Question

An oil drop of radius 2 mm with a density $3 \,g\,cm ^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} V m ^{-1}$ in the Millikan's oil drop experiment. What is the number of excess electrons that the oil drop will possess? (consider $g =9.81 \,m / s ^{2}$ )

• A. $48.8 \times 10^{11}$
• B. $1.73 \times 10^{10}$
• C. $17.3 \times 10^{10}$
• D. $1.73 \times 10^{12}$

Answer 1

Answer: (B)

$qE = Mg$
$neE =\rho\left(\frac{4}{3} \pi r ^{3}\right) \times g$
$n \times 1.6 \times 10^{-19} \times 3.55 \times 10^{5}$
$=3 \times 10^{3} \times \frac{4}{3} \times \pi \times\left(2 \times 10^{-3}\right)^{3} \times 9.81$
$n =173 \times 10^{(3-9-5+19)}$
$n =1.73 \times 10^{10}$