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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In steady state, potential across $C_{1}=$ potential across $C_{2}$

i.e, $\frac{Q_{1}}{C_{1}}=\frac{Q_{2}}{C_{2}}=\frac{Q_{2}}{2 C_{1}} \Rightarrow Q_{2}=2 Q_{1}$

Also total charge, $Q=Q_{1}+Q_{2}=Q_{1}+2 Q_{1}=3 Q_{1}$

$\therefore Q_{1}=\frac{Q}{3}, Q_{2}=\frac{2 Q}{3}$

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