In many introductory books on electrostatics, you can find the statement that the field inside a conducting shell is zero if there are no charges within the shell. For example, if we place an uncharged conducting sphere in a uniform electric field we would get something like this:
For simplicity, I have drawn a uniform external field and a spherical shell, but the internal field is zero for any external field and any shape of the shell.
We know the field in the metal of the conductor is zero, so the external field must induce a charge separation in the conductor and this charge separation produces a field that cancels out the external field within the metal of the shell (shaded blue in the diagram). However, there is no obvious reason why the field should also be canceled to zero in the interior of the shell.
Although it is widely stated in introductions to electrostatics that the field is zero in the interior of the shell I cannot find proof of this. So the question is how we prove that field inside the shell is always zero whatever the external field and shape of the shell?
Best Answer
Here's a self-contained proof, with no hand-waving.
Let the surface $S$ be the boundary between the empty cavity and the conducting medium that surrounds it. Things may be discontinuous at the surface $S$, but that doesn't matter. The only thing that matters is that the conductor imposes a boundary condition on the electric field $\mathbf{E}$ inside the cavity: if we start somewhere inside the cavity and approach a point on $S$, then the component of $\mathbf{E}$ parallel to $S$ must go to zero. (Otherwise, it would create a current in the conductor.)
Inside the cavity, Maxwell's equations give $\nabla\times\mathbf{E}=0$ and $\nabla\cdot\mathbf{E}=0$. The first equation implies $\mathbf{E}=\nabla \phi$ for some scalar function $\phi$, and the second one implies $\nabla^2\phi=0$.
Altogether, the only conditions we need are:
$\nabla^2\phi=0$ everywhere inside the cavity.
$\nabla\phi$ is orthogonal to the boundary at every point on the boundary.
The boundary condition implies that $\phi$ must be equal to a constant $k$ on the boundary $S$, because otherwise $\nabla\phi$ would have a component parallel to $S$. Define $\phi'\equiv\phi-k$. Then $\phi'=0$ on $S$, and $\nabla^2\phi'=0$ everywhere. Integration-by-parts gives $$ \int_\text{cavity} (\nabla\phi')\cdot(\nabla\phi') \propto \int_S \mathbf{n}\cdot (\phi'\nabla\phi') $$ where $\mathbf{n}$ is the unit normal to $S$. The right-hand side is zero because of the boundary condition $\phi'=0$ on $S$, and therefore the left-hand side must also be zero. The integrand on the left-hand side is non-negative, so the integrand itself must be zero. This implies $\nabla\phi'=0$, which implies that the electric field $\nabla\phi$ is zero everywhere inside the cavity.