$A$ condenser of capacity $C_1$ is charged to a potential $V_0$. The electrostatic energy stored in it is $U_0$. It is connected to another uncharged condenser of capacity $C_2$ in parallel. The energy dissipated in the process is

$A$ $10 \ \mu F$ capacitor is charged to a potential difference of $50 \ V$ and connected in parallel to another uncharged capacitor. The common potential difference becomes $20 \ V$. The capacitance of the second capacitor is ........ $\mu F$.

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 $C_{1}$ of capacitance $5\,\mu F$ is charged to a potential of $30\,V$ using a battery. The battery is then removed and the charged capacitor is connected to an uncharged capacitor $C_{2}$ of capacitance $10\,\mu F$ as shown in the figure. When the switch is closed,charge flows between the capacitors. At equilibrium,the charge on the capacitor $C_{2}$ is . . . . . . $\mu C$.

Two circuits $(a)$ and $(b)$ have charged capacitors with capacitances and charges as shown in the figures. The switches are initially open. On closing the switches, what happens to the charge flow?

$A$ capacitor of $10\,\mu F$ charged up to $250\,V$ is connected in parallel with another capacitor of $5\,\mu F$ charged up to $100\,V$. The common potential is.....$V$.