# Homework Help: Gauss's Law Problem

1. Oct 17, 2008

### MichaelT

1. A thick spherical shell has an inner radius Rsub1 and outer radius Rsub2, and for Rsub1 is less than r which is less than Rsub2, a volume charge density rho=C/r^3, where C is a constant. Find Electric Field for r=3Rsub2

My problem here is determining the excess charge of the shell. So I have figured out the volume of the shell to be (4/3)pi(Rsub2^3 - Rsub1^3). I have also realized that to find the excess charge of the shell I can set Rsub2 equal to r. Now, the integration. That is where I am stuck. I am still left with the Rsub1 inside the integral, as well as 1/r^3. What do I do here?

Am I correct that the other integral would be the integral of 1/r dr and become ln(r)?

Please help, I really appreciate it. (Sorry for the mess, I know nothing about Latex)

2. Oct 17, 2008

### Staff: Mentor

Q = ∫ρ dv. Since ρ depends on r, as does v, you can't separate out the volume integral. (You could if the charge density was constant, not variable.)

That will come in handy.

3. Oct 17, 2008

### MichaelT

ok. so I did take some constants out to clean it up a little. I have got:

4piC(ln(r) - integral[(1/r^3)(Rsub1^2)dr])

I really can't think of what to do here. I know that I can't take the Rsub1 out of the integral.

Thank you!

4. Oct 17, 2008

### Staff: Mentor

:yuck:

You have to start over and redo the integral from scratch, like I suggested:

$$Q = \int \rho dv = \int \frac{C}{r^3} dv = \int (\frac{C}{r^3}) (4\pi r^2) dr$$

5. Oct 17, 2008

### MichaelT

Aha! And then set the limits of integration from Rsub2 to Rsub1, correct?

6. Oct 17, 2008

### Staff: Mentor

R_1 to R_2. (In the direction of increasing r.)

7. Oct 17, 2008

### MichaelT

Sorry, that is what I meant. Wow, I feel stupid now LOL. I was approaching it from the wrong direction. Thank you so much, I think I got it now.

Electric field at r=3Rsub2

E= [C(ln(Rsub2/Rsub1)]/9(Rsub2)^2