Question

A conducting liquid bubble of inner radius $a$ and thickness $t$ where $a$ = $81\text{cm}$ and $t$ = $0.001\text{cm}$ is charged to potential $\text{V}$. If the bubble collapses to a droplet, find the potential on the droplet in terms of $\text{V}$.

• A. $10\text{V}$
• B. $20\text{V}$
• C. $30\text{V}$
• D. $40\text{V}$

Answer 1

The correct option is C $30\text{V}$

Given that,
Inner radius of bubble, $a=81\text{cm}$
Thickness of the bubble, $t=0.001\text{cm}$
Electric potential due to charged bubble = $V$


The electric potential due charged bubble is given by
$V=\frac{kq}{a}$
$\Rightarrow q=\frac{aV}{k}$

If the bubble collapses to droplet the volume remains the same.

Volume of bubble = $\frac{4}{3}\pi \{{(a+t)}^3-a^3\}$
Let the radius of the droplet be $r$
$\Rightarrow \frac{4}{3}\pi \{{(a+t)}^3-a^3\}=\frac{4}{3}\pi r^3$
$\Rightarrow {(a+t)}^3-a^3=r^3$
$\Rightarrow t^3+3a{t}^2+3a^{2}t=r^3$

As $t$ is very small we can neglect its higher powers.
So $r^3=3a^{2}t$
$r={\left(3a^{2}t\right)}^{\frac{1}{3}}$

The electric potential on the droplet is
$V^{\prime }=\frac{kq}{r}$
$\Rightarrow V^{\prime }=k\times \frac{aV}{k}\times \frac{1}{{\left(3a^{2}t\right)}^{\frac{1}{3}}}$
$\Rightarrow V^{\prime }=\frac{a}{a^{\frac{2}{3}}{\left(3t\right)}^{\frac{1}{3}}}V$
$\Rightarrow V^{\prime }={\left(\frac{a}{3t}\right)}^{\frac{1}{3}}V$
$\Rightarrow V^{\prime }={\left(\frac{81}{3\times 0.001}\right)}^{\frac{1}{3}}V$
$\Rightarrow V^{\prime }={(27\times {10}^3)}^{\frac{1}{3}}$
$\Rightarrow V^{\prime }=30\text{V}$

Hence, option (c) is correct.