Question

There are $n$ electrons of charge $e$ on a spherical drop of oil of density $\rho$. It is in equilibrium in air in the electric filed $E$. Then radius of drop is

• A. ${\left[\frac{2neE}{\pi \rho g}\right]}^{\frac{1}{3}}$
• B. ${\left[\frac{2neE}{4\pi \rho g}\right]}^{\frac{1}{2}}$
• C. ${\left[\frac{neE}{\rho g}\right]}^{\frac{1}{2}}$
• D. ${\left[\frac{3neE}{4\pi \rho g}\right]}^{\frac{1}{3}}$

Answer 1

The correct option is D ${\left[\frac{3neE}{4\pi \rho g}\right]}^{\frac{1}{3}}$


The spherical drop can be in equilibrium, when electrostatic force balancing its weight,

$W=F$

$mg=Eq[\because q=ne]$

$\frac{4}{3}\pi r^3\rho g=Ene$

$r^3=\frac{3neE}{4\pi \rho g}$

<!--td {border: 1px solid #ccc;}br {mso-data-placement:same-cell;}--> Hence, (d) is the correct answer.
$r={\left[\frac{3neE}{4\pi \rho g}\right]}^{\frac{1}{3}}$