The capacitance of the system of parallel plate capacitor shown in the figure is

$\frac{2 \varepsilon_0 A_1 A_2}{\left(A_1+A_2\right) d}$ $\frac{2 \varepsilon_0 A_1 A_2}{\left(A_2-A_1\right) d}$ $\frac{\varepsilon_0 A_1}{d}$ $\frac{\varepsilon_0 A_2}{d}$

$\frac{\varepsilon_0 A_1}{d}$

Since the electric field between the parallel charge plates is uniform and independent of the distance, neglecting the edge effect, the effective area of the plate of area A2 is A1. Thus the capacitance between the plates is $C=\frac{\varepsilon_0 A_1}{d}$ ∴ (C)