Four metallic plates of equal surface area $A$ are arranged, as shown in figure. The equivalent capacitance between $A$ and $B$ is,

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Aicharya · Accepted Answer

The equivalent circuit for the given combination of plates is shown in figure. The effective capacitance of the series combination of $C_{1}$ and $C_{2}$ is

$\begin{array}{c}
\frac{1}{C_{s}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{1}{\frac{\varepsilon_{0} A}{d}}+\frac{1}{\frac{\varepsilon_{0} A}{d}} \\
C_{s}=\frac{\varepsilon_{0} A}{2 d}
\end{array}$
$\mathrm{C}_{\mathrm{s}}$ and $\mathrm{C}_{3}$ are parallel.
$C_{A B}=C_{s}+C_{3}=\frac{\varepsilon_{0} A}{2 d}+\frac{\varepsilon_{0} A}{d}$
$\Rightarrow C_{A B}=\frac{3}{2} \frac{\varepsilon_{0} A}{d}$ *

Four metallic plates of equal surface area \(A\) are arranged, as shown in figure. The equivalent…

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