# Gauss Law - Conducting cylindrical shell

MichaelT
1. A thick conducting cylindrical shell has an inner radius Rsub1 and outer radius Rsub2. It has a net excess charge = Q, and it is L long. Find electric field at certain points given.

Ok, when r is less than Rsub 1, the electric field is zero. And when r is greater than Rsub2, it is easy to calculate using Gauss's Law. Now, when r is between Rsub1 and Rsub2...what is the excess charge inside the Gaussian cylinder? I know that any net charge in a conductor must be on one of the surfaces. But on which one, and how much on each?

This might be a silly guess, but maybe Q/2 ?

Mentor
What's the field within the material of the conducting shell? (Between R_1 and R_2.)

MichaelT
The electric field inside the material of a conducting shell is zero when the conductor is in electrostatic equilibrium. Does that still hold even though this conductor has a net charge of Q?

Mentor
Absolutely.

MichaelT
Ok, that's cool. A little strange to me though. Intuitively it does make sense, but when you use Gauss's Law, wouldn't there be some excess charge inside a Gaussian cylinder with r between R_1 and R_2? Or would the excess charge Q remain on the outer surface of the conductor?

Mentor
The excess charge can only exist on the surface of the conducting cylinder. Since there's no charge contained in the cavity (at r < R_1), there's no charge on the inner surface. (Convince yourself of this using Gauss's law.) Thus all excess charge is on the outer surface.