= Solution
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 $\epsilon_i$ denotes absolute <permittivity>; replace it by $\epsilon_0\epsilon_{ri}$ if dielectric constants are relative.
In the steady conducting state, <charge conservation> makes the normal <current density> the same in both layers:
$$
J=\sigma_1E_1=\sigma_2E_2,\qquad V=E_1d_1+E_2d_2.
$$
Writing $\mathcal D=\sigma_2d_1+\sigma_1d_2$, the <steady fields in two leaky dielectric layers> are
$$
\boxed{E_1=\frac{\sigma_2V}{\mathcal D},\qquad E_2=\frac{\sigma_1V}{\mathcal D},\qquad J=\frac{\sigma_1\sigma_2V}{\mathcal D}.}
$$
The signed interfacial free <surface charge density> is the jump of $D_n$:
$$
\boxed{\sigma_f=\frac{(\epsilon_2\sigma_1-\epsilon_1\sigma_2)V}{\mathcal D}.}
$$
The total free-plus-bound sheet charge instead follows from the vacuum form of <Gauss's law>,
$$
\boxed{\sigma_{\rm total}=\epsilon_0(E_2-E_1)
=\frac{\epsilon_0(\sigma_1-\sigma_2)V}{\mathcal D}.}
$$
The bound contribution is $\sigma_{\rm total}-\sigma_f$. 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.
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