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The equivalent capacitance of the arrangement shown in figure, if A is the area of each plate, is

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

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

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

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

answer is B.

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Detailed Solution

$C = C_{L H S} + C_{R H S}$

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

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

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