In an isolated parallel plate capacitor of capacitance $C$, the four surface have charges $Q_1, Q_2,Q_3\text{ and }{Q}_4$as shown. The potential difference between the plates is

Plane conducting surfaces facing each other must have equal and opposite charge densities. Here as the plate areas are equal, $Q_2=-Q_3.$

The charge on a capacitor means the charge on the inner surface of the positive plate (here it is  Q₂)

Potential difference between the plates  $=\frac{\text{ charge }}{\text{ capacitance}}=\frac{Q_2}{C}=\frac{2Q_2}{2C}=\frac{Q_2-\left(-Q_2\right)}{2C}=\frac{Q_2-Q_3}{2C}\text{. }$