$A$ $4 \mu F$ capacitor is charged by a $200 \ V$ battery. It is then disconnected from the supply and connected to another uncharged $2 \mu F$ capacitor. During the process,the loss of energy (in $J$) is:

$A$ capacitor of capacitance $6\ \mu F$ is charged to $100\, V$. It is then connected to another uncharged capacitor of capacitance $14\ \mu F$. Find the ratio of charges on the $6\ \mu F$ and $14\ \mu F$ capacitors and the potential across the $6\ \mu F$ capacitor.

Two identical capacitors have same capacitance $C$. One of them is charged to the potential $V$ and other to the potential $2V$. The negative ends of both are connected together. When the positive ends are also joined together,the decrease in energy of the combined system is:

$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a capacitor of capacity $C_2$ as shown in the figure. The ratio of the total energy before and after the connection of switch $S$ 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 $C$ is charged to a potential $V$. It is then disconnected and connected in parallel with an identical uncharged capacitor. Find the charge on each capacitor.