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Q.
$N$ identical spherical drops are charged to the same potential $V$. They combine to form a bigger drop. The potential of the big drop will be:
Jharkhand CECEJharkhand CECE 2002
Solution:
Volume of $N$ small drops is equal to volume of single big drop.
Volume of $N$ small drops = volume of one big drop i.e.,
$\left(\frac{4}{3} \pi r^{3}\right) N=\frac{4}{3} \pi R^{3}$
$\Rightarrow R=N^{1 / 3} r$
Also electric potential on big drop is
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{N q}{R}$
$=\frac{1}{4 \pi \varepsilon_{0}} \frac{N q}{N^{1 / 3} r} $
$\Rightarrow V=N^{2 / 3} V $
$\left(a s V=\frac{1}{4 \pi \varepsilon_{0}} \frac{q}{r}\right)$