`N` drops of mercury of equal radii and possessing equal charges combine to from a big drop. Compare the charge, capacitance and potential of bigger drop with the corresponding quantities of individual drops.

Answer

`N ; N^(1//3) ; N^(2//3)`

Explanation

Let `r` be the radius of each small drop carrying charge `q` and potential `V`. When `N` such drops combine, there is no loss of charge.
`:.` Charge on bigger drop `= Q = Nq or (Q)/(q) = N`
As volume of big drop = Volume of `N` small drops,
`:. (4)/(3) pi R^(3) = N xx (4)/(3) pi r^(3) or R = N^(1//3) r`
As capacitance `prop` radius, therefore
`(C')/(C) = (R)/(r) = N^(1//3)`
`= N^(2//3) xx` pot. of small drop