Question

A charged dust particle of radius $5\times {10}^{-7}m$ is located in a horizontal electric field having an intensity of $6.28\times {10}^{5}V{m}^{-1}$ . The surrounding medium is air with coefficient of viscosity $\eta =1.6\times {10}^{-5}N-s{m}^{-2}$ . If this particle moves with a uniform horizontal speed $0.02m{s}^{-1}$ . Find the number of electrons on it.

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

Answer: (C)

The horizontal force on dust particle $=qE$.
If $v$ is speed of particle in air, then
Viscous force $=6\pi \eta rv$ (Stoke's law)
For uniform speed $v$, we have
$qE=6\pi \eta rv$
If $n$ is the number of excess electrons, then $q=ne$
$\therefore neE=6\pi \eta rv$
or $n=\frac{6\pi \eta rv}{eE}$
Substituting given values
$n=\frac{6\times 3.14\times 1.6\times 10^{-5}\times 5\times 10^{-7}\times 0.02}{1.6\times 10^{-19}\times 6.28\times 10^5}$
$=30$