Electric field inside hollow conductor boundary value problem

[Tomkat]

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]

\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.

 

[inempty]

Considering the laplace equation, We know the whole conductor is an equipotential, including the surface of it.

 

[Tomkat]

That makes some more sense. Thank you.

 

[sardar]

Hi Tomkat,

Thank you for your question. The proof for the electric field being zero inside a hollow conductor is based on the fact that the potential inside the conductor must obey Laplace's equation and that the boundary of the region is an equipotential.

In this case, the boundary value problem is set up in such a way that the potential on the surface of the conductor is held constant at a value of phi. This means that the potential inside the conductor can only take on values that are equal to phi, as any other value would not satisfy the boundary condition.

If we were to consider a solution where the potential inside the conductor is constant but not equal to phi, this would not satisfy the boundary condition and therefore would not be a valid solution. The uniqueness theorem states that there can only be one solution that satisfies the given boundary conditions, and in this case, that solution is a constant potential equal to phi throughout the volume of the conductor.

I hope this helps clarify the mathematical step that was confusing to you. If you have any further questions, please don't hesitate to ask.

Best,