One of the two identical capacitors having the same capacitance $C$ is charged to a potential $V_1$ and the other is charged to a potential $V_2$. If they are connected with their like plates together,then the decrease in the electrostatic potential energy of the combined system is

$A$ parallel plate capacitor of capacitance '$C$' is connected to a battery and charged to a potential '$V$'. Another capacitor of capacitance '$3C$' is charged to a potential '$3V$'. The charging battery is then disconnected and both the capacitors are connected in parallel to each other such that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is (in $CV^2$)

Two capacitors of capacitances $C$ and $2C$ are charged to potential differences $V$ and $2V$,respectively. These are then connected in parallel in such a manner that the positive terminal of one is connected to the negative terminal of the other. The final energy of this configuration is $.....CV^2$.

Two capacitors of capacitance $3 \ \mu F$ and $6 \ \mu F$ are charged to a potential of $12 \ V$. They are then connected by joining their positive plates together and their negative plates together. What will be the potential difference across each capacitor in $V$?

An uncharged parallel plate capacitor having a dielectric of constant $K$ is connected to a similar air-cored parallel capacitor charged to a potential $V$. The two share the charge and the common potential is $V'$. The dielectric constant $K$ is

$A$ capacitor of $2\,\mu F$ is charged as shown in the diagram. When the switch $S$ is turned to position $2,$ the percentage of its stored energy dissipated is ......$\%$