In the circuit shown,the steady state voltage across capacitor $C$ is a fraction of the battery e.m.f. $V$. The fraction is decided by:

$A$ 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 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 given circuit,what are the electric potentials at points $A$ and $B$?

$A$ parallel plate capacitor of capacity $100 \mu F$ is charged by a battery of $50 \text{ V}$. The battery remains connected and if the plates of the capacitor are separated so that the distance between them becomes double the original distance,the additional energy given by the battery to the capacitor in joules is

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

$512$ identical drops of mercury are charged to a potential of $2 \ V$ each. The drops are joined to form a single drop. The potential of this drop is ......... $V.$