**Answer **

According to the question, the acceleration due to gravity is

**g = 9.8 m/s ^{2}**

The radius of the uncharged drop is

**r = 2.0 × 10 ^{-5} m**

The density of the uncharged drop is

**ρ = 1.2 × 10 ^{3} 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 :

And the expression of the viscous force on the drop is as follows :

**F = 6πηrv**

**F = 6 x 3.14 x 1.8 × 10 ^{-5}x2 x 10^{-5} x 5.8 10^{-2}**

**f = 3.91 x 10**

^{-10}N