# Homework Help: Gauss' Law Multiple Choice

1. Apr 16, 2013

### pietastesgood

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

1. Positive charge Q is placed on a conducting spherical shell with inner radius R1 and outer
radius R2. A point charge q is placed at the center of the cavity. The magnitude of the electric
field produced by the charge on the inner surface at a point in the interior of the conductor, a
distance r from the center, is:
A. 0
B. Q/4vπε0(R1)2
C. Q/4πε0(R2)2
D. q/4πε0r^2
E. Q/4πε0r^2

2. Relevant equations

E=kq/r^2
flux=q/ε0

3. The attempt at a solution

I don't quite understand why there is an electric field within the conductor itself, which is what answer choice D is saying. Sure, there's a charge within the cavity, but since the net charge has to equal zero, there will be an equal but opposite charge on the inner cavity surface, which allows no flux to leave through a Gaussian surface within the conductor at a radius greater than the inner cavity.

Unless I'm actually just reading the problem wrong, and it's talking about just the charge on the inner surface, and ignoring the charge within the cavity. In that case, wouldn't the charge -q cause a field of magnitude q/4πε0(r-r1)^2? Any help would be appreciated!

2. Apr 16, 2013

### Dick

Yeah, I think you have to read it carefully. There is no net electric field inside of the conductor. The charge at the center will create an electric field corresponding to answer D. The charges on the conductor will rearrange to cancel that. But the inner charge of -q is not a point charge. Be careful what field you think it makes.

3. Apr 16, 2013

### pietastesgood

Oops, you're right. If you take the inner charge -q and use Gauss' law, you get EA=q/ε0, and A=4πr^2, so E=q/4πε0r^2. Thanks for the help!