Statement $(A)$: Inside a charged hollow metal sphere,$E = 0$ and $V \neq 0$. ($E$ = electric field,$V$ = electric potential).
Statement $(B)$: The work done in moving a positive charge on an equipotential surface is zero.
Statement $(C)$: When two like charges are brought closer,their mutual electrostatic potential energy will increase.

Find $V_A - V_B$ for the following given circuit in volts.

$A$ capacitor,$C_1 = 6 \ \mu F$ is charged to a potential difference of $V_0 = 5 \ V$ using a $5 \ V$ battery. The battery is removed and another capacitor,$C_2 = 12 \ \mu F$ is inserted in place of the battery. When the switch $S$ is closed,the charge flows between the capacitors for some time until an equilibrium condition is reached. What are the charges ($q_1$ and $q_2$) on the capacitors $C_1$ and $C_2$ when the equilibrium condition is reached?

In the circuit shown in the figure,the total charge is $750\, \mu C$ and the voltage across capacitor $C_{2}$ is $20\, V$. If $C_{3} = 8\, \mu F$,then the charge on capacitor $C_{2}$ is $....\mu C$.

The resultant capacitance between $A$ and $B$ in the following figure is equal to.....$\mu F$

Three capacitors $C_1, C_2$ and $C_3$ are connected as shown in the figure to a battery of $V$ volt. If the capacitor $C_3$ breaks down electrically,the change in total charge on the combination of capacitors is