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Q.

Four metallic plates, each with a surface area of one side A, are placed at a distance d from each other. The plates reconnected as shown in the fig.

Then the capacitance of the system between A and B is

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a

$\frac{3 ε_{0} A}{d}$

b

$\frac{2 ε_{0} A}{d}$

c

$\frac{2}{3} \frac{ε_{0} A}{d}$

d

$\frac{3}{2} \frac{ε_{0} A}{d}$

answer is D.

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Detailed Solution

detailed_solution_thumbnail

The equivalent arrangement of given combination is shown in fig.

Question Image

Capacitance between a and D is given by
$\frac{1}{C_{ab}} = \frac{1}{C} + \frac{1}{C} = \frac{2}{C}$ or $C_{ab} = \frac{C}{2} = \frac{1}{2} \frac{ε_{0} A}{d}$
$∴ C_{AB} = C + \frac{1}{2} \frac{ε_{0} A}{d} = \frac{ε_{0} A}{d} + \frac{1}{2} \frac{ε_{0} A}{d} = \frac{3}{2} \frac{ε_{0} A}{d}$

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