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If the potential of a capacitor having capacity 6 µF is increased from 10 V to 20 V, then increase in its energy will be
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- 4 × 10-4J
- 4 × 10-4J
- 9 × 10-4J
- 12 × 10-6J
Correct Option: C
Capacitance of capacitor (C) = 6 µF = 6 ×10-6 F; Initial potential (V1) = 10 V and final potential (V2) = 20 V. The increase in energy (∆U) .
| = | C(V2² + V1²) | |
| 2 |
| = | × (6 × 10-6) × [(20)² - (10)²] | |
| 2 |
= (3 × 10-6) × 300 = 9 × 10-4J
