# Homework Help: Gauss' Law MC

1. May 7, 2010

### keemosabi

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

http://img685.imageshack.us/img685/9501/10953060.png" [Broken]

2. Relevant equations

3. The attempt at a solution
I know you use Gauss' Law, but why wouldn't the charges outside of the shell induce a charge on the shell, which would then affect the field at P?

Last edited by a moderator: May 4, 2017
2. May 7, 2010

### kuruman

The charges outside the shell affect the charge distribution on the outer shell only. Likewise, the charge inside the shell affects the charge distribution on the inside surface of the shell only. What you do to the outside charges cannot be communicated to the cavity inside because the field in the conducting material is always zero and that decouples the two regions of space.

You can see this from Gauss's Law. If you draw a Gaussian surface so that it is entirely inside the conducting material, the field on it is zero everywhere which means that the flux through it is zero which means that the net charge is zero always and no matter what happens outside the shell.

3. May 9, 2010

### keemosabi

I pretty much understand, I'm just wondering how Gauss' Law proves this. Can't a point outside of a Gaussian Surface still affect the E-field at a point on the surface? It just wouldn't be in included in the integral, right? So how would this say that the charge on the inner shell does not affect the E-field at point P?

4. May 9, 2010

### kuruman

Not if the Gaussian surface is entirely inside the conducting material. The integral is zero because the E-field is zero everywhere on the Gaussian surface regardless of what kind of charge you put outside or inside the shell anywhere you please. This then says that the net charge enclosed by the Gaussian surface (as defined above) is zero no matter what.

5. May 9, 2010

### keemosabi

So it's like the conducting shell prevents the outer charges from affecting the E-field inside of the shell?

6. May 9, 2010

Yup.