Question

An oil drop of $n$ excess electrons is held stationary under a constant electric field $E$ in Millikan’s oil drop experiment. The density of oil is $\rho$ . The radius of the drop is

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

Answer 1

Answer: (C)

In Millikan’s oil drop experiment, the charged oil drop remains suspended (in equilibrium) when downward weight of drop is balanced by upward electrostatic force and charge on drop $q=ne$, i.e.
$qE=mg$
$\Rightarrow neE=mg$
If $r$ is radius of oil drop, then mass $m=\frac{4}{3}\pi r^3\rho$
$neE=\frac{4}{3}\pi r^3\rho g$
$r={\left[\frac{3neE}{4\pi \rho g}\right]}^{1/3}$