Eight drops of mercury, each of the same radius and having the same charge, coalesce to form a single larger drop. The capacitance of the larger drop compared to each individual smaller drop is how many times greater?

$A$ cylindrical capacitor has inner and outer conductors whose radii are in the ratio $10:1$. The inner conductor is replaced by a wire whose radius is half of the original inner conductor. To obtain the same capacitance as the original capacitor,by what ratio should the length of the wire be increased?

Two spherical conductors $A$ and $B$ of radii $a$ and $b$ $(b > a)$ are placed concentrically in air. The two are connected by a copper wire as shown in the figure. Then the equivalent capacitance of the system is

Consider a spherical drop of mercury of radius $R$ with capacitance $C = 4 \pi \epsilon_0 R$. If two such droplets combine to form a larger one,what would be its capacitance in terms of $C$?

The Earth is assumed to be a charged conducting sphere having volume $V$ and surface area $A$. The capacitance of the Earth in free space is ($\epsilon_{0} =$ permittivity of free space).

Eight drops of mercury of equal radii combine to form a big drop. The capacitance of a bigger drop as compared to each smaller drop is (in $\times$)