Question

A conducting spherical bubble of radius $a$ and thickness $t$$(t<<a)$ is charged to a potential $V$. Now it collapses into a spherical droplet. Find the potential of the droplet.

• A. ${\left(\frac{a}{3t}\right)}^{\frac{1}{3}}V$
• B. $\left(\frac{a}{t}\right)V$
• C. ${\left(\frac{2a}{t}\right)}^{\frac{1}{2}}V$
• D. $V$

Answer 1

The correct option is A ${\left(\frac{a}{3t}\right)}^{\frac{1}{3}}V$

Let, $\rho$ be the density of conducting bubble and $r$ be the radius of the resulting drop.

As mass and charge of the bubble remain conserved, so

$\text{mass of the droplet=mass of the bubble}$

$\Rightarrow \left(\frac{4}{3}\pi r^3\right)\rho =\left(4\pi a^{2}t\right)\rho$
$\Rightarrow r=(3a^{2}t{)}^{\frac{1}{3}}$

If $q$ is the charge of the bubble, then the potential of the bubble

$V=\frac{1}{4\pi {\epsilon }_0}\frac{q}{a}$,

$\Rightarrow q=4\pi {\epsilon }_{0}aV$

Now, the potential of the resulting drop

$V^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{q}{r}$

Substitutimg the value of $q$ and $r$,

$\Rightarrow V^{\prime }=\frac{1}{4\pi {\epsilon }_0}\frac{4\pi {\epsilon }_{0}aV}{(3a^{2}t{)}^{\frac{1}{3}}}$

$\Rightarrow V^{\prime }={\left(\frac{a}{3t}\right)}^{\frac{1}{3}}V$

Hence, option (a) is the correct answer.