Question

A conducting bubble of radius $a$, thickness $t(t<<a)$ has potential V. Now the bubble collapses into a droplet. Find the potential of the droplet.

• A. $V\left(\frac{a}{3t}\right)$
• B. $V(\frac{a}{3t}{)}^{1/2}$
• C. $V(\frac{a}{3t}{)}^{1/3}$
• D. None of these

Answer 1

Answer: (C)

$V(\frac{a}{3t}{)}^{1/3}$

Potential at the surface of bubble , $V=\frac{kq}{a}$ or $q=\frac{Va}{k}$

Since bubble collapses into droplet of radius R. So $\left(4\pi a^2\right)t=\frac{4}{3}\pi R^3$

or $R=(3a^{2}t{)}^{1/3}$

Now the potential of the droplet is $V^{\prime }=\frac{kq}{R}=\frac{Va}{(3a^{2}t{)}^{1/3}}=V{\left(\frac{a}{3t}\right)}^{1/3}$