# An oil drop of radius

Question:

An oil drop of radius $2 \mathrm{~mm}$ with a density $3 \mathrm{~g}$

$\mathrm{cm}^{-3}$ is held stationary under a constant electric field $3.55 \times 10^{5} \mathrm{~V} \mathrm{~m}^{-1}$ in

the Millikan's oil drop experiment. What is the number of excess electrons

that the oil drop will possess?

(consider $\mathrm{g}=9.81 \mathrm{~m} / \mathrm{s}^{2}$ )

1. (1) $48.8 \times 10^{11}$

2. (2) $1.73 \times 10^{10}$

3. (3) $17.3 \times 10^{10}$

4. (4) $1.73 \times 10^{12}$

Correct Option: , 2

Solution:

(2)

$\mathrm{qE}=\mathrm{Mg}$

$\mathrm{neE}=\rho\left(\frac{4}{3} \pi \mathrm{r}^{3}\right) \times \mathrm{g}$

$\mathrm{n} \times 1.6 \times 10^{-19} \times 3.55 \times 10^{5}$

$=3 \times 10^{3} \times \frac{4}{3} \times \pi \times\left(2 \times 10^{-3}\right)^{3} \times 9.81$

$\mathrm{n}=173 \times 10^{(3-9-5+19)}$

$\mathrm{n}=1.73 \times 10^{10}$