Two identical thin metal plates are given charges q₁ and q₂ (q₂ < q₁) respectively. If they are now brought close together to form a parallel plate capacitor with a capacitance 'C' then the potential difference between the plates is:

$\frac{q_1-q_2}{2C}$

After they are brought close together to form a parallel plate capacitor then charges are rearranged and redistribution of charge takes place.

New charges on the sides of capacitors facing each other is $q_1 - q_2$ and $ q_2 -  q_1$

Hence Potential difference is $ V = \frac{q}{C} = \frac{q_1 - q_2}{2C}$