$N$ identical spherical drops,each charged to the same potential $V$,are combined to form a single big drop. What will be the potential of the new big drop?

Consider two identical metallic spheres of radius $R$ each having charge $Q$ and mass $m$. Their centers have an initial separation of $4 R$. Both the spheres are given an initial speed of $u$ towards each other. The minimum value of $u$, so that they can just touch each other is: (Take $k=\frac{1}{4 \pi \epsilon_0}$ and assume $k Q^2 > G m^2$ where $G$ is the Gravitational constant)

$A$ 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

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

Two capacitors $C_1 = 2\,\mu F$ and $C_2 = 6\,\mu F$ in series are connected in parallel to a third capacitor $C_3 = 4\,\mu F$. This arrangement is then connected to a battery of $e.m.f. = 2\,V$,as shown in the figure. How much energy is lost by the battery in charging the capacitors?

$A$ capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. $A$ charge $+Q$ is now given to its positive plate. The potential difference across the capacitor is now: