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 $C_{1}$ is first charged to a potential difference of $10\, V$ using a battery. Then the battery is removed and the capacitor is connected to an uncharged capacitor $C_{2}$ of $8\, \mu F$. The charge in $C_{2}$ on equilibrium condition is $\ldots\, \mu C$. (Round off to the Nearest Integer)

Two capacitors of capacitance $2 \ \mu F$ and $5 \ \mu F$ are charged to $2 \ V$ and $10 \ V$ respectively. Find the ratio of their charges after they are connected by a wire.

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

$A$ capacitor of capacitance $C_1 = 1 \ \mu F$ is charged using a $9 \ V$ battery. $C_1$ is then removed from the battery and connected to capacitors $C_2$ and $C_3$ of $2 \ \mu F$ and $3 \ \mu F$ respectively,as shown in the figure. Find the charge on $C_3$ after equilibrium is reached.