Q. A charged dust particle of radius $5 \times 10^{- 7}$m is located in a horizontal electric field having an intensity of $6.28 \times 10^{5}\ Vm^{- 1}$. The surrounding medium in air with coefficient of viscosity $\eta = 1.6 \times 10^{- 15}Nsm^{- 2}.$ If this particle moves with a uniform horizontal speed of 0.01 $ms^{- 1},$ the number of electrons on it will be
A) 20
B) 15
C) 25
D) 30
Explanation
(b) $n\ e\ E = 6\pi\ \eta\ r\ v\ $ or $n = \frac{6\pi\ \eta\ r\ v}{e\ E}$
$= \frac{6 \times 3.14 \times 1.6 \times 10^{- 5} \times 5 \times 10^{- 7} \times 0.01}{1.6 \times 10^{- 19} \times 6.28 \times 10^{5}} = 15$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.