Question

A capacitor of 10 µF, which is charged upto

• A. 400 V
• B. 200 V
• C. 100 V

Answer 1

Answer: (B)

200 V

Answer 2

The potential difference $V$ is found from

$$V = \frac{Q}{C}.$$

Here,

$$Q = 2 \times 10^{-3}\,{\rm C},\quad C = 10 \times 10^{-6}\,{\rm F}.$$

Thus,

$$V = \frac{2 \times 10^{-3}}{10 \times 10^{-6}} = \frac{2 \times 10^{-3}}{10^{-5}} = 200~{\rm V}.$$