The space between the plates of a parallel plates capacitor is filled consecutively with two dielectrics layers $1$ and $2$ having the thickness $d_1$ and $d_2$ respectively and permittivities $\epsilon_1$ and $\epsilon_2$. The area of each plate is equal to $S$ find:

1. The capacitance of the capacitor

2. The density $\sigma'$ of the bound charges on the boundary plane if the voltage across the capacitor equals $V$ and the electric field is directed from layer $1$ to layer $2$.

I managed to find $C_{eq}$ easily, for the second part I don't understand that word " *boundary plane*", I only know $\sigma'=\sigma\bigg(1-\dfrac{1}{\epsilon}\bigg)$ holds when there would have been just one plate, how to work on two such consecutive plates. Please help.

Probably you have found the capacitance by modelling the two-dielectric capacitor as two separate single-dielectric capacitors in series, with each dielectric sandwiched between metal plates. There are surface charges $-\sigma_1, +\sigma_2$ on the two dielectrics adjacent to the inner connected plates. When the dielectrics are adjacent in the same two-layer capacitor the same surface charges exist on the inner surfaces.

I think $\sigma'$ must be the algebraic sum of these two surface charges, ie $\sigma_2-\sigma_1$.

edited Oct 28 by sammy gerbil