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 $1.26\, g\, cm^{-3}$. Radius of the oil drop is
(Take, $g= 9.8 \,ms^{-2}$, $e = 1.6 \times 10^{-19}\, C$)

• 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, $n = 10$, $E = 3.65 \times 10^4\, N\, C^{-1}$
$\rho_{oil}= 1.26\, g \,cm^{-3} = 1.26 \times 10^3\, kg \,m^{-3}$
As the droplet is stationary,
weight of droplet = force due to electric field
or $\frac{4}{3}\pi\,r^{3}\,\rho g=neE\, \therefore r^{3}=\frac{3neE}{4\pi\rho g}$
$\therefore r^{3}=\frac{3\times10\times1.6\times10^{-19}\times3.65\times10^{4}}{4\times3.14\times1.26\times10^{3}\times9.8}$
$=1.13\times10^{-18}$
or $r=\left(1.13\times10^{-18}\right)^{1/3}$
$=1.04\times10^{-6}\,m$