Question

The gap between the plates of a parallel-plate capacitor is filled up with two dielectric layers $1$ and $2$ with thicknesses $d_1$ and $d_2$, permittivities ${ϵ}_1$ and ${ϵ}_2$, and resistivities ${\rho }_1$ and ${\rho }_2$. A dc voltage $V$ is applied to the capacitor, with electric field directed from layer $1$ to layer $2$. Find $\sigma$, the surface density of extraneous charges at the boundary between the dielectric layers, and the condition under which $\sigma =0$.

Answer 1

We have, $E_1{d}_1+E_2{d}_2=V$

and by current conservation

$\frac{1}{{\rho }_1}{E}_1=\frac{1}{{\rho }_2}{E}_2$

Thus, $E_1=\frac{{\rho }_{1}V}{{\rho }_1{d}_1+{\rho }_2{d}_2}$,

$E_2=\frac{{\rho }_{2}V}{{\rho }_1{d}_1+{\rho }_2{d}_2}$

At the boundary between the two dielectrics,

$\sigma =D_2-D_1={ϵ}_0{ϵ}_2{E}_2-{ϵ}_0{ϵ}_1{E}_1$

$\frac{{ϵ}_{0}V}{{\rho }_1{d}_1+{\rho }_2{d}_2}({ϵ}_2{\rho }_2-{ϵ}_1{\rho }_1)$.