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Q.
$n$ identical drops, each of capacitance $C$ and charged to a potential $V$, coalesce to form a bigger drop. Then the ratio of the energy stored in the big drop to that in each small drop is
KEAMKEAM 2010Electrostatic Potential and Capacitance
Solution:
Volume of big drop $ =n\times $ volume of small drop
$ \frac{4}{3}\pi {{R}^{3}}=n\times \frac{4}{3}\pi {{r}^{3}} $
$ R={{n}^{1/3}}r $
Capacitance of small drop, $ C=4\pi {{\varepsilon }_{0}}r $
Capacitance of big drop, $ C=4\pi {{\varepsilon }_{0}}R $
$ =4\pi {{\varepsilon }_{0}}{{n}^{1/3}}r $
$ C={{n}^{1/3}}C $
The potential of small drop $ V=\frac{q}{C}=\frac{q}{4\pi {{\varepsilon }_{0}}r} $
The potential of big drop $ V=\frac{nq}{(4\pi {{\varepsilon }_{0}}){{n}^{1/3}}r} $
$ V={{n}^{2/3}}V $
$ \therefore $ Energy of small drop $ =\frac{1}{2}C{{V}^{2}} $
Energy of big drop $ =\frac{1}{2}CV{{}^{2}} $
$ =\frac{1}{2}{{n}^{1/3}}C{{({{n}^{2/3}}V)}^{2}} $
$ ={{n}^{5/3}}\frac{1}{2}C{{V}^{2}} $
$ \therefore $ $ \frac{Energ{{y}_{(big\,drop)}}}{Energ{{y}_{(small\,drop)}}}=\frac{{{n}^{5/3}}}{1} $