Question

$27$ drops of the same size are charged at $220V$ each. They combine to form a bigger drop. Calculate the potential of the bigger drop.

Answer 1

The radius of the bigger drop will be

$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$

$R=3r$

the potential of the small drop

= $\frac{KQ}{r}$ =$v$

The potential of the big drop

$Q=nq=27q$

$V^{\prime }=\frac{KQ}{R}$ = $\frac{K\times 27q}{3r}$

= $9v$

$9\times 220$

= $1980V$