Question

In Millikans experiment, an oil drop of radius $1.64$ am and density $0.85\, g / cm ^{3}$ 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 $g=9.8\, m / s ^{2}$
$\because E=\frac{f}{q}$
$\Rightarrow E=\frac{m g}{q}$
$m=\frac{4}{3} \pi r^{3} \cdot \rho$
$q=\frac{m g}{E}$
$q=\frac{\frac{4}{3} \pi r^{3} \cdot \rho \cdot g}{E}$
$q=\frac{4 \times 3.14 \times\left(1.64 \times 10^{-6}\right)^{3} \times 0.85 \times 10^{3} \times 9.8}{3 \times 1.9 \times 10^{5}}$
$q=\frac{461.49 \times 10^{-20}}{5.7} C$
$q=-\frac{461.49 \times 10^{-20}}{5.7 \times 1.6 \times 10^{-19}} e$
$q=-\frac{461.49 \times 10^{-1}}{9.12} e$
$=-50.6 \times 10^{-1} e$
$=-5.0 e=-5 e$