`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. `(V)/(n)`
B. `V n`
C. `V n^(1//3)`
D. `V n^(2//3)`

Answer

4

Explanation

`V = (1)/(4 pi in_(0)) .(q)/(r )`
Since volume remains same
`n . (4)/(3) pi r^(3) = (4)/(3) pi R^(3) rArr R = n^(1//3) r`

`V' = (1)/(4 pi in_(0)) (nq)/( R) = (1)/( 4 pi in_(0)) (nq)/(n^(1//3) r) = n^(2//3) V`