Question

Example 1 In a Millikan's oil drop experiment, a drop is observed to fall with a terminal speed 1.4 mm s7 in the
absence of electric filed. When a vertical electric field of 4.9 x 105 V m1 is applied, the droplet is observed to
continue to move downward at a lower terminal speed 1.21 mm s 1. Calculate the charge on the drop. (density of
oil = 750 kg m3, viscosity of air = 1.81 x10-5 kg m s 1, density of air = 1.29 kg m']
[HSEB 2072]​

Answer 1

Answer:

Charge on the oil drop is approximately $6.245 * 10^{-19}$ Coulombs, which is approximate 4 electrons.

Explanation:

First of all we'll find radius of the oil drop.

According to the equation

$V_{T} = \frac{2}{9} \frac{r^{2}g}{n}(d_{oil}-d_{air})\\ \\r = 3.9 * 10^{-6} m$

Now, When there's no electric field,

mg = 6πηrv

And, when electric field is present,

mg = 6πηrv' + qE

⇒ 6πηrv = 6πηrv' + qE

$q = \frac{6\pi nr(v-v')}{E}$

Abbreviations:

v is terminal velocity when there was no electric field

v' is terminal velocity when there was the applied electric field

n or η is viscosity of air.

q is charge on oil drop

E is applied electric field.