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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A'}{d'}$, where $A'$ is the area and $d'$ is the separation.Given

Vedclass · Accepted Answer

$\frac{11\epsilon_0A}{18d}$

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(C) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A'}{d'}$, where $A'$ is the area and $d'$ is the separation.Given that the area of each capacitor is $A' = A/3$.The capacitances of the three capacitors are:$C_1 = \frac{\epsilon_0 (A/3)}{d} = \frac{\epsilon_0 A}{3d}$$C_2 = \frac{\epsilon_0 (A/3)}{2d} = \frac{\epsilon_0 A}{6d}$$C_3 = \frac{\epsilon_0 (A/3)}{3d} = \frac{\epsilon_0 A}{9d}$Since they are connected in parallel, the equivalent capacitance $C_{eq}$ is the sum of individual capacitances:$C_{eq} = C_1 + C_2 + C_3$$C_{eq} = \frac{\epsilon_0 A}{3d} + \frac{\epsilon_0 A}{6d} + \frac{\epsilon_0 A}{9d}$Taking the common denominator as $18d$:$C_{eq} = \frac{6\epsilon_0 A + 3\epsilon_0 A + 2\epsilon_0 A}{18d} = \frac{11\epsilon_0 A}{18d}$.

Three parallel plate air capacitors are connected in parallel. Each capacitor has plate area $A/3$ and separation between the plates is $d$, $2d$, and $3d$ respectively. The equivalent capacity of the combination is ($\epsilon_0$ is the permittivity of free space).

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