= Solution
For <steady fields in two leaky dielectric layers>, set $\mathcal D=\sigma_2d_1+\sigma_1d_2$. Then $E_1=\sigma_2V/\mathcal D$, $E_2=\sigma_1V/\mathcal D$, $J=\sigma_1\sigma_2V/\mathcal D$. The interfacial free and total charges are $(\epsilon_2\sigma_1-\epsilon_1\sigma_2)V/\mathcal D$ and $\epsilon_0(\sigma_1-\sigma_2)V/\mathcal D$, respectively.
Full derivation: <past-exam-of-the-physics-b-course-of-the-university-of-cambridge/2010/paper-2a/section-b/b6/solution>.
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