An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10^-17 kg. It is falling freely in air with terminal speed…

Q: An oil drop carrying a charge of 2 electrons has a mass of 3.2 × 10-17 kg. It is falling freely in air with terminal speed. The electric field required to make the drop move upwards with the same speed is

(a) 2 × 103 V/m

(b) 4 × 103 V/m

(c) 3 × 103 V/m

(d) 8 × 103 V/m

Ans:(a)

Sol: Without electric field when oil drop is moving downwards,

mg = 6 π η r v …(i)

Under electric field of strength E , oil drop is to move upwards

F – mg = 6 π η r v

F – mg = mg …using (i)

(2 e)E = 2 mg

E = mg/e

$\displaystyle E = \frac{3.2 \times 10^{-17}\times 10}{1.6 \times 10^{-19}} $

= 2 × 103 V /m