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

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U i n i t i a l = Q 2 2 C 1 U f i n a l = Q 2 2 C 2 Work done = U f i n a l - U i n i t i a l = Q 2 2 C 2 - Q 2 2 C 1 = 5 × 10 - 6 2 2 × 2 × 10 - 6 - 5 × 10 - 6 2 2 × 5 × 10 - 6 = 25 4 - 5 2 × 10 - 6 = 15 4 × 10 - 6 = 3.75 × 10 - 6 J

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