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Q. $64$ small drops of mercury, each of radius $r$ and charge $q$ coalesce to form a big drop. The ratio of the surface density of charge of each small drop with that of the big drop is

KCETKCET 2002Electrostatic Potential and Capacitance

Solution:

Here $R^3 = (4r)^3 \, \, i.e. \, R = 4r$ $\frac{\sigma_1}{\sigma_2} = \frac{r^2}{4r^2} = \frac{1}{4}$