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Gauss's Law Problem

  1. Oct 17, 2008 #1
    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. jcsd
  3. Oct 17, 2008 #2

    Doc Al

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    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. :wink:
  4. Oct 17, 2008 #3
    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! :smile:
  5. Oct 17, 2008 #4

    Doc Al

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    Staff: Mentor


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

    [tex]Q = \int \rho dv = \int \frac{C}{r^3} dv = \int (\frac{C}{r^3}) (4\pi r^2) dr[/tex]
  6. Oct 17, 2008 #5
    Aha! And then set the limits of integration from Rsub2 to Rsub1, correct?
  7. Oct 17, 2008 #6

    Doc Al

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    Staff: Mentor

    R_1 to R_2. (In the direction of increasing r.)
  8. Oct 17, 2008 #7
    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
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