Four metallic plates, each with surface area of one side, A, are placed at a distance d from each other. The plates are connected as shown in fig. Then the capacitance of the system between P and Q is

 

The equivalent circuit is shown in fig. The condensers Y and Z are in parallel. Hence, their equivalent capacitance is 2 C. This combination is in series with capacitor X. Hence the equivalent capacitance between P and Q is given by

${\mathrm{C}}_{\mathrm{PQ}}=\frac{\mathrm{C}\times 2\mathrm{C}}{\mathrm{C}+2\mathrm{C}}=\frac{2}{3}\mathrm{C}=\frac{2}{3}\frac{{\mathrm{\epsilon }}_0\mathrm{A}}{\mathrm{d}}$