# Homework Help: Flux through a nonconducting shell

1. Feb 25, 2010

### reising1

A particle of charge q is inside a nonconducting spherical shell of uniformly spread charge of Q. What is the flux through a spherical Gaussian surface concentric with the shell if the radius of this Gaussian surface is less than the shell's radius?

I know we can use the formula for Gauss' Law:
Flux = charge enclosed / Epsilon not.

However I am not sure specifically what the charge enclosed is. Certainly there is charge enclosed of size q, however is there not a portion of the Q charge enclosed since the Q is uniformly distributed?

2. Feb 25, 2010

### kuruman

Yes there is. Can you calculate what fraction of Q is enclosed by a Gaussian shell of radius r < R (R = radius of entire charge distribution)?

3. Feb 25, 2010

### reising1

Sure, I could integrate to find that. However the answer appears to be that the only charge that contributes is the q, meaning that the answer is simply q/Epsilon not.

4. Feb 25, 2010

### reising1

And hence the answer states that the charge Q does not play a role unless you are at a radius greater than or equal to R (where R = radius of entire charge distribution)

5. Feb 25, 2010

### kuruman

What answer appears to be saying that? Do you know what the answer is?

6. Feb 25, 2010

### reising1

Yes, this is a homework problem that I have the book's answer to.

7. Feb 25, 2010

### reising1

Quoted from the answer: "If the radius of the G-sphere is less than that of the shell then the only charge enclosed by the G-sphere is the charge of the particle"

8. Feb 25, 2010

### kuruman

I now see what the problem is. A shell has two radii, one larger than the other.
Let a = smaller radius and b = larger radius.

If your Gaussian surface has radius r < a, then the only charge enclosed is q.
If your Gaussian surface has radius a < r < b, then the charge enclosed is q plus a fraction of Q that depends on r.
If your Gaussian surface has radius r > b, then the charge enclosed is q + Q.

Last edited: Feb 25, 2010
9. Feb 25, 2010

### reising1

Oh I understand. I assumed the "less than the shell's radius" was referring to the outer radius. The wording was a bit ambiguous. But yes, if it refers to the inner radius, this makes much more sense.