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Question

If C1=14 μF and C2=38 μF, what is the charge on C2 in the steady state?

A
5500 μC
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B
0
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C
4160 μC
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D
2080 μC
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Solution

The correct option is B 0
Due to the presence of conducting wire AB, no charge will be accumulated on the plates of the capacitors.

AB path is a short circuit path, so all the charges bypass through AB.

Hence, the charges on both the capacitors will be 0

Why this Question ?
Key point: Whenever a short-circuit path is introduced in the direction of charge flow, entire charge passes through the short-circuit path.

These type of circuits are also called as bypass circuits.

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