Question:
A parallel plate capacitor has \(1\ \mu\)F capacitance. One of its two plates is given \(+2\ \mu\)C charge and the other plate, \(+4\ \mu\)C charge. The potential difference developed across the capacitor is:-
A parallel plate capacitor has \(1\ \mu\)F capacitance. One of its two plates is given \(+2\ \mu\)C charge and the other plate, \(+4\ \mu\)C charge. The potential difference developed across the capacitor is:-
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Only the charge on the inner (facing) surfaces matters; it equals half the difference of the plate charges.
Updated On: Nov 5, 2026
- 1V
- 5V
- 2V
- 3V
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The Correct Option is A
Solution and Explanation
For two parallel plates with charges \(Q_1\) and \(Q_2\), the facing surfaces carry \(\pm\frac{Q_1 - Q_2}{2}\) while each outer surface carries \(\frac{Q_1 + Q_2}{2}\).
Here the outer surfaces each carry \(\frac{2 + 4}{2} = 3\ \mu\)C, and the inner surfaces carry \(\mp\frac{4 - 2}{2} = \mp 1\ \mu\)C.
The capacitor charge is \(Q = 1\ \mu\)C, so \(V = \frac{Q}{C} = \frac{1\ \mu\text{C}}{1\ \mu\text{F}} = 1\) V.
Answer: (1) 1V
Here the outer surfaces each carry \(\frac{2 + 4}{2} = 3\ \mu\)C, and the inner surfaces carry \(\mp\frac{4 - 2}{2} = \mp 1\ \mu\)C.
The capacitor charge is \(Q = 1\ \mu\)C, so \(V = \frac{Q}{C} = \frac{1\ \mu\text{C}}{1\ \mu\text{F}} = 1\) V.
Answer: (1) 1V
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