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A 10 μF capacitor is charged to a potential difference of 50 V and is connected to another uncharged capacitor in parallel. Now the common potential difference becomes 20 V . What is the capacitance value (in μ F ) of the second capacitor?

Correct answer

15

Step-by-step solution

q 1 = 10 × 50 = 500   μC ,   C 1 = 10   μF ,   C 2 = ? ,   q 2 = 0 As, V = q 1 + q 2 C 1 + C 2 C 1 + C 2 = q 1 + q 2 V = 500 + 0 20 = 25   μF C 2 = 25 - C 1 = 25 - 10 = 15   μF

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