`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)piR^(3)=nxx(4)/(3)pir^(3)`
`therefore" "r=n^(1//3)r`
New potentail, `V'=(nq)/(4pi epsilon_(0)r)=n^(2//3)V`