$n$ small drops of the same size are charged to $V$ volts each. If they coalesce to form a single large drop,then its potential will be:

Three capacitors $A, B$ and $C$ are connected to a battery of $25 \text{ V}$ as shown in the figure. The ratio of charges on capacitors $A, B$ and $C$ will be $-$

Consider two charged metallic spheres $S_{1}$ and $S_{2}$ of radii $R_{1}$ and $R_{2},$ respectively. The electric fields $E_{1}$ (on $S_{1}$) and $E_{2}$ (on $S_{2}$) on their surfaces are such that $E_{1} / E_{2} = R_{1} / R_{2}.$ Then the ratio $V_{1} / V_{2}$ of the electrostatic potentials on each sphere is:

$64$ identical mercury drops,each charged to a potential of $10\, V$,are combined to form a single larger drop. The potential of the new larger drop is........$V$.

$n$ small drops, each of capacitance $C$, coalesce to form a single large drop. What is the ratio of the energy stored in the large drop to the energy stored in each small drop?

$A$ parallel plate capacitor of capacitance $C$ is connected to a battery and is charged to a potential difference $V$. Another capacitor of capacitance $2C$ is connected to another battery and is charged to a potential difference $2V$. The charging batteries are now disconnected and the capacitors are connected in parallel to each other in such a way that the positive terminal of one is connected to the negative terminal of the other. The final energy of the configuration is