_{y}} +v_{i_y})t</math>
|}
 
If you think about throwing a ball, you can always align your x axis to follow the shadow (projection) of the ball. This means you only need 2D to describe projectile motion. We do not consider a z axis.
 
As usual, the axes are always independent. ''Time'' is the parameter which links the axes. 
==Derivation of the Equation for Horizontal Range==

You are throwing a ball with initial velocity of magnitude <math>v_i</math> and direction <math>\theta</math> above the horizontal. You are asked to calculate the ''Horizontal Range''. The Range is a special term for the x displacement when the ball comes down to the same level from which it is thrown. That is to say <math>\Delta y(t) = 0</math>. 
[[File:HorizontalRange.gif|thumb|center|x500px|alt=ProjectileMotion|Projectile Motion]] 
Then the x and y components of the initial velocity are: 

<math> v_{i_{x}} = v_{i} cos\theta</math> 
<math> v_{i_{y}} = v_i sin\theta</math> 

We also note that <math>\Delta x(t) = R</math>, =the horizontal Range; and <math>\Delta y(t) = 0</math>. We will substitute these values into the equations of kinematics for projectile motion. 
First you can create a table of all the information you have and what you are looking for, with x and y separated out (because they are independent). 

{| class="wikitable"
|-
! Parameter !! Values for Horizontal (X) Axis !! Values for Vertical (Y) Axis
|-
| Displacement <math>\Delta d</math>|| <math>\Delta x=R=?</math> || <math>\Delta y=0 m</math>
|-
| <math>v_i</math>|| <math> v_{i_{x}} = v_{i} cos\theta</math> || <math> v_{i_{y}} = v_i sin\theta</math>
|-
| <math>v_f</math>|| <math> v_{f_{x}} = v_{i} cos\theta</math> || <math> v_{f_{y}} = ?</math>
|-
| <math>a</math>|| <math> a_x = 0ms^{-2} </math> || <math> a_y = -g </math>
|-
| <math>t</math>||colspan="2" style="text-align: center;"| <math>t=?</math>
|}
 
You are looking for something along the x axis, R, so find an expression for time using the y axis. 
The equation you should use is: 
<math>y_f-y_i = v_{i_y}t + \frac{1}{2}a_yt^2</math> 
Substituting in from the above table: 
<math>0 = v_i (sin\theta)t + \frac{1}{2}(-g) t^2</math> 
Isolate t :

<math>0 = v_i (sin\theta)t - \frac{g}{2}t^2</math> 
The roots are <math> t = \frac{2v_i sin\theta}{g}</math> and <math> t = 0</math>. 
The <math> t = 0</math> root corresponds to the initial conditions. 
 
Now in the x direction the only equation you need is: 
<math>x_f-x_i = v_{i_x}t </math> 
Therefore,
<math>R = v_i (cos\theta) t</math> 

and substitute for t : 

<math>R = v_i (cos\theta)\frac{2v_i sin\theta}{g} = \frac{2v_i^2 sin\theta cos\theta}{g}</math> 
Almost there! 
You are going to manipulate this equation into a slightly different form to get some interesting conclusions. 
We note the trigonometric identity: 
<math>sin2\theta = 2 sin\theta cos\theta</math>. 
You can substitute for <math>sin\theta cos\theta</math> in the equation for the range and obtain: 

<math>R = \frac{2v_i^2 sin2\theta /2}{g} = \frac{v_i^2 sin2\theta}{g}</math> 

So, in general: 

<math>R = \frac{v_i^2 sin2\theta}{g}</math> 
 
'''!!!!Be very careful with this equation: It can only be used when the y displacement is zero, that is to say the object comes back to the initial y position!!!'''
 
For what angle is R maximum? 
That will occur when <math> sin2\theta=1</math> or rather when <math> 2\theta=90^\circ</math> or rather when 
<math> \theta=45^\circ</math> 
This means to maximize the range of a ball throw, you should throw it at an angle equal to <math> \theta=45^\circ</math>. 
Also look at the velocity term <math>v_i^2 </math>. It is a square relationship. So by increasing <math>v_i </math> you increase R by alot. But you probably knew that already! 
How does g affect the range? How far would your ball go on the moon, all the other parameters held constant? 

 
See how this plays out in the following video. Pay attention to the 45 degree case. 

<li>[http://www.youtube.com/v/N0H-rv9XFHk Range of Projectile]</li>
See how the horizontal range of a projectile depends on the initial angle of incidence. The initial velocity is the same in each case.

:::::<youtube>N0H-rv9XFHk</youtube>
 

===Solving Projectile Motion Problems===
Projectile problems are easily solved if you stick to a logical procedure. Your very first task is to separate the motion into x (horizontal) and y (vertical) parts (because they are independent). The problems are almost always solved in two parts, one involving horizontal motion, the other involving vertical motion. 
The part you attack first depends upon the particular problem you are solving. But in general if you are asked to find a parameter on the horizontal axis, e.g. the horizontal displacement, you should start with the vertical to find a value, or an expression for the time (using the equation of motion for that axis), and then substitute into the equation of motion for the other axis. 
To repeat: The general procedure is that you use one of the independent motions (horizontal or vertical) to find the time in flight, then use that time to determine the parameter of interest in the other axis. 
<ol>
[http://www.youtube.com/watch?v=Q53HHMMWtf0&feature=related Video of Problem Solving Strategy] 
:::::<youtube>Q53HHMMWtf0</youtube>
[http://www.youtube.com/watch?v=XDIvbf9jVLU Video of Problem Solving Strategy] 
Note in this example, the object is going straight off the cliff. In this case the initial y velocity is zero. 
:::::<youtube>XDIvbf9jVLU</youtube>
[http://www.youtube.com/watch?v=XDIvbf9jVLU More Problem Solving Strategy] 
This video is quite detailed. One thing that this person recommends is to split problems up into up and down sections- to avoid quadratic equations. Do not show this fear! 
:::::<youtube>uBZCDMIT8ms</youtube>
</ol>

==Exercises on Projectile Motion==
[http://gauss.vaniercollege.qc.ca/pwiki/index.php/Exercises_on_Projectile_Motion Exercises on Projectile Motion]