I am trying to prove that given a conductor with zero net charge on it then the charges necessarily distributes on its surface so that the density is zero at every point.

We can assume the definition of conductor so that an electron moves freely inside of it (i.e. it does not need to develop work in the process) and we can also assume stationary conditions and all the immediate consequences (e.g.: zero field inside the conductor, equipotential surface, Coulumb's Theorem, ...). Moreover we can also assume that, given the potential on the surface, there exists a unique distribution which satisfies this condition (i.e. we are assuming that the solution of Laplace's Equation with a Boundary Condition exists and is unique for a conductor), notice however that my problem is that the charge is given, not the potential.


The reason behind my request is that it could allow me to conclude that the charge on a conductor determines its potential (otherwise the same charge would give different potentials and using the principle of superposition I can reach a contradiction by using the above statement).

Any comment or answer is welcome and let me know if I can explain myself any clearer!


1 Answer 1


The assumption of zero net charge is not necessary. Also for a conductor with finite conductivity, this is valid only after relaxation, ie the stationary state.

This can be seen using the local Ohm’s law: $$ E=\sigma j $$ with $\rho$ the resistivity, $E$ electric field and $j$ current density. Injecting it into the continuity equation: $$ \partial_t\rho+\nabla\cdot j=0 $$ with $\rho$ charge density and using Gauss’ law: $$ \nabla \cdot E=\frac{\rho}{\epsilon_0} $$ you get: $$ \partial_t\rho+\frac{\sigma}{\epsilon_0}\rho=0 $$ Thus, assuming $\sigma$ is independent of $\rho$ (which is not the case in general), you get an exponential relaxation to: $$ \rho=0 $$ as you wanted.

In fact, you can view this as the continuous analogue of an RC circuit. More precisely, you can model a real conductor by a discrete mesh of leaky capacitors. As the mesh gets finer, you’ll retrieve the continuum limit.

Hope this helps.


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