Electric field and curvature

Question

My physic teacher said that

In a conductor the electric field, which is non-zero only on the surface, is stronger where the curvature is bigger*.

But he did not provide a mathematical proof for this. Furthermore, I don't know the correct statement of this proposition: does curvature mean Gaussian curvature? And which are the hypotheses? Is the statement true for every equipotential surface?

I am asking for

1. The correct and complete statement;
2. The proof (or a reference for the proof).

Thanks in advance.

EDIT

• That doesn't mean that the field is proportional to curvature. It means: if we pick two points A and B, and in A the curvature is bigger than in B, then the electric field in A will be stronger than in B.

Answer 1

The relation between the electric field and the curvature of the conductur is not very easy to state, as you can see in [1], and the idea can be formalized in many ways:

Green (see: [2]) proved that

$\frac{dE}{dn}=-2kE$

where $k$ is the mean curvature, $E$ is the electric field on the conductor and $n$ is the normal to the conductor.

This equation, if integrated, would give the relation between the E.F. and curvature, but this operation is very hard, if not impossible in the general setting:

In general there is no unique relationship between conductor curvature and surface charge density. However by restricting attention to situations for which the potential is a function of a single variable, the authors demonstrate that the magnitude of the surface charge density at any point of the conductor surface is proportional to the fourth root of the magnitude of the Gaussian curvature at this location.

Mc Allister

In the specific case of ellipsoids, hyperboloids and paraboloids of rotation, the relation (founded by Liu and McAllister, see [3] and [4]) was:

$\sigma\propto k_g^{\frac{1}{4}}$

Note that $k_g$ is the gaussian curvature, not the mean curvature