' n ' identical drops, each of capacitance c and charged to potential V, coalesce to from a bigger d

Question

' n ' identical drops, each of capacitance c and charged to potential V, coalesce to from a bigger drop. Then the ratio of the energy stored in the big drop to the each small drop is\

Professor PC Thomas Classes · Answer

Let $r$ be radius of each small drop and $R$ us of bigger drop\ As the volume remains constant\ $	herefore \frac{4}{3} \pi \mathrm{R}^{3}=\mathrm{n} 	imes \frac{4}{3} \pi \mathrm{r}^{3}$\ $\mathrm{R}=\mathrm{n}^{1 / 3} \mathrm{r}$\ Capacitance of each small drop, $\mathrm{C}=4 \pi \varepsilon_{0} \mathrm{r}$\ Capacitance of bigger drop, $\mathrm{C}^{\prime}=4 \pi \varepsilon_{0} \mathrm{R}$\ $=4 \pi \varepsilon_{0} n^{1 / 3} r=n^{1 / 3} C$\ Charge on each small drop, $\mathrm{Q}=\mathrm{CV}$\ Charge on bigger drop, $\mathrm{Q}^{\prime}=\mathrm{nQ}$\ Potential of bigger drop, $\mathrm{V}^{\prime}=\frac{\mathrm{Q}^{\prime}}{\mathrm{C}^{\prime}}=\frac{\mathrm{nQ}}{\mathrm{n}^{1 / 3} \mathrm{C}}=\mathrm{n}^{2 / 3} \mathrm{~V}$\ Energy stored in each small droop. $=\frac{1}{2} \mathrm{CV}^{2}$\ $=\frac{1}{2} \mathrm{n}^{1 / 3} \mathrm{C}\left(\mathrm{n}^{2 / 3} \mathrm{~V}ight)^{2}=\frac{1}{2} \mathrm{n}^{5 / 3} \mathrm{CV}^{2}$\ $	herefore \frac{\mathrm{U}^{\prime}}{\mathrm{U}}=\frac{1}{2} \mathrm{n}^{5 / 3} \mathrm{CV}^{2} 	imes \frac{2}{\mathrm{CV}^{2}}$\