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 the capacitors will be:

Two isolated conducting spheres $S_{1}$ and $S_{2}$ of radius $\frac{2}{3} R$ and $\frac{1}{3} R$ have $12\, \mu C$ and $-3\, \mu C$ charges,respectively,and are at a large distance from each other. They are now connected by a conducting wire. $A$ long time after this is done,the charges on $S_{1}$ and $S_{2}$ are respectively:

In the circuit shown in the figure,initially $K_1$ is closed and $K_2$ is open. What are the charges on each capacitor? Then $K_1$ was opened and $K_2$ was closed (order is important),what will be the charge on each capacitor now? [Given: $C = 1 \,\mu F$,$C_1 = 6C$,$C_2 = 3C$,$C_3 = 3C$,$E = 9 \, V$]

$A$ capacitor of capacitance $C$ charged by an amount $Q$ is connected in parallel with an uncharged capacitor of capacitance $2C$. The final charges on the capacitors are

Consider the situation shown in the figure. Capacitor $A$ has a charge $q$ on it,whereas $B$ is uncharged. The charge appearing on capacitor $B$ a long time after the switch $S$ is closed is

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential difference $V$ by connecting it to a battery. Another capacitor of capacitance $2C$ is charged to a potential difference $2V$ by connecting it to another battery. Now,the charging batteries are removed and the capacitors are connected in parallel such that the positive plate of one is connected to the negative plate of the other and the negative plate of the first is connected to the positive plate of the second. Find the final energy of this configuration.