# Electric potential at the center of a cylinder enclosed in a cylindrical shell

1. Oct 28, 2012

### aftershock

1. The problem statement, all variables and given/known data

A solid metal cylinder of radius R and length L, carrying a charge Q, is surrounded by a thick coaxial metal shell of inner radius a and outer radius b. The shell carries no net charge. Find the potential at the center, using r=b as the reference point.

2. Relevant equations

V= (1/4πεo)q/r

V = -∫E dot dl

3. The attempt at a solution

The cylinder is not infinite but I assume I'd have to ignore end effects.

I want to approach this problem using gauss's law to find the field and then integrating. I don't even think I can use the first equation since my reference point is not at infinity.

This is easy enough but I'm having problems because I don't know the electric field at the boundaries. I can find the field easily when r<R when R < a < b and when a < r < b

I don't know what the field is exactly at the surfaces?

2. Oct 28, 2012

### TSny

Hello, aftershock.

The fact that E is discontinuous at the surfaces will not cause any problem. You do not need to worry about the value of E at a surface. E will be a piecewise continuous function. When you integrate a piecewise continuous function, you do not need to know the value of the function at the discontinuities.

3. Oct 28, 2012

### aftershock

Thank you, but I'm still confused. I actually have the solution but I can't make sense of it.

the solution tells me V(b)-V(a) = ∫Q/(2πoLr) integrated from a to b

Why am I integrating the electric field in the region R < r < a from a to b. Isn't the electric field zero from a to b since it's inside a conductor?

Then it goes on to claim V(0) = V(a)

How does that happen?

4. Oct 28, 2012

### TSny

You're right, the integration should be from R to a. The integration form a to b would be zero since E is zero in that region.

5. Oct 28, 2012

### aftershock

So the answer is the integral of the field from R to a?

And b does not appear in the solution even though it's the reference point?

6. Oct 28, 2012

### TSny

Yes. Technically you should be integrating from R to b, but the part of the integral form a to b is zero. So, you just need to integrate from R to a.

[Edit: Actually, technically, you should be integrating from r = 0 to r = b. But the integral from 0 to R is also zero.]

Last edited: Oct 28, 2012
7. Oct 28, 2012

### aftershock

Thank you!

It's actually integrating from a to R though, right? Or are you reversing the order and throwing away the negative sign?

8. Oct 28, 2012

### TSny

Yes, that's right. I was reversing the limits to get rid of the negative sign. Good.