If n drops of equal size and same potential V merge into a big drop, the new potential of the big drop will be

Options

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

Answer

D

Explanation

`(4)/(3)piR^(3)=n(4)/(3)pir^(3)" "R=n^(1//3)r`
Potential of the small drop`=(1)/(4piepsi_(0))(q)/(r)`
Potential of the big drop`=(1)/(4piepsi_(0))(nq)/(R)`
`therefore V'=(1)/(4piepsi_(0))(q)/(r)[(n)/(n^(1//3))]=n^(2//3)V`