$A$ capacitor of capacity $C_1$ is charged up to $V$ volt and then connected to an uncharged capacitor of capacity $C_2$. The final potential difference across each will be:

$A$ capacitor of capacitance $C_1 = 1 \ \mu F$ is charged using a $9 \ V$ battery. $C_1$ is then removed from the battery and connected to capacitors $C_2$ and $C_3$ of $2 \ \mu F$ and $3 \ \mu F$ respectively,as shown in the figure. Find the charge on $C_3$ after equilibrium is reached.

Energy $E$ is stored in a parallel plate capacitor $C_1$. An identical uncharged capacitor $C_2$ is connected to it,kept in contact with it for a while,and then disconnected. The energy stored in $C_2$ is

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and charged to a potential difference $V$. Another capacitor of capacitance $2C$ is similarly charged to a potential difference $2V$. The charging batteries are then removed,and the capacitors are connected in parallel such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is:

Two capacitors having capacitance $C_{1}$ and $C_{2}$ respectively are connected as shown in the figure. Initially,capacitor $C_{1}$ is charged to a potential difference $V$ volt by a battery. The battery is then removed and the charged capacitor $C_{1}$ is now connected to uncharged capacitor $C_{2}$ by closing the switch $S$. The amount of charge on the capacitor $C_{2}$ after equilibrium is

$A$ capacitor of capacitance $0.2 \, \mu F$ is charged to a potential of $600 \, V$. After removing the battery,it is connected in parallel to an uncharged capacitor of $1.0 \, \mu F$. The final potential across the capacitors will be.........$V$.