Question

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

• A. $V_{droplet}=V{\left[\frac{a}{3t}\right]}^{\frac{1}{3}}$
• B. $V_{droplet}=V{\left[\frac{a}{t}\right]}^{\frac{1}{3}}$
• C. $V_{droplet}=V{\left[\frac{3a}{t}\right]}^{\frac{1}{3}}$
• D. $V_{droplet}=V{\left[\frac{3a}{4t}\right]}^{\frac{1}{3}}$

Answer 1

Answer: (A)

Let $q$ be the charge on the bubble, then
$\begin{array}{l}V=\frac{Kq}{a}\left(\text{ Here }K=\frac{1}{4\pi {\epsilon }_0}\right)\\ \therefore q=\frac{Va}{K}\end{array}$
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
$V^{\prime }=\frac{Kq}{R}$
Substituting the values, we have
$V^{\prime }=\frac{(K)\left(\frac{Va}{K}\right)}{{\left(3a^{2}t\right)}^{1/3}}$
or $V^{\prime }=V{\left(\frac{a}{3t}\right)}^{1/3}$