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Homework Help: Flux through a nonconducting shell

  1. Feb 25, 2010 #1
    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. jcsd
  3. Feb 25, 2010 #2

    kuruman

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    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)?
     
  4. Feb 25, 2010 #3
    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.
     
  5. Feb 25, 2010 #4
    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)
     
  6. Feb 25, 2010 #5

    kuruman

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    What answer appears to be saying that? Do you know what the answer is?
     
  7. Feb 25, 2010 #6
    Yes, this is a homework problem that I have the book's answer to.
     
  8. Feb 25, 2010 #7
    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"
     
  9. Feb 25, 2010 #8

    kuruman

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    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
  10. Feb 25, 2010 #9
    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.
     
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