- From physics we know that given a charged conductor put in vacuum ( no external electric fields) , the charge distribution on its surface is approximately proportional to the curvature of the surface on that point: I want to understand more about it.
- I formulate it in terms of PDE:
- Let $C$ be a connected bounded domain of $\mathbb R^3$ with smooth boundary.
- We can assume $C$ has no cavity ( i.e. the complement of $C$ is connected).
- Let $D=\mathbb R^3-\bar C$. WLOG assume the potential is $1$ on $C$ and $0$ at infinity.
- Then the induced potential outside is the solution $u$ of the Dirichlet problem on $D$: \begin{align} \Delta u & = 0\text{ on } D \\ u & = 1\text{ on } \partial D \\ u & \to 0 \text{ near } \infty \end{align}
- The charge density on $\partial D$ is proportional to the normal derivative $\dfrac{\partial u}{\partial n}$.
Is there any information we can obtain about this derivative $?$.
EDIT: Some attempts: Let the Green's function on $D$ be $G(x,y)$. We have $G(x,y)=G(y,x)$. Then $$u(x)=\int_{\partial D}{u(y)\dfrac{\partial}{\partial n_y}G(x,y)}\mathrm{d}y=\int_{\Gamma}{\dfrac{\partial}{\partial n_y}G(x,y)}\mathrm dy,$$ where $\Gamma$ is the inner boundary of $D$. It follows that $$\dfrac{\partial u(x)}{\partial n_x}=\int_{\Gamma}{\dfrac{\partial}{\partial n_x}\dfrac{\partial}{\partial n_y}G(x,y)}\mathrm dy.$$
The integrand looks very symmetric, and possibly there are some good properties I did not see.
I also attempt to apply divergence theorem formally. $$\dfrac{\partial u(x)}{\partial n_x}=\int_{\Gamma}{\dfrac{\partial}{\partial n_y}(\dfrac{\partial}{\partial n_x}G(x,y)})\mathrm dy=\int_D{\Delta_y\dfrac{\partial}{\partial n_x}G(x,y)\mathrm dy}-\lim_{r\to\infty}{\int_{|y|=r}{\dfrac{\partial}{\partial n_x}\dfrac{\partial}{\partial n_y}G(x,y)}\mathrm dy}.$$
The first term is $\int_D {\dfrac{\partial}{\partial n_x} \delta_x(y)\mathrm dy}$. But the first term involves some distribution theory that is far beyond my scope.
My guess is that the statement from physics is not entirely true. Imagine arbitrarily shaped conductor enclosed in conducting sphere. Than all the charge is distributed on the sphere and none on the weirdly shaped conductor inside.(Ok I know it does not satisfy the assumption with no cavity, but you can make small puncture to the sphere and the situation won't change dramatically)
The physics statement is true for sphere, where it is easy to find exact solution of Laplace equation. $$ u(x) = \frac{R}{\|x\|} $$ $$ \nabla u(x) = -R\frac{x}{\|x\|^3} $$ This gives us answer: $$ \|\nabla u(R)\| = \frac1{R} $$ which is exactly curvature.
And physicists once again abuse result such as this to use it on everything!