Answer to Question #97490 in Mechanics | Relativity for Ak

Question

In an experiment, the tiny droplets of the negatively electrified oil are dropping through a vac-
uum. Magnitude of the electric field is 5.92 × 104 N C−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 15.3 cm in 0.275 s. 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

So,

$$q_1E=mg$$

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

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

then according to question,

$$s=ut+\frac{1}{2}at^2\\.153=0.5a\times(.275^2) \\or\\a=4.05\ m\ sec^{-2}$$

balancing force,

$$ma=mg-q_2E$$

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

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

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