A parallel plate capacitor has 1µF capacitance. One of its two plates is given + 2µC charge and the other plate, + 4µC charge. The potential difference developed across the capacitor is :

Engineering

Physics

Circuit Analysis Having Capacitor Only

Question

2 V

1 V

3 V

5 V

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Charges at inner plates are 1 µC and –1 µC.

$∴$ Potential difference across capacitor

$= \frac{q}{c} = \frac{1 μC}{1 μF} = \frac{1 \times 10^{- 6} C}{1 \times 10^{- 6} Farad} = 1 V$