Two parallel plates of areas \( A_1 \) and \( A_2 \) \( (A_1\lt A_2) \) are placed a distance d apart, as shown in the figure. Neglecting edge effects, the capacitance of the system is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \( \frac{\varepsilon_0 A_1}{d} \)
The capacitance of a parallel plate capacitor is \( C=\varepsilon_0A/d \), where \( A \) is the area over which the two plates face each other. With edge effects neglected, the field exists only between the facing parts of the plates. The charge on the larger plate gathers on the part that faces the smaller plate.
From the figure, the smaller plate \( A_1 \) lies fully above the larger plate \( A_2 \), a distance \( d \) away. So the area common to both plates is \( A_1 \). The rest of \( A_2 \) has no plate facing it and adds nothing to the capacitance.
\( \begin{aligned} C &= \frac{\varepsilon_0A}{d} \\ &= \frac{\varepsilon_0A_1}{d} \end{aligned} \)