Question

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

• A. 10
• B. 20
• C. 30
• D. 40

Answer 1

The correct option is C 30

Given that, particle moves with constant speed and so net force is zero.
from the free body diagram we can see that
$F_v=F_e$
$6\pi \eta rv=neE$ ( where n = No. of electrons)
we know,
$\eta =1.6\times {10}^{-5}Ns{m}^{-2}$
$r=5\times 10m$,
v = 0.02 $m{s}^{-1}$
$E=6.28\times {10}^5\frac{V}{m}$,
e= $1.602\times {10}^{-19}$ Coulomb
substitute these values and solve

n = $\frac{\left(6\right)\times \left(3.14\right)\times (1.6\times {10}^{-5})\times \left(50\right)\times \left(0.02\right)}{(6.28\times {10}^5)\times (1.602\times {10}^{-19})}=30$

So, the answer is option (C).