Electric field inside hollow conductor boundary value problem

  • Thread starter Tomkat
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Hi,

I am in Purcell's E&M book at the section explaining why the field is zero inside a hollow conductor of any shape. The proof given is that the potential function inside the conductor must obey Laplace's equation, and that the boundary of the region (in this case a rectangular metal box) is an equipotential. It calls the potential on the box some constant phi, then states that one solution (i.e. value of the potential inside the box) is that the potential is constant throughout the volume and equal to phi, the potential on the surface. Then it states by the uniqueness theorem that this is the only solution.

I understand all of this except why one couldn't say: one solution is that the potential is constant throughout the volume and *not equal to* phi (the potential on the surface of the box still being equal to phi of course). Therefore this is the only solution. The proof would come out the same, but it is this mathematical step that confuses me. Thanks,

Tomkat
 

Born2bwire

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\phi here is the boundary condition for the Laplace equation. If you define the interior to be some constant other than \phi, then there must be a discontinuity at the boundary. This would not satisfy Laplace's equation since we would have to take the derivative at the discontinuity causing the Laplacian to be nonzero.
 
Considering the laplace equation, We know the whole conductor is an equipotential, including the surface of it.
 
That makes some more sense. Thank you.
 

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