# Charge within a cavity inside a conducting material

1. Jan 10, 2014

### kmm

Suppose I have a charge inside a conductor as shown in the image I've attached. For any charge distribution: $$\oint \mathbf{E} \cdot d \mathbf{a} = 0$$

I can see that if I took some path from the charge, through the conductor, and back to the charge, the integral would be zero still.

Now, that charge is in the center of the cavity, but now imagine the charge is shifted to the right by some amount. Let's say I took a path clockwise by going straight left from the charge, then through the conductor, then came out of the conductor to go straight left again to meet the charge. Inside the conductor $\mathbf{E} = 0$ so that doesn't contribute to the integral. But the path to the left leaving the charge is longer than the path on the right that comes back to the charge. This seems to give a nonzero integral which is a violation.

I know the charge would shift the charge on the surface of the cavity, but the E fields due to those charges would cancel since the total E field inside the cavity would be zero without the charge on the inside. I'm not sure how the integral would equal zero then.

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Last edited: Jan 10, 2014
2. Jan 10, 2014

### thegreenlaser

If there's a charge +Q inside the cavity, there has to be a negative charge -Q on the surface of the cavity because of Gauss's law and the fact that E=0 inside the conductor. That negative charge on the surface of the cavity will alter the electric field inside the cavity, ensuring that $\oint \mathbf{E}\cdot d\mathbf{l} = 0$ is satisfied.

3. Jan 10, 2014

### kmm

For that to happen, wouldn't the field inside the cavity due to the charges on the surface of the cavity have to be non zero? In a shell with uniform charge distribution, the E field on the interior of the shell is zero. I suppose if the charge was unevenly distributed along the surface of a shell, say more to one side, the field wouldn't be zero. I'm not sure if I'm thinking correctly about that though.

4. Jan 13, 2014

### thegreenlaser

In a spherical shell with uniform charge distribution, the E field inside the shell will be zero. In an arbitrarily shaped shell of uniformly distributed charge, that's not necessarily true.

You're exactly right. There's no reason to believe that the charge is uniformly distributed unless there's some sort of symmetry going on (e.g., spherical cavity with a charge exactly in the centre). If there's no symmetry (e.g., spherical cavity but the charge isn't exactly in the centre), then the charge will, in general, have a non-uniform distribution. If the charge is really close to one side of the cavity, then you'll find charge will tend to build up more on that side. So even in a spherical cavity, you'll have a non-uniform charge distribution leading to an altered E field in the cavity.

In general, you'll have a non-symmetrically shaped cavity and a non-uniform charge distribution, meaning you'll almost certainly have an altered E field inside the cavity.

5. Jan 13, 2014

### kmm

Thanks for helping me clear that up!