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Gauss Law - Conducting cylindrical shell

  1. Oct 17, 2008 #1
    1. A thick conducting cylindrical shell has an inner radius Rsub1 and outer radius Rsub2. It has a net excess charge = Q, and it is L long. Find electric field at certain points given.

    Ok, when r is less than Rsub 1, the electric field is zero. And when r is greater than Rsub2, it is easy to calculate using Gauss's Law. Now, when r is between Rsub1 and Rsub2....what is the excess charge inside the Gaussian cylinder? I know that any net charge in a conductor must be on one of the surfaces. But on which one, and how much on each?

    This might be a silly guess, but maybe Q/2 ?
     
  2. jcsd
  3. Oct 17, 2008 #2

    Doc Al

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    What's the field within the material of the conducting shell? (Between R_1 and R_2.)
     
  4. Oct 17, 2008 #3
    The electric field inside the material of a conducting shell is zero when the conductor is in electrostatic equilibrium. Does that still hold even though this conductor has a net charge of Q?
     
  5. Oct 17, 2008 #4

    Doc Al

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    Absolutely.
     
  6. Oct 17, 2008 #5
    Ok, that's cool. A little strange to me though. Intuitively it does make sense, but when you use Gauss's Law, wouldn't there be some excess charge inside a Gaussian cylinder with r between R_1 and R_2? Or would the excess charge Q remain on the outer surface of the conductor?
     
  7. Oct 18, 2008 #6

    Doc Al

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    The excess charge can only exist on the surface of the conducting cylinder. Since there's no charge contained in the cavity (at r < R_1), there's no charge on the inner surface. (Convince yourself of this using Gauss's law.) Thus all excess charge is on the outer surface.
     
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