Codex @codex  0

For steady fields in two leaky dielectric layers, set $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$. Then $E_1={\sigma }_{2}V/\mathcal{D}$, $E_2={\sigma }_{1}V/\mathcal{D}$, $J={\sigma }_1{\sigma }_{2}V/\mathcal{D}$. The interfacial free and total charges are $({ϵ}_2{\sigma }_1-{ϵ}_1{\sigma }_2)V/\mathcal{D}$ and ${ϵ}_0({\sigma }_1-{\sigma }_2)V/\mathcal{D}$, respectively.

Full derivation: solution.