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

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 \varepsilon_{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(3 a ^{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(3 a ^{2} t \right)^{1 / 3}} $
or $V ^{\prime}= V \left(\frac{ a }{3 t }\right)^{1 / 3}$