A charged dust particle of radius 5x10^-7 m is located in a horizontal electric field having an intensity of 6.28 x 10⁵ Vm^-1. The surrounding medium is air with coefficient of viscosity $\mathrm{\eta }=1.6\times {10}^{-5} {\mathrm{Nsm}}^{-2}$. If this particle moves with a uniform horizontal speed 0.02 ms^-1, find the number of electrons in it.

The horizontal force on dust particle is F = qE
If v is speed of particle in air, then by Stoke's Law,
Viscous force $=6\mathrm{\pi \eta rv}$
For uniform speed v, we have
 $\mathrm{qE}=6\mathrm{\pi \eta rv}$
If n is the number of excess electrons, then $\mathrm{q}=\mathrm{ne}$
$\begin{array}{l}\Rightarrow \mathrm{neE}=6\mathrm{\pi \eta rv}\\ \Rightarrow \mathrm{n}=\frac{6\mathrm{\pi \eta rv}}{\mathrm{eE}}\end{array}$
Substituting given values 
$\mathrm{n}=\frac{6\times 3.14\times 1.6\times {10}^{-5}\times 5\times {10}^{-7}\times 0.02}{1.6\times {10}^{-19}\times 6.28\times {10}^5}$
$\Rightarrow \mathrm{n} = 30$