Question

In Millikan's oil drop experiment, what is the terminal speed of an uncharged drop of radius 2.0 x 10^-5 m and density 1.2 x 10³ kg m^-3. Take the viscosity of air at the temperature of the experiment to be 1.8 x 10^-5 Pa s. How much is the viscous force on the drop at that speed? Neglect buoyancy of the drop due to air.

Answer 1

Here,

r = 2 x 10^-5 m

ρ = 1.2 x 10³ kg m^-3

η = 1.8 x 10^-5 Pa s

Density of air(d) = 1.293 kg m^-3

v = 2/9.{r²(ρ -d)g}/{η}

= {2 x (2 x 10^-5) x (1.2 x 10³ - 1.293) x 9.8}/{9 x 1.8 x 10^-5}

= 0.058 ms^-1

= 5.8 cm s^-1

Also, viscous force on the drop,

F_v = 6π rv η

= 6 x 3.14 x 2 x 10^-5 x 0.058 x 1.8 x 10^-5

= 3.93 x 10^-10 N