Question

$n$ small drops of same size are charged to V volts each. If they coalesce to form a signal large drop, then its potential will be :

• A. $V/n$
• B. $Vn$
• C. $V{n}^{1/3}$
• D. $V{n}^{2/3}$

Answer 1

The correct option is D $V{n}^{2/3}$

Let potential of $1$ drop = $\frac{Kq}{r}=V$

Volume of $n$ drops = volume of bigger drop

i.e., $n\frac{4}{3}\pi r^3=\frac{4}{3}\pi R^3$

Therefore, $R=n^{\frac{1}{3}}r$

Also charge remains conserved

Charge on $1$ drop = $q$

Charge on big drop containing n drops = $nq$

Therefore potential on big drop = $K\frac{nq}{n^{\frac{1}{3}}r}$= $n^{\frac{2}{3}}\frac{Kq}{r}$ = $n^{\frac{2}{3}}V$