Question

In Millikans experiment, an oil drop of radius $1.64$ am and density $0.85g/c{m}^3$ is suspended when a downward electric field of $1.9\times 10^{5}N/C$ is applied. What is the charge on the drop in terms of $e$ ?

• A. -9e
• B. -7e
• C. -5e
• D. -3e

Answer 1

Answer: (C)

The given, $r=1.64\mu m=1.64\times 10^{-6}m$,
$\rho =0.85\times 10^{3}kg/m^3$,
$E=1.9\times 10^{5}N/C$
and $g=9.8m/s^2$
$\because E=\frac{f}{q}$
$\Rightarrow E=\frac{mg}{q}$
$m=\frac{4}{3}\pi r^3\cdot \rho$
$q=\frac{mg}{E}$
$q=\frac{\frac{4}{3}\pi r^3\cdot \rho \cdot g}{E}$
$q=\frac{4\times 3.14\times {\left(1.64\times 10^{-6}\right)}^3\times 0.85\times 10^3\times 9.8}{3\times 1.9\times 10^5}$
$q=\frac{461.49\times 10^{-20}}{5.7}C$
$q=-\frac{461.49\times 10^{-20}}{5.7\times 1.6\times 10^{-19}}e$
$q=-\frac{461.49\times 10^{-1}}{9.12}e$
$=-50.6\times 10^{-1}e$
$=-5.0e=-5e$