The equivalent capacitance between P & Q in the figure is

$\frac{A \varepsilon_0}{2 d}$ $\frac{3 A \varepsilon_0}{2 d}$ $\frac{2 A \varepsilon_0}{d}$ $\frac{5}{3} \frac{A \varepsilon_0}{d}$

$\frac{5}{3} \frac{A \varepsilon_0}{d}$

Equivalent circuit is shown in figure $C_{eq}=C+\frac{C \times 2 C}{C+2 C}$ $=C+\frac{2}{3} C=\frac{5}{3} C=\frac{5 A \varepsilon_0}{3 d}$ ∴ (D)