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Let the unit normal point from layer 1 to layer 2. The electrostatic electric field boundary condition is $\mathbf{n}\times ({\mathbf{E}}_2-{\mathbf{E}}_1)=0$, while Gauss's law for the electric displacement field gives $\mathbf{n}\cdot ({\mathbf{D}}_2-{\mathbf{D}}_1)={\sigma }_f$. Here ${ϵ}_i$ denotes absolute permittivity; replace it by ${ϵ}_0{ϵ}_{ri}$ if dielectric constants are relative.

In the steady conducting state, charge conservation makes the normal current density the same in both layers:Writing $\mathcal{D}={\sigma }_2{d}_1+{\sigma }_1{d}_2$, the steady fields in two leaky dielectric layers areThe signed interfacial free surface charge density is the jump of $D_n$:The total free-plus-bound sheet charge instead follows from the vacuum form of Gauss's law,The bound contribution is ${\sigma }_{\mathrm{total}}-{\sigma }_f$. These are signed quantities for the chosen normal and voltage direction; their magnitudes are obtained by absolute values. They describe the long-time DC state, not the initial capacitive transient. Positive conductivities suffice for the derivation; if both vanish, conductivity alone does not determine the field division, which also depends on the initial free interfacial charge.

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.