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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.For the first capacitor, area $A_1 = A/2$ and separation $d

Vedclass · Accepted Answer

$\frac{3A\epsilon_0}{4d}$

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(B) The capacitance of a parallel plate capacitor is given by $C = \frac{\epsilon_0 A}{d}$.For the first capacitor, area $A_1 = A/2$ and separation $d_1 = d$. Thus, $C_1 = \frac{\epsilon_0 (A/2)}{d} = \frac{\epsilon_0 A}{2d}$.For the second capacitor, area $A_2 = A/2$ and separation $d_2 = 2d$. Thus, $C_2 = \frac{\epsilon_0 (A/2)}{2d} = \frac{\epsilon_0 A}{4d}$.Since the capacitors are connected in parallel, the equivalent capacitance is $C_{eq} = C_1 + C_2$.$C_{eq} = \frac{\epsilon_0 A}{2d} + \frac{\epsilon_0 A}{4d} = \frac{2\epsilon_0 A + \epsilon_0 A}{4d} = \frac{3\epsilon_0 A}{4d}$.

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

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