Using a battery,a $100 \ pF$ capacitor is charged to $60 \ V$ and then the battery is removed. After that,a second uncharged capacitor is connected to the first capacitor in parallel. If the final voltage across the second capacitor is $20 \ V$,its capacitance is: (in $pF$)

$A$ 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:

Energy of a charged capacitor is $U$. Now it is removed from a battery and then connected to two other identical uncharged capacitors in parallel. What will be the energy of each capacitor?

$A$ $20 \mu F$ capacitor is charged to $5 V$ and isolated. It is then connected in parallel with an uncharged $30 \mu F$ capacitor. The decrease in the energy of the system will be (in $\text{J}$)

The circuit shows two capacitors $A$ and $B$ of capacitances $C$ and $2C$ respectively. When they are fully charged,the cell is removed and the capacitors are connected with their plates of opposite polarities touching each other. Then:
$(a)$ Charge on $A$ is $\frac{4CE}{9}$
$(b)$ Charge on $B$ is $\frac{8CE}{9}$
$(c)$ Loss of energy in this process is $\frac{CE^2}{3}$
The correct statement$(s)$ is/are:

$A$ capacitor of capacitance $50 \; pF$ is charged by a $100 \; V$ source. It is then connected to another uncharged identical capacitor. The electrostatic energy loss in the process is $\dots \; nJ$.