Two capacitors $C_{1}$ and $C_{2}$ are joined in series where $C_{2} = 4 C_{1}$ . They are connected in a circuit with a switch between them as shown in figure. When the switch is open, net charge on $C_{1}$ is $2 Q$
Now, the switch is closed. At steady state, the charge on each capacitor will be

2 Q, 4 Q

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2 Q, 8 Q

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\frac{2 Q}{5}, \frac{4 Q}{5}

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\frac{2 Q}{5}, \frac{8 Q}{5}

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Solution

The correct option is D $\frac{2 Q}{5}, \frac{8 Q}{5}$
In steady state cahrge on $C_{1}$ is
$Q_{1} = (\frac{C_{1}}{C_{1} + C_{2}}) \times 2 Q = \frac{2 Q}{10 Q} \times 2 Q = \frac{2}{5} Q$
Steady state cahrge on $C_{2}$ is
$Q_{2} = (\frac{C_{2}}{C_{1} + C_{2}}) \times 2 Q = \frac{8 Q}{10 Q} \times 2 Q = \frac{8}{5} Q$

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