Field inside a cavity inside a conductor

  1. In the context of properties of conductor & 1st Uniqueness theorem, Griffiths proves that field inside a cavity( empty of charge) within a conductor is 0.
    Is the result same if we place a +q & a -q (so that Q(enc)=0)
    suspended in air inside the cavity?
  2. jcsd
  3. siddharth

    siddharth 1,191
    Homework Helper
    Gold Member

    I would say no, because Laplace's equation won't be satisfied everywhere, because of the point charges. So you can't apply the first uniqueness theorem.
  4. I agree with you about the result but cannot be satisfied with your argument.As Poisson's eqn. takes into consideration Rho(r)...& still satisfies Uniqueness theorems. Lastly I think it is solved:we will have a unique V(r) function from which E follows.This V(r) will not satisfy properti--es of Laplace eqn.In boundary, V(r)=V(0),Following properties of a conductor...Otherwise the system I'm talking of will not exist at all.It will collapse immediately after we place them together within the cavity,following Earnshaw's theorem.Any conceptual mistake?Please help!
  5. siddharth

    siddharth 1,191
    Homework Helper
    Gold Member

    Yeah, you're right, the uniqueness theorem still holds and the potential can be uniquely determined, but it won't be constant inside the conductor.

    Also, I don't understand why you say V(R) = V(0)?
    Last edited: Jun 17, 2006
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