$A$ parallel plate capacitor of capacitance $10 \mu F$ is charged by a $220 \text{ V}$ supply. The capacitor is then disconnected from the supply and is connected to another uncharged parallel plate capacitor of capacitance $12 \mu F$. The loss of electrostatic energy in this process is (in $\text{ mJ}$)

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 $.....CV^2$.

Circuits $(a)$ and $(b)$ have charged capacitors with capacitances $C, 2C$ and $3C$ as shown,with an open switch $S$. When the switch is closed,determine the direction of charge flow for each circuit.

$A$ parallel plate capacitor of capacitance $C$ is charged to a potential $V$ by a battery. Another capacitor of capacitance $2C$ is charged to a potential $2V$ by another battery. After removing the batteries,they are connected in parallel such that the positive plate of one is connected to the negative plate of the other. Calculate the final energy of the system.

$A$ $20 \text{ F}$ 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 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