AP Physics C: Electricity & Magnetism 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 other0089

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 conductor0173

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 and0364

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/ R²0387

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 potential0424

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 conductor0780

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 dimensions0974

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 field0990

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