# Field inside a cavity inside a conductor

1. Jun 16, 2006

### Kolahal Bhattacharya

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. Jun 16, 2006

### siddharth

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.

3. Jun 16, 2006

### Kolahal Bhattacharya

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!

4. Jun 17, 2006

### siddharth

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