Answer
According to the question, the acceleration due to gravity is
g = 9.8 m/s2
The radius of the uncharged drop is
r = 2.0 × 10-5 m
The density of the uncharged drop is
ρ = 1.2 × 103 kg m-3
The viscosity of air is
η = 1.8 × 10-5 Pa s
To avoid taking into account air buoyancy, we set the density of air to zero.Therefore terminal velocity (v) is :
$ v=\frac{2{{r}^{2}}g\rho }{9\eta } $
$ v=\frac{2{{(2\times {{10}^{-5}})}^{2}}\times 9.8\times 1.2\times {{10}^{3}}}{9\times 1.8\times {{10}^{-5}}} $
$ v=5.8c{{m}^{-1}} $
And the expression of the viscous force on the drop is as follows :
F = 6πηrv
F = 6 x 3.14 x 1.8 × 10-5x2 x 10-5 x 5.8 10-2
f = 3.91 x 10-10 N