Question

An oil drop of $10$ excess electrons is held stationary under a constant electric field of $3.65\times 10^{4}N{C}^{-1}$ in Millikan's oil drop experiment. The density of oil is $1.26gc{m}^{-3}$. Radius of the oil drop is
(Take, $g=9.8m{s}^{-2}$, $e=1.6\times 10^{-19}C$)

• A. $1.04\times 10^{-6}m$
• B. $4.8\times 10^{-5}m$
• C. $4.8\times 10^{-18}m$
• D. $1.13\times 10^{-18}m$

Answer 1

Answer: (A)

Here, $n=10$, $E=3.65\times 10^{4}N{C}^{-1}$
${\rho }_{oil}=1.26gc{m}^{-3}=1.26\times 10^{3}kg{m}^{-3}$
As the droplet is stationary,
weight of droplet = force due to electric field
or $\frac{4}{3}\pi r^3\rho g=neE\therefore r^3=\frac{3neE}{4\pi \rho g}$
$\therefore r^3=\frac{3\times 10\times 1.6\times 10^{-19}\times 3.65\times 10^4}{4\times 3.14\times 1.26\times 10^3\times 9.8}$
$=1.13\times 10^{-18}$
or $r={\left(1.13\times 10^{-18}\right)}^{1/3}$
$=1.04\times 10^{-6}m$