Question

In the figure shown find the equivalent capacitance between terminals 'A' and 'B'. The letters have their usual meaning capacitance is $\frac{x{\epsilon }_{0}A}{10d}$ then x is?

Answer 1

This is the case where $C_1$ & $C_3$ were in parallel and $C_2$ is in series.

$C_1=\frac{{\epsilon }_{0}A/2}{d/2},C_3=\frac{4{\epsilon }_{0}A/2}{d/2},C_2=\frac{{\epsilon }_{0}A}{d/2}$

$=$ $C_1=\frac{{\epsilon }_{0}A}{d}$

$C_2=\frac{2{\epsilon }_{0}A}{d}$

$C_3=\frac{4{\epsilon }_{0}A}{d}$

$\frac{1}{ceff}=\frac{1}{C_1+C_3}+\frac{1}{C_2}$

$=\frac{1}{ceff}=\frac{1}{\frac{{\epsilon }_{0}A}{d}+\frac{4{\epsilon }_{0}A}{d}}+\frac{1}{\frac{2{\epsilon }_{0}A}{d}}$

$=\frac{1d}{5{\epsilon }_{0}A}+\frac{d}{2{\epsilon }_{0}A}=\frac{d}{{\epsilon }_{0}A}[\frac{1}{5}+\frac{1}{2}]$

$\frac{1}{ceff}=\frac{7d}{10{\epsilon }_{0}A}$