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A capacitor with capacitance 5 μF is charged to 5 μC. If the plates are pulled apart to reduce the capacitance to 2 μF, how much work is done?

1. 3.75 × 10⁻⁶ J
2. 6.25 × 10⁻⁶ J
3. 2.55 × 10⁻⁶ J
4. 2.16 × 10⁻⁶ J

Correct answer: 3.75 × 10⁻⁶ J

Solution

Charge Q=5 uC is constant. U = Q^2/2C, so U2 = (25e-12)/(2*2e-6) = 6.25e-6 J and U1 = (25e-12)/(2*5e-6) = 2.5e-6 J. Work done = U2 - U1 = 3.75e-6 J.

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