Question

An oil drop of $12$ excess electrons is held stationary under a constant electric field of $2.55\times {10}^{4}V{m}^{-1}$ in Millikan's oil drop experiment. The density of the oil is $1.26gc{m}^{-3}$. Estimate the radius of the drop. Given that $g=9.81m{s}^{-2}$ and $e=1.6\times {10}^{-19}C$

• A. $6382\times {10}^{-4}mm$
• B. $7.82\times {10}^{-4}mm$
• C. $9.82\times {10}^{-4}mm$
• D. $10.82\times {10}^{-4}mm$

Answer 1

Answer: (C)

$9.82\times {10}^{-4}mm$

Number of electrons, $n=12$

$Electricfield,E=2.55\times {10}^{4}V/m,density,\rho =1.26\times {10}^{-3}\times {10}^{6}kg/m^{3}atequilibrium,qE=mgneE=\frac{4}{3}\pi r^3\times \rho \times g{r}^3=\frac{3neE}{4\pi \rho g}r=\sqrt{\frac{3\times 12\times 1.6\times {10}^{-19}\times 2.55\times {10}^4}{4\pi \times 1.26\times {10}^3\times 9.81}}\Rightarrow r=9.82\times {10}^{-4}mm$