Question

A charged dust particle of radius $5\times {10}^{-7}m$ is located in a horizontal electric field having an intensity of $6.28\times {10}^{5}V/m$. The surrounding medium is air with a 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, find the number of electrons on it.

Answer 1

As the charged dust particle moves with uniform speed, hence net force acting on it is zero.
So, Backward viscous force = Forward electric force
If n = Number of electrons on the dust particle, then
$6\pi \eta rv=neE\Rightarrow n=\frac{6\pi \eta rv}{eE}=\frac{6\times 22\times 1.6\times {10}^{-5}\times 5\times {10}^{-7}\times 0.02}{7\times 1.6\times {10}^{-19}\times 6.28\times {10}^5}=30$