A conducting soap bubble having a radius $a$ , thickness $t (t<<a)$ is charged to a potential $V$. If now, the bubble collapses and forms droplet, then the potential of the droplet is

Let q be the charge on the bubble, then  $V=\frac{Kq}{a}(HereK=\frac{1}{4\pi {\epsilon }_0})$ $\therefore q=\frac{Va}{K}$
Let after collapsing, the radius of droplet becomes R, then equating the volume, we have $\left(4\pi a^2\right)t=\frac{4}{3}\pi R^3$ $\therefore R={\left(3a^{2}t\right)}^{1/3}$ 

Now, potential of droplet will be
$\mathrm{V}\text{'}=\frac{kq}{R}$

Substituting the values, we have

 $\mathrm{V}\text{'}=\frac{\left(K\right)\left(\frac{Va}{K}\right)}{{\left(3a^{2}t\right)}^{1/3}}$   or  $\mathrm{V}\text{'}=\mathrm{V}{\left(\frac{a}{3t}\right)}^{1/3}$