Answer to Question #97566 in Mechanics | Relativity for Munawar Hussain

Question

In an experiment, the tiny droplets of the negatively electrified oil are dropping through a vacuum. Magnitude of the electric field is 5.92×104 NC−1. Direction of the electric field is assumed to be downward. (a)An

experimenter observes a one particular droplet and that droplet remains suspended against gravity. Assume the mass of the droplet to be 2.93×10−15 kg. What is the charge carried by the droplet? (b) Now another droplet of same mass falls, in the vacuum, from rest to 15.3cm in 0.275s. What is the charge on the droplet in this case

Answer 1

(a) Let $q_1$ be charge on droplet

Gravitational force due to weight = electrostatic force due to electric field

Then,

$$q_1E=mg$$

$$or\\\\\\\\ \\q_1=\frac{mg}{E}=\frac{2.93\times 10^{-15}\times9.8}{5.92\times 10^4}=4.85\times 10^{-19}\ C$$

(b) Let $q_2$ ​ be charge on this droplet

then according to given question,

$$s=ut+\frac{1}{2}at^2\\0.153=0.5a\times(.275^2) \\or\\a=4.05\ m\ sec^{-2}$$

balancing the force

we get,

$$ma=mg-q_2E$$

$$m=2.93\times 10^{-15}\ kg\\E=5.92\times 10^{4}$$

$$(2.93\times 10^{-15}\times 4.05)=(2.93\times 10^{-15}\times 9.8)-(q_2\times 5.92\times 10^4)$$

$$Or\\q_2=2.84\times 10^{-19} \ C$$