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Field inside a cavity inside a conductor

by Kolahal Bhattacharya
Tags: cavity, conductor, field, inside
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Kolahal Bhattacharya
#1
Jun16-06, 01:03 PM
P: 140
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?
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siddharth
#2
Jun16-06, 01:25 PM
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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.
Kolahal Bhattacharya
#3
Jun16-06, 07:22 PM
P: 140
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!

siddharth
#4
Jun17-06, 08:43 AM
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Field inside a cavity inside a conductor

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)?


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