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Q.
If $n$ drops of potential $V$ merge, find new potential on the big drop.
Rajasthan PMTRajasthan PMT 2006Electrostatic Potential and Capacitance
Solution:
Volume of big drop $=n \times$ volume of a small drop
$\Rightarrow \frac{4}{3} \pi R^{3} =n \frac{4}{3} \pi r^{3} $
$\Rightarrow R =n^{1 / 3} r\ldots$ (i)
Potential of small drop $V=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}$
$\Rightarrow q=V \times 4 \pi \varepsilon_{0} r \ldots$(ii)
So, potential of big drop $=\frac{1}{4 \pi \varepsilon_{0}} \times \frac{n q}{R}$
$\Rightarrow V'=\frac{4 \pi \varepsilon_{0} \times V \times r}{4 \pi \varepsilon_{0} \times n^{1 / 3} r}=n^{2 / 3} V$