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 $ 0.02\,m{{s}^{-1}} $ . Find the number of electrons on it.

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The Correct Option is C

The horizontal force on dust particle $=q E$.
If $v$ is speed of particle in air, then
Viscous force $=6 \pi \eta r v$ (Stoke's law)
For uniform speed $v$, we have
$q E=6 \pi \eta r v$
If $n$ is the number of excess electrons, then $q=n e$
$\therefore n e E=6 \pi \eta r v$
or $n=\frac{6 \pi \eta r v}{e E}$
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}} $
$=30$