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If the area of each plate is S and the successive separations are d, 2d and 3d then the equivalent capacitance across A and B is :

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$\frac{3 ε_{0} A}{4 d}$

$\frac{ε_{0} A}{4 d}$

$\frac{5 ε_{0} A}{4 d}$

answer is B.

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Detailed Solution

Between A and B two capacitor (with separation d and 3d) are in series

$C = \frac{C_{1} C_{2}}{C_{1} + C_{2}}$

$\frac{(\frac{ε_{0} A}{d}) (\frac{ε_{0} A}{3 d})}{(\frac{ε_{0} A}{d}) + (\frac{ε_{0} A}{3 d})}$

Hence, $= \frac{ε_{0} A}{4 d}$

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