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Conducting sphere in a conducting spherical shell

  1. Aug 31, 2004 #1
    A conducting sphere that carries a total charge of 6 micro C is placed at the center of a conducting spherical shell that also carries a total charge of 6 micro C .
    (a) Determine the charge on the inner surface of the shell.
    (b) Determine the total charge on the outer surface of the shell.

    this seems like it shouldn't be too hard, but i guess i just don't understand the concepts well enough to figure it out...

    I was thinking about where I would put field lines in this to understand what's going on- but all i can come up with is that inside a spherical shell there's no electric field....but there's a conducting sphere in there which throws that off. I'm not sure that would help anyway. I guess I'm just thinking- that since they both have + charges, the inside part of the shell would be...i dunno, -6 micro Coulombs...since the electrons would move to the charge??....and that maybe, since the charges are both radiating out- the outer surface would be 12 micro C? I am probably very wrong but this is why i am looking for help. anything would be much appreciated.
  2. jcsd
  3. Aug 31, 2004 #2

    Doc Al

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    Staff: Mentor

    Gauss's Law

    Welcome to PF!
    Depends what you mean by "inside the shell". Inside the cavity the field will be zero only if there's no charge inside. Inside the conducting material of the shell itself--that's where the static electric field will always be zero. So imagine a Gaussian surface inside the conducting material.
    Well, whether you know it or not you are exactly correct. Since the field within the conducting material is zero, the total charge within that Gaussian surface must be zero. That tells you the charge on the inner surface. And you know what the total charge is on the shell, thus you can deduce what the outer charge must be. Think it over.
  4. Aug 31, 2004 #3
    Ah ha! that's a big help- that's basically what i was thinking but i couldn't find anything in my book that said it clearly. Thanks so much for your help!
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