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Field inside a cavity inside a conductor 
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#1
Jun1606, 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? 


#2
Jun1606, 01:25 PM

HW Helper
PF Gold
P: 1,198

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
Jun1606, 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 properties 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!



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