Two spherical conductors each of capacity $C$ are charged to potentials $V$ and $-V$. These are then connected by means of a fine wire. The loss of energy will be

$A$ 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 $CV^2$)

$A$ $2 \ \mu F$ capacitor is charged as shown in the figure. The percentage of its stored energy dissipated after the switch $S$ is turned to position $2$ is (in $\%$)

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 $\mu C$?

$A$ $4 \;\mu F$ capacitor is charged by a $200 \;V$ supply. It is then disconnected from the supply and connected to another uncharged $2 \;\mu F$ capacitor. How much electrostatic energy of the first capacitor is lost in the form of heat and electromagnetic radiation?

$A$ $400\, pF$ capacitor is charged with a $100\, V$ battery. After disconnecting the battery,this capacitor is connected to another uncharged $400\, pF$ capacitor. Find the energy loss.