# induced charge density at boundary surface

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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 $\sigma_2-\sigma_1$.
Initial charges on plates $\dfrac{\epsilon_oS}{d_1+d_2}V=q=\sigma S$ this must be on both plates since these can be treated as in series combination. $\sigma_1=\sigma\bigg(1-\dfrac{1}{\epsilon_1}\bigg)$ and $\sigma_2=-\sigma\bigg(1-\dfrac{1}{\epsilon_2}\bigg)$, on adding gives $\dfrac{\epsilon_oV}{d_1+d_2}\cdot\bigg(\dfrac{\epsilon_2-\epsilon_1}{\epsilon_2\epsilon_1}\bigg)$, which is not in accordance with answer at back.