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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.

asked Oct 28, 2018 in Physics Problems by n3 (348 points)
edited Jan 2 by sammy gerbil
I think the 'boundary plane' which is referred to is the interface between the two dielectric layers.

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.
You seem to be assuming that charge on capacitor plates is same with or without the dielectric layers. Voltage is same, capacitance is different therefore charge on plates is also different.
Thank you so much, I got the correct answer.

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