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{3ne E}{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}$