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 $1.26g{cm}^{-3}$. 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

The correct option is A $1.04\times {10}^{-6}m$

Here $n=10,E=3.65\times {10}^{4}N{C}^{-1},{\rho }_{oil}=1.26g{cm}^{-3}=1.26\times {10}^{3}kg{m}^{-3}$

As the droplet is stationary, wheight 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 $r={(1.13\times {10}^{-18})}^{1/3}=1.04\times {10}^{-6}m$