Two conducting shells of radius $a$ and $b$ are connected by a conducting wire as shown in the figure. The larger shell is grounded. The capacitance of the system is:

$A$ parallel plate capacitor of capacity $C_0$ is charged to a potential $V_0$. $(i)$ The energy stored in the capacitor when the battery is disconnected and the plate separation is doubled is $E_1$. (ii) The energy stored in the capacitor when the charging battery is kept connected and the separation between the capacitor plates is doubled is $E_2$. Then,the value of $E_1 / E_2$ is:

In the circuit shown in the figure,four capacitors are connected to a battery. $A$ parallel plate capacitor has an electric field of $10^5 \ V/m$ between its plates. If the charge on the capacitor plate is $1 \ \mu C$,then the force on each capacitor plate is.....$N$.

Two capacitors of capacitances $1 \ \mu F$ and $2 \ \mu F$ are charged to potential differences $20 \ V$ and $15 \ V$ as shown in the figure. If terminals $B$ and $C$ are connected together and terminals $A$ and $D$ are connected together,then find the final charge on the $1 \ \mu F$ capacitor.

For the given circuit,calculate the charge on each capacitor in steady state in $\mu C$.

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.