$A$ 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.........$V$.

$A$ capacitor of capacity $4 \, \mu F$ is charged to $80 \, V$ and another capacitor of capacity $6 \, \mu F$ is charged to $30 \, V$. When they are connected, the energy lost by the $4 \, \mu F$ capacitor is: (in $ \, mJ$)

In the circuit shown here $C_1 = 6\,\mu F$,$C_2 = 3\,\mu F$ and battery $B = 20\,V$. The switch $S_1$ is first closed. It is then opened and afterwards $S_2$ is closed. What is the charge finally on $C_2$ in $\mu C$?

$A$ 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.

Two identical capacitors $C_{1}$ and $C_{2}$ of equal capacitance are connected as shown in the circuit. Terminals $a$ and $b$ of the key $k$ are connected to charge capacitor $C_{1}$ using a battery of $emf$ $V$ volt. Now,disconnecting $a$ and $b$,the terminals $b$ and $c$ are connected. Due to this,what will be the percentage loss of energy?

$A$ capacitor of capacity $C_1$ is charged to a potential of $V_0$. After disconnecting it from the battery,it is connected to a neutral capacitor of capacity $C_2$ as shown in the adjoining figure. The ratio of the energy of the system before and after the connection of switch $S$ will be