Gauss's Law Problem

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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)
 

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  • #2
Doc Al
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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?
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.)

Am I correct that the other integral would be the integral of 1/r dr and become ln(r)?
That will come in handy. :wink:
 
  • #3
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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:
 
  • #4
Doc Al
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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])
:yuck:

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]
 
  • #5
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Aha! And then set the limits of integration from Rsub2 to Rsub1, correct?
 
  • #6
Doc Al
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R_1 to R_2. (In the direction of increasing r.)
 
  • #7
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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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