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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

1. 4 × 10⁻⁴ J
2. 8 × 10⁻⁴ J
3. 9 × 10⁻⁴ J
4. 12 × 10⁻⁴ J

Correct answer: 8 × 10⁻⁴ J

Solution

The energy stored in a capacitor is given by U = (1/2)CV². The initial energy is U₁ = (1/2)(6 × 10⁻⁶)(10²) = 3 × 10⁻⁴ J, and the final energy is U₂ = (1/2)(6 × 10⁻⁶)(20²) = 12 × 10⁻⁴ J. The increase in energy is ΔU = U₂ - U₁ = 12 × 10⁻⁴ - 3 × 10⁻⁴ = 8 × 10⁻⁴ J.

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