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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When $S$ is open, the all capacitors will charge.
$C_{e q}=\frac{C(C+C)}{C+(C+C)}=\frac{2}{3} C$
and $Q_{e q}=C_{e q} E=\frac{2}{3} C E$
When S is closed, the capacitor $C$ on the right will not charge and positive charge will flow out of the positive terminal of cell.
now $C_{e q}^{\prime}=C+C=2 C$ and $Q_{e q}^{\prime}=C_{e q}^{\prime} E=2 C E$
The charge flow through the cell $=Q_{e q}^{\prime}-Q_{e q}=2 C E-\frac{2}{3} C E=\frac{4}{3} C E$
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