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A parallel-plate capacitor of capacitance C has initial charges on its plates as shown in a circuit figure. At t = 0 the switch S is closed. Which of the following statements about the steady state are correct? (Select all that apply.)

1. In steady state the charges on the outer surfaces of plates A and B are equal in magnitude and have the same sign.
2. In steady state the charges on the outer surfaces of plates A and B are equal in magnitude and opposite in sign.
3. In steady state the charges on the inner surfaces of plates A and B are equal in magnitude and opposite in sign.
4. The work done by the battery by the time steady state is reached is 5*epsilon²*C/2.

Correct answer: In steady state the charges on the outer surfaces of plates A and B are equal in magnitude and have the same sign.

Solution

A standard result for a two-plate capacitor in a circuit is that the outer surfaces of the two plates always carry the same charge (magnitude and sign), while the inner surfaces carry equal and opposite charges. Option A and C are correct; the exact work done by the battery (option D) cannot be verified without the figure.

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