`n` small drops of same size are charged to `V` volts each .If they coalesce to from a single large drop, then its potential will be -

Options

A. Vn
B. `Vn^(-1)`
C. `Vn^(1//3)`
D. `Vn^(2//3)`

Answer

D

Explanation

As `(4)/(3) pi R^(3) = n xx (4)/(3) pi r^(3) :. R = n^(1//3)r`.
New potential `V. = (nq)/(4pi epsilon_(0)R) = (nq)/(4pi epsilon_(0)(n^(1//3)r)) = n^(2//3) (q)/(4pi epsilon_(0)r) = n^(2//3) V`