Question

$n$ 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

• A. ${{n}^{5/3}}:1$
• B. ${{n}^{4/3}}:1$
• C. $n:1$
• D. ${{n}^{3}}:1$

Answer 1

Answer: (A)

${{n}^{5/3}}:1$

Answer 2

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 $=\frac{1}{2}CV{{}^{2}}$ $=\frac{1}{2}{{n}^{1/3}}C{{({{n}^{2/3}}V)}^{2}}$ $={{n}^{5/3}}\frac{1}{2}C{{V}^{2}}$ $\therefore$ $\frac{Energ{{y}_{(big\,drop)}}}{Energ{{y}_{(small\,drop)}}}=\frac{{{n}^{5/3}}}{1}$