For charge $Q_1$, electric field is

$E_1=\frac{Q_1}{2{\epsilon }_{0}A}$

where $A$ is area.

For charge $Q_2$ electric field is

$E_2=\frac{Q_2}{2{\epsilon }_{0}A}$

Resultant electric field

$E=E_1-E_2=\frac{(Q_1-Q_2)}{2{\epsilon }_{0}A}$

Potential difference between plates when they are brought close together to form a parallel plate capacitor is

$V=Ed=\frac{(Q_1-Q_2)}{2{\epsilon }_{0}A}d$

Since, $C=\frac{{\epsilon }_{0}A}{d}$ for parallel plate capacitor.

$V=\frac{(Q_1-Q_2)}{2C}$