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Question:

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

3.75×10⁻⁶ J

2.55×10⁻⁶ J

2.16×10⁻⁶ J

6.25×10⁻⁶ J

Solution:

Correct option is A. 3.75×10⁻⁶ J
Work done = ΔU = ΔU = Uf − Ui = q²/2Cf − q²/2Ci = (5×10⁻⁶)²/2(2×10⁻⁶) − (5×10⁻⁶)²/2(5×10⁻⁶) = 3.75 × 10⁻⁶ J