Two capacitors $C_1$ and $C_2 = 2C_1$ are connected in a circuit with a switch between them as shown in the figure. Initially the switch is open and $C_1$ holds charge $Q$. The switch is closed. At steady state, the charge on each capacitor will be
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(b) In steady state charge on $C_1$ is ${Q_1} = \left( {\frac{{{C_1}}}{{{C_1} + {C_2}}}} \right) \times Q = \frac{Q}{3}$
and charge on $C_2$ is ${Q_2} = \left( {\frac{{{C_2}}}{{{C_1} + {C_2}}}} \right).Q = \frac{2}{3}Q$
and charge on $C_2$ is ${Q_2} = \left( {\frac{{{C_2}}}{{{C_1} + {C_2}}}} \right).Q = \frac{2}{3}Q$
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