For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

For more information, please see full course syllabus of AP Physics C: Electricity & Magnetism

## Discussion

## Study Guides

## Download Lecture Slides

## Table of Contents

## Transcription

## Related Books

### Conductors

- Charges are free to move in conductors.
- At electrostatic equilibrium, there are no moving charges in a conductor, therefore there is no net force, and the electric field inside the conductor must be zero.
- All excess charge on a conductor lies on the surface of the conductor.
- The electric field on the surface of a conductor must be perpendicular to the surface.
- The electric field inside any hollow conductor is zero (assuming there are no charges in the hollow region). This allows hollow conductors to be utilized to isolate regions completely from electric fields. In this configuration, the hollow conductor is known as a Faraday Cage.

### Conductors

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

- Intro
- Objectives
- Charges in a Conductor
- Charge is Free to Move Until the
- All Charge Resides at Surface
- Field Lines are Perpendicular to Surface
- Electric Field at the Surface of a Conductor
- Hollow Conductors
- Draw Hollow Conductor and Gaussian Surface
- Applying Gaussian Law
- Any Hollow Conductor Has Zero Electric Field in Its Interior
- Faraday Cage
- Electric Field and Potential Due to a Conducting Sphere
- Example 1
- Example 2

- Intro 0:00
- Objectives 0:08
- Charges in a Conductor 0:32
- Charge is Free to Move Until the
- All Charge Resides at Surface
- Field Lines are Perpendicular to Surface
- Electric Field at the Surface of a Conductor 3:04
- Looking at Just the Outer Surface
- Large Electric Field Where You Have the Largest Charge Density
- Hollow Conductors 4:22
- Draw Hollow Conductor and Gaussian Surface
- Applying Gaussian Law
- Any Hollow Conductor Has Zero Electric Field in Its Interior
- Faraday Cage
- Electric Field and Potential Due to a Conducting Sphere 6:03
- Example 1 7:31
- Example 2 12:39

### AP Physics C: Electricity and Magnetism Online Course

I. Electricity | ||
---|---|---|

Electric Charge & Coulomb's Law | 30:48 | |

Electric Fields | 1:19:22 | |

Gauss's Law | 52:53 | |

Electric Potential & Electric Potential Energy | 1:14:03 | |

Electric Potential Due to Continuous Charge Distributions | 1:01:28 | |

Conductors | 20:35 | |

Capacitors | 41:23 | |

II. Current Electricity | ||

Current & Resistance | 17:59 | |

Circuits I: Series Circuits | 29:08 | |

Circuits II: Parallel Circuits | 39:09 | |

RC Circuits: Steady State | 34:03 | |

RC Circuits: Transient Analysis | 1:01:07 | |

III. Magnetism | ||

Magnets | 8:38 | |

Moving Charges In Magnetic Fields | 29:07 | |

Forces on Current-Carrying Wires | 17:52 | |

Magnetic Fields Due to Current-Carrying Wires | 24:43 | |

The Biot-Savart Law | 21:50 | |

Ampere's Law | 26:31 | |

Magnetic Flux | 7:24 | |

Faraday's Law & Lenz's Law | 1:04:33 | |

IV. Inductance, RL Circuits, and LC Circuits | ||

Inductance | 6:41 | |

RL Circuits | 42:17 | |

LC Circuits | 9:47 | |

V. Maxwell's Equations | ||

Maxwell's Equations | 3:38 | |

VI. Sample AP Exams | ||

1998 AP Practice Exam: Multiple Choice Questions | 32:33 | |

1998 AP Practice Exam: Free Response Questions | 29:55 |

### Transcription: Conductors

*Hello, everyone, and welcome back to www.educator.com.*0000

*I'm Dan Fullerton and in this lesson we are going to talk about conductors.*0003

*Let us start with some of our objectives.*0007

*To begin with, we are going to try and understand the nature of electric fields and electric potential in a round conductor.*0009

*We are going to explain how the all the axis charge on a conductor resides on the conductor's surface not in its bulk.*0015

*Explain why a conductor must be an equal potential.*0022

*And finally, graph the electric field and electric potential inside and outside a charge conducting sphere.*0025

*As we start talking about charges in a conductor, we have to remember that charge in a conductor is free to move until there is no E field.*0031

*Charge is free to move until the electric field is equal to 0.*0041

*If we have all these free charges in a conductor and they are freedom to move,*0052

*they are going to migrate to the area of the lowest possible energy state,*0055

*and as far apart as possible and that usually puts them on the surface.*0060

*If they are moving, they are no longer in equilibrium.*0064

*If there are electric fields inside the conductor, the charges would move.*0066

*When you have a conductor that is in equilibrium, all the charges have already distributed themselves so that they are not going to move anymore.*0069

*There is no net internal electric field, otherwise they will keep moving.*0075

*The electric field inside a conductor is 0.*0081

*You can think about that if we draw some sort of random conducting shape here.*0084

*If we have some charge + Q that we put somewhere on that surface, the charge is are going to repel each other*0089

*and they are going to go and move as far apart as they possibly can, in order to minimize their potential energy that puts them all on the outer surface.*0095

*We can also go and draw a Gaussian surface inside a conductor.*0109

*Let us put our Gaussian surface something like that.*0116

*As we draw our Gaussian surface, we can look at it from Gausses law perspective of the integral / the close surface of E ⋅ DA.*0118

*It is the enclosed charge ÷ ε₀, but we just said that the electric field must be 0.*0128

*Therefore, the enclosed charge is going to be 0.*0135

*All the charge has to reside at the surface not in the bulk of the conductor.*0139

*Now field lines are all going to run perpendicular to the surface.*0152

*If they are not perpendicular to the surface, there will be some component that is parallel to the surface and the charges would then move.*0163

*As we just said, the charges already moved in the conductor*0173

*until the electric field is 0 so electric field lines have to intersect that conductor at 90° angles.*0176

*They have to be perpendicular.*0180

*Let us look some more at the electric field at the surface of a conductor.*0184

*If we look at just the outer surface, let us make a note of that.*0187

*Just the outer surface, let us take, using the Gausses law, the integral / the closed loop of E ⋅ DA is our enclosed charge ÷ ε₀.*0197

*Let us assume a surface where we have some symmetry.*0212

*We will say that our charge is σ A, therefore, electric field × the area is going to be our enclosed charge σ A / ε₀.*0215

*Or the electric field is equal to the surface charge density σ ÷ the constant ε₀.*0229

*What this really means that, is you are going to have the largest electric field where you have the largest charge density,*0238

*which is common sense as well but probably we are stating explicitly here.*0254

*Let us talk for a minute about hollow conductors.*0260

*In a hollow conductor, you can determine the location of charge by utilizing Gauss’s law.*0264

*Choosing Gaussian surfaced in the metal of a hollow conductor.*0269

*Making out that the electric field inside that conductor must be 0.*0273

*Let us start by drawing our hollow conductor, there it is.*0277

*And inside of that we are going to draw our Gaussian surface.*0283

*We know the electric field inside there must be 0.*0288

*Applying Gauss’s law, the integral / the closed surface of E ⋅ DA is our enclosed charge / ε 0.*0291

*Again, we know the electric field must be 0 so our enclosed charge is going to be 0.*0304

*Therefore, the charge must remain on that outer surface,*0312

*the entire conductor is at equal potential with the same electric potential, the same voltage,*0315

*and field lines must run perpendicular to that conducting sphere.*0320

*Any hollow conductor has 0 electric field in its interior.*0324

*That allows you to use hollow conductors as a way to isolate regions completely from electric fields.*0329

*When you do that, the device we are using a hollow conductor to isolate from electric fields is known as Faraday cage.*0335

*Hollow conductor completely closed so great way to isolate items from electric fields.*0352

*Alright, let us take a look at the electric field and electric potential due to a conducting sphere.*0359

*If this is our conducting sphere down at the bottom and*0364

*we want to plot the electric field as a function of position, we know inside our conductor the electric field must be 0.*0367

*Outside, right at the surface we said the electric field is equal to the surface charge density ÷ ε 0.*0379

*And then that is going to fall off our proportional to 1/ R²*0387

*as if we are treating it all as a point charge located at the center, when we are outside that sphere.*0393

*We have done this derivation a few times.*0398

*But looking at the potential, there is something you have to be careful of.*0400

*Just because the electric field is 0 inside the conductor does not mean the potential is 0.*0403

*What it means is the potential is constant, it is not changing.*0409

*Our potential graph from the outside as we come from infinity toward the edge, we have Q / 4 π ε₀ R.*0412

*Again, the same as if we had a point charge located in the center of our sphere.*0420

*Once we get to touching the sphere however, the entire sphere is in equal potential so we have a constant potential*0424

*that is the same value we had right at the radius of the sphere, which we are calling here R.*0429

*Take note that potential inside a conducting sphere does not have to be 0.*0439

*It is what you have right at the edge to the entire sphere is in equal potential.*0444

*Let us see if we can do a problem or two with this.*0451

*We have two conducting spheres A and B, they are placed at large distance from each other.*0454

*By large distance, what we are saying is that the electric field from one does not affect the electric field from the other.*0459

*The radius of sphere A is 5 cm and the radius of sphere B over here is 20 cm, that is 4 × larger radius.*0465

*A charge Q of 200 nC is placed on sphere A, our sphere B is uncharged.*0472

*The spheres are then connected by a wire.*0478

*Find the charge on each sphere after that wire is connected.*0480

*A couple of things that may help us here.*0484

*First, I note that the total charge Q must be equal to QA + QB because we have a closed system.*0486

*We have a total charge that is going to remain constant.*0494

*It is going to be 200 nC.*0497

*Finally, if we rearrange this a little bit, QB = Q – QA.*0500

*That will be helpful as well.*0507

*Once we connect these by a wire, we have one large single conducting object.*0510

*Once everything rearranges itself, all the charges move to wherever they are going to.*0515

*The entire thing is it in equal potential.*0519

*We know that the potential in A and the potential at B must be the same once they are connected by the wire.*0522

*The sum of the charges on each sphere must equal the total charge it had initially by the law of conservation of charge which is going to be 200 nC.*0529

*Let us see if we can solve this.*0538

*Looking at sphere A first, let us write that the potential at A we know is going to be the charge on A ÷ 4 π ε₀ RA.*0542

*We can also look at B and say that the potential at B is going to be the charge on B QB/ 4 π ε₀ radius of B.*0557

*And because they must be at equal potential, VA = VB.*0569

*We can state then that QA/ 4 π ε₀ RA must be equal to QB/ 4 π ε₀ RB.*0575

*We can do a little simplification here.*0598

*We have got 4 π ε₀ in the denominator on both sides and let us just simplify this with one more step.*0600

*We can rewrite this now as QA/ RA must equal QB/ RB.*0605

*We also said on the previous slide that QB must equal Q -QA by the law of conservation of charge.*0616

*We can write this again as QA/ RA will equal Q –QA, substituting that in for QB, ÷ RB.*0626

*And now with a little bit of algebra, we can start to solve for our charges.*0638

*Let us do some cross multiplication QA RB must equal Q RA-QA RA.*0643

*Or rearranging these, QA RA + QA RB will equal Q RA.*0655

*If I factor out a QA from this left hand side, QA × RA + RB must equal Q RA.*0674

*Therefore, solving for just QA, QA will equal Q × RA/(RA+ RB).*0688

*Now I can substitute in the values that I know.*0700

*Q is 200 nC, radius of A we said was 5 cm and RA is 5 cm again + RB 20 cm.*0703

*That is going to be 5/ 25 or 20% of 200 nC is just going to be 40 nC.*0718

*If that is the charge on the sphere A and our total is 200, that means that the charge on B must be 200 -40 or 160 nC by the law of conservation of charge.*0732

*QA is 40 nC, QB must be 160 nC, our total charge is still 200 nC.*0748

*Alright let us take a look at an AP style problem, going back to the 2004 APC E and M exam free response number 1.*0758

*Take a minute, you can pull up off the site, find it from the Internet, download it, print it out.*0768

*Take a minute to give it a try and we will see what happens when we try and do it together here.*0773

*As you look at this problem, we are given a hollow conductor*0780

*that has a line of linear charge density parallel to the axis but it is a little bit off center.*0785

*We are asked to sketch the electric field lines and use ± signs to show any charge that is used on the conductor.*0792

*Let us try drawing it here first.*0799

*Looks like it is kind of something like that and then we have got our line of charge + λ there.*0805

*And we know that we have, this is a positive charge + λ.*0812

*The first thing is to sketch the electric field lines in these different regions.*0816

*Let us start here at the middle.*0820

*As we draw electric field lines, because this is closer to the edge over here, we are going to have more field lines over here.*0822

*As I draw these, I'm trying very hard to draw them so that they intersect the conductor at a 90° angle.*0834

*They have to remain perpendicular.*0840

*We are going to have more of these lines over on the right hand side than the left hand side because we are close to the conductor there.*0842

*Now inside the conductor, we are not going to have any electric field.*0854

*The electric field inside the conductor is 0.*0859

*But outside in region 3, now we can draw our electric field lines here.*0861

*Of course, we are going to have them roughly equally spaced going radially outward.*0866

*Looking at the charges, if this is a + λ then we are going to have negative charges attracted right at the surface of the conductor.*0880

*By conservation of charge, we have to have the same opposite charge on the outside of our hollow conductor.*0892

*That I think would do pretty well for getting us through part A.*0904

*For part B, it asks us to rank the electric potentials at A, B, C, D, and E from the highest to the lowest.*0909

*Without a doubt, as I look at our diagram there, it looks to me like the highest has to be D because it is going to be closest to the charge.*0918

*I would say that we are going to have, VD is going to be the highest.*0926

*And next, it looks like we have VC, and then we get into B and E, that are in the conductor themselves.*0931

*B and E because they are in the conductor must be at equal potential.*0943

*We can write that we have VB equal VE.*0947

*All of those must be greater than our lowest potential which is going to be VA, way out there in region 3,*0951

*the farthest from our line of charge.*0959

*That would be how I would rank the electric potentials.*0962

*Going on to part C, let us give ourselves more room here on the next page.*0967

*For part C, we are replacing the shell by another cylindrical shelling that has the same dimensions*0974

*but it is non conducting and has some uniform volume charge density +ρ.*0978

*The infinite line of charge is still there but now it is in the center of the shell.*0984

*We have got something nice and symmetric.*0988

*We are asked to use Gauss’s law to find the magnitude of the electric field*0990

*as a function of distance from the center of the shell for the different regions.*0994

*Alright let us start as we look inside our shell.*0998

*As we go in that region, we will use Gauss’s law, integral / the close surface of E ⋅ DA = Q enclosed / ε₀.*1002

*The left hand side is going to become EA again, where A if we are for careful about how we choose this,*1017

*we are going to pick a cylindrical shape for our Gaussian surface.*1024

*That is going to be 2 π R, the circumference of our Gaussian cylinder × its length, let us call that L.*1027

*Our charge enclosed is just going to be the linear charge density λ × its length L ÷ ε₀.*1038

*Therefore, our electric field strength is going to be λ L / 2 π RL ε₀ or just λ / 2 π ε₀ R.*1047

*Moving onto region 2, between R1 and R2, find the electric field in the region between R1 and R2.*1069

*As we look there between R1 and R2, the electric field is going to be.*1085

*How are we going to do that?*1091

*Let us take a look, use Gauss’s law again, integral / the close surface of E ⋅ DA = Q enclosed/ ε₀,*1093

*which implies then that E × 2 π RL =, as we look at our charge enclosed, we are going to have the charge from that line λ L ,*1108

*+ we have to take into account the volume that we have in our cylinder, the portion that is inside our Gaussian surface.*1121

*That is going to be π R² L × ρ - π R1 1² L × that volume charge density / ε₀.*1131

*That is what is going to give us our enclosed charge when we are in that region between R1 and R2.*1151

*A little bit of math here, our electric field then is going to be.*1157

*We have got, we can take the L out, we can factor that of all the sides.*1163

*We will have λ + ρ π R² – ρ π R1² all over 2 π ε₀ R.*1167

*Or just factoring this a little bit, let us call that λ/ 2 π ε₀ R.*1183

*The contribution we have from that internal line + factoring out a ρ/ 2 ε₀ R² – R1².*1190

*And that gives you the electric field in the region between R1 and R2.*1212

*The region outside R2 by now should be pretty straightforward to use.*1218

*I’m going to leave that one to you, for you guys do on your own.*1222

*At the same place where you found this problem, you can check your answer there.*1226

*Thank you so much for your time and for watching www.educator.com.*1229

*We will see you at the next lesson.*1232

*Make it a great day everyone.*1234

1 answer

Last reply by: Professor Dan Fullerton

Fri Mar 17, 2017 6:25 AM

Post by Mark Sim on March 16, 2017

Why is a charge free to move until the E=0? Why can't they move when E=o?