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:
Answer & explanation
Correct answer: option 3
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}$