# Homework Help: E field of a conductor with an arbitrary shape enclosing charge

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1. Jun 16, 2017

### weezy

In this image charge q is enclosed within an arbitrary shaped conductor. I'm asked to find what if the Electric field in regions 1,2,3,4 are uniform and if the surface charge densities on the interior wall of the arbitrary shaped conductor and the hollow sphere conductor are uniform.

My analysis tells me that since the inner conductor is arbitrary the E field outside it won't be radially symmetric and thus should induce a non-uniform charge density on the interior of the hollow sphere. On the other hand since we know E field can't exist inside the hollow conductor, by Gauss's law we must have a uniform field in region 4.

What I am having trouble with is the following fact:

If I am to take a gaussian sphere of radius R in region 2 (it encloses the arbitrary shape) and apply Gauss's theorem, I should find the integral give me $E = \frac{q}{4 \pi \epsilon R^2}$ which is just the E field produced by point charge located at the origin. However this conflicts the fact that the E field is not symmetric because I expect the arbitrary conductor to have uniform charge density on it's surface but due to it's arbitrary shape it'll have a non-uniform E field in region 2. What is happening here?

EDIT #1: Also I'd like to know if changing the position of q changes the E field in region 2. I think it does not if the arbitrary conductor has some thickness. I'm not entirely sure about this.

Last edited: Jun 16, 2017
2. Jun 16, 2017

### Staff: Mentor

Is the conductor between 1 and 2? If yes, what is 3, another conductor?
Right.
That is not true in general, assuming "inside" means vacuum inside the hollow conductor.
That is not inside.

Only if you assume that the electric field has the same magnitude everywhere, otherwise you don't get this result.
Make a wrong assumption and you get a wrong result.
What do you know about the potential of the conductor at its surface? How does that change if you move the charge?

3. Jun 16, 2017

### weezy

3 is a hollow conductor sphere

Oh my! this is what I overlooked!

The surface must be equipotential so that means in any case the field shouldn't vary. Yet the answer key had my answer marked wrong.