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Four metallic plates each with a surface area of one side A are placed at a distance d from each other. The plates are connected as shown in the circuit diagram. Then, the capacitance of the system between points a and b is $\frac{x ε_{0} A}{2 d}$ . The value of x is

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answer is 3.

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Detailed Solution

In the given figure, the metallic plates form a combination of two capacitors in series with one capacitor in parallel. Let the capacity of each capacitor is C, then equivalent capacitance between a and b

$C_{a b} = \frac{C}{2} + C = \frac{3 C}{2} ∵ C = \frac{ε_{0} A}{d} ∴ C_{a b} = \frac{3 ε_{0} A}{2 d}$

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