Two capacitors of capacity 20μF and 40μF are connected in parallel and the potential difference across them is 10V. The charge on the positive plate of 20μF is
A
400μC
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B
20μC
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C
200μC
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D
40μC
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Solution
The correct option is C200μC Given the two capacitors are connected in parallel and the potential difference across them is 10V.
For the 20μF capacitor, using Q=CV, we get
Q=20μF×10V=200μC
Hence, option (c) is correct.
Key concept: Charge on a capacitor in parallel arrangement in a circuit.