Parallel plate capacitor is constructed using three different dielectric materials as shown in fig.

The parallel plates, across which a potential difference is applied are of area A cm² and separated by a distance d cm. The capacitance across AB is

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$\frac{ε_{0} A}{d} [\frac{2}{K_{1}} + \frac{K_{2} K_{3}}{K_{1} + K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{K_{1}}{2} + \frac{(K_{2} + K_{3})}{K_{2} K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{2}{K_{1}} + \frac{(K_{2} + K_{3})}{K_{1} K_{3}}]$

$\frac{ε_{0} A}{d} [\frac{K_{1}}{2} + \frac{K_{2} K_{3}}{K_{2} + K_{3}}]$

answer is A.

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Detailed Solution

$\begin{aligned} C_{1} = \frac{K_{1} ε_{0} (\frac{A}{2})}{d} = \frac{K_{2} ε_{0} A}{d} \\ C_{2} = \frac{K_{2} ε_{0} (\frac{A}{2})}{(\frac{d}{2})} = \frac{K_{2} ε_{0} A}{d} \end{aligned}$
and $C_{3} = \frac{K_{3} ε_{0} A}{d}$
Now $C = C_{1} + \frac{C_{2} C_{3}}{C_{2} + C_{3}}$