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. $n^3:1$
• E. $n^{2/3}:1$

Answer 1

Answer: (A)

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, $C=4\pi {\epsilon }_{0}r$
Capacitance of big drop, $C=4\pi {\epsilon }_{0}R$
$=4\pi {\epsilon }_0{n}^{1/3}r$
$C=n^{1/3}C$
The potential of small drop $V=\frac{q}{C}=\frac{q}{4\pi {\epsilon }_{0}r}$
The potential of big drop $V=\frac{nq}{(4\pi {\epsilon }_0)n^{1/3}r}$
$V=n^{2/3}V$
$\therefore$ Energy of small drop $=\frac{1}{2}C{V}^2$
Energy of big drop $=\frac{1}{2}CV{}^2$
$=\frac{1}{2}{n}^{1/3}C{(n^{2/3}V)}^2$
$=n^{5/3}\frac{1}{2}C{V}^2$
$\therefore$ $\frac{Energ{y}_{(bigdrop)}}{Energ{y}_{(smalldrop)}}=\frac{n^{5/3}}{1}$