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(D) When $S$ is open,the two capacitors on the left are in parallel,and this combination is in series with the capacitor on the right.The equivalent c

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$A$ and $C$ both

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(D) When $S$ is open,the two capacitors on the left are in parallel,and this combination is in series with the capacitor on the right.The equivalent capacitance is $C_{eq} = \frac{C(C+C)}{C+(C+C)} = \frac{2}{3}C$.The initial charge on the circuit is $Q_{eq} = C_{eq}E = \frac{2}{3}CE$.When $S$ is closed,the capacitor on the right is short-circuited and does not store any charge. The two capacitors on the left remain in parallel across the cell.The new equivalent capacitance is $C'_{eq} = C + C = 2C$.The new charge on the circuit is $Q'_{eq} = C'_{eq}E = 2CE$.The charge that flows out of the positive terminal of the cell is $\Delta Q = Q'_{eq} - Q_{eq} = 2CE - \frac{2}{3}CE = \frac{4}{3}CE$.Since $\Delta Q > 0$,positive charge flows out of the positive terminal of the cell.

In the circuit shown,each capacitor has a capacitance $C$. The emf of the cell is $E$. If the switch $S$ is closed:

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