bottom surfaces of the film are connected by a wire that maintains a zero voltage difference between them. This geometry is commonly adopted to minimize the effects of depolarizing electric fields. We thus have two kinds of charge: a bound charge $\rho^\mathrm{b}(z) = - \partial_z P_\mathrm{tot}(z)$ coming from a sum of atomic and lattice polarizations, and the external charges $\rho^\mathrm{ext}(z)$ on the top and bottom electrodes.

The electric field in equation  is obtained from these charges via Gauss’ law, $$\epsilon_{0} \frac{d}{dz} E(z) = \rho^\mathrm{b}(z) + \rho^\mathrm{ext}(z).$$ We break the polarization into lattice and atomic pieces, $P(z)$ and $\epsilon_0 \alpha E(z)$ respectively, with $\alpha$ the atomic polarizability, and defining the optical dielectric constant $\epsilon_\infty=\epsilon_0(1+\alpha)\approx 5.5 \epsilon_0$ [@raslan17; @zollner00], we obtain the usual expression $$\label{Gauss}
\frac{d}{dz} \left[ \epsilon_{\infty} E(z) + P(z) \right] = \rho^\mathrm{ext}(z),$$ which can be integrated to find $E(z)$.

The charge density in the top and bottom electrodes is written as $$\rho^\mathrm{ext}(z) = \frac{en}{a^{2}} [ \delta(z) - \delta(z-L) ],$$ where $L$ is the film thickness, and $n$ is the positive charge per 2D unit cell on the top electrode. Integrating equation  gives $$\label{E_Gauss}
\epsilon_{\infty} E(z) = - P(z) + \frac{en}{a^{2}}.$$ A second integration, of equation  across the thickness of the 