Answer to Question #97399 in Mechanics | Relativity for Ali Hamza

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) gravitational force = electrostatic force

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

(b)

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

so,

$$ma=mg-q_2E$$

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

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