Question

How do you calculate capacitance with two dielectrics?

Answer 1

Find the capacitance of the parallel plate without dielectrics

Assume a parallel-plate capacitor having area $A$ of each plate and $d$ be the spacing between them.

The capacitance of the parallel plate without dielectrics is given by $C_0=\frac{{\epsilon }_{0}A}{d}$, where ${\epsilon }_0$ is the permittivity of the free space.

Find the capacitance, when the two dielectrics are kept side by side

When the two dielectrics are kept side by side between the parallel plates, it is termed parallel capacitors as shown in the figure,

Then the net-capacitance is given by $C=C_1+C_2$, with $C_1=\frac{k_1{\epsilon }_0{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{d}$ and $C_2=\frac{k_2{\epsilon }_0{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{d}$ , where $k$ is the dielectric constant of the medium, then

$\begin{array}{rcl}C& =& C_1+C_2\\ & =& \frac{k_1{\epsilon }_0{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{d}+\frac{k_2{\epsilon }_0{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{d}\\ & =& \frac{{\epsilon }_0{\displaystyle \raisebox{1ex}{$A$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}{d}\left(k_1+k_2\right)\end{array}$

Find the capacitance when the two dielectrics are one above another.

When the two dielectrics are kept in the form of a stack upon each other, between the parallel plates, then it forms a series connection shown in the figure,

Then the is given by with $C_1=\frac{k_1{\epsilon }_0{\displaystyle A}}{{\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ and $C_2=\frac{k_2{\epsilon }_0{\displaystyle A}}{{\displaystyle \raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}}$ then

$\begin{array}{rcl}\frac{1}{C}& =& \frac{1}{C_1}+\frac{1}{C_2}\\ \frac{1}{C}& =& \frac{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{{\displaystyle k_1{\epsilon }_{0}A}}+\frac{\raisebox{1ex}{$d$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{{\displaystyle k_2{\epsilon }_{0}A}}\\ \frac{1}{C}& =& \frac{d}{2{\epsilon }_{0}A}\left(\frac{1}{{\displaystyle k_1}}+\frac{1}{{\displaystyle k_2}}\right)\\ \frac{1}{C}& =& \frac{d}{2{\epsilon }_{0}A}\left(\frac{k_2+k_2}{{\displaystyle k_1{k}_2}}\right)\\ C& =& \frac{2k_1{k}_2}{\left(k_2+k_2\right)}\frac{{\epsilon }_{0}A}{d}\\ C& =& \frac{2k_1{k}_2}{{\displaystyle k_1+k_2}}{C}_0\end{array}$