Question

A conductor with resistivity $$\rho$$ bounds on a dielectric with permittivity $$\epsilon$$. At a certain point $$A$$ at the conductor's surface the electric displacement equals $$D$$, the vector $$D$$ being directed away from the conductor and forming an angle $$\alpha$$ with the normal of the surface. Find the surface density of charges on the conductor at the point $$A$$ and the current density in the conductor in the vicinity of the same point.

Answer 1

The dielectric ends in a conductor. It is given that on one side (the dielectric side) the electric displacement $$D$$ is as shown. Within the conductor, at any point $$A$$, there can be no normal component of electric field. For it there were such a field, a current will flow towards depositing charge there which in turn will set up countering electric field causing the normal component to vanish. Then by Gauss theorem, we easily derive $$\sigma = D_{n} = D\cos \alpha$$ where $$\sigma$$ is the surface charge density at $$A$$.
The tangential component is determined from the circulation theorem
$$\oint \vec {E} \cdot d\vec {r} = 0$$
It must be continuous across the surface of the conductor. Thus, inside the conductor there is a tangential electric field of magnitude,
$$\dfrac {D\sin \alpha}{\epsilon_{0} \epsilon}$$ at $$A$$
This implies a current, by Ohm's law, of
$$j = \dfrac {D \sin \alpha}{\epsilon_{0} \epsilon \rho}$$.