Question

In Millikan’s oil drop experiment, an uncharged drop of radius 2.0 × 10^–5 m is falling with terminal velocity in air. The density of air is negligible relative to oil density, 1.2 × 10³ kg m^–3 . The viscosity co efficient for air at the temperature of working day is 1.8 × 10 ^–5 Pa s. 

Determine : (a) the terminal speed of the drop and (b) the viscous force acting on the drop.

Answer 1

The terminal velocity as given by

(b) The viscous force is given by Stokes law.

As another method we may have also used the equilibrium condition that weight is balanced by viscous force. Hence

viscous force weight of drop = 4/3 πr³ρg

= 4/3 × 3.14 × (2.0 × 10 –5 ) 3 × (1.2 × 10 3 ) × 9.8 N

= 3.9 × 10 ^–10 N.