Two capacitors of $2\ \mu F$ and $3\ \mu F$ are charged to $300\ V$ and $500\ V$ respectively. When they are connected in parallel, the energy lost is: (in $J$)

Two capacitors of capacitances $3 \mu F$ and $6 \mu F$ are charged to a potential of $12 V$ each. They are now connected to each other,with the positive plate of each joined to the negative plate of the other. The potential difference across each will be $volt$.

Two metallic charged spheres whose radii are $20\,cm$ and $10\,cm$ respectively,have each $150\,\mu C$ positive charge. The common potential after they are connected by a conducting wire 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 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

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

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