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Q. Eight charged water drops, each with a radius of $1\, mm$ and a charge of $10^{-10} C$, coalesce to form a single drop. The potential of the big drop is

Electrostatic Potential and Capacitance

Solution:

Volume is conserved.
$\therefore \frac{4}{3} \pi r^{3}=8 \times \frac{4}{3} \pi r^{3}$
or $R^{3}=2^{3} r^{3}$ or $R=2 r$
$V=\frac{1}{4 \pi \varepsilon_{0}} \frac{8 q}{2 r}$
$=\frac{4 \times 9 \times 10^{9} \times 10^{-10}}{1 \times 10^{-3}}$ volt
$=3600$ volt