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Question

Two capacitors of capacity 20 μF and 40 μF 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 C 200μC
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×10 V=200μC

Hence, option (c) is correct.
Key concept: Charge on a capacitor in parallel arrangement in a circuit.

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