Electrostatics Part 2

Given, $n=27$ $V_{1}=10\ \mathrm{V}$ Let $q_{1}$ be the charge of one drop, $r_{1}$ be its radius, $r$ be the radius of bigger drop and $q$ be its charge. As, volume remains constant. $\;\text{Electric potential}\;(V)=\frac{3 k q_{1}^{2}}{5 r}$ $\therefore\;\frac{4}{3}\pi r^{3}\;=\frac{4}{3}\pi r_{1}^{3}\times 27$ $\Rightarrow\;r^{3}\;=27 r_{1}^{3}$ $\Rightarrow\;r\;=3 r_{1}$ where, $k$ is Coulomb constant. Therefore, potential energy of small drop, $U_{s}=\frac{3}{5}\frac{k q_{1}^{2}}{r_{1}}$ . . . (i) and potential energy of bigger drop, $U_{B}=\frac{3}{5}\frac{k(q_{1}^{2}\times 27)^{2}}{r}$ $U_{B}=\frac{3}{5}\frac{k(27)^{2} q_{1}^{2}}{3 r_{1}}$. . . (ii) On dividing Eq. (ii) by Eq. (i), we get $\frac{U_{B}}{U_{s}}=\frac{27\times 27}{3}=243$ $\;U_{B}=243 U_{s}$ Hence, potential energy of big drop is 243 times of small drop.