Two identical thin metal plates are given charges \( q_1 \) and \( q_2 \) (\( q_2 \lt q_1 \)) 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
The correct answer is Option (3) → \( \displaystyle\frac{q_1-q_2}{2C} \)
When two charged plates face each other, the charge on each plate spreads over its two faces so that the field inside each metal plate is zero. This gives the standard distribution:
\( \text{outer faces: } (q_1+q_2)/2 \text{ each} \)
\( \text{inner faces: } +(q_1-q_2)/2 \text{ and } -(q_1-q_2)/2 \)
The charges on the outer faces produce equal and opposite fields between the plates, which cancel. The field in the gap comes only from the charges on the inner faces, which behave like a capacitor holding charge \( Q = (q_1-q_2)/2 \).
\( \displaystyle V = \frac{Q}{C} = \frac{q_1-q_2}{2C} \)