Millikan Oil Drop Experiment

Determining the Electron Charge by the Millikan Oil Drop Experiment

This is a digram of the laboratory setup for the Millikan oil drop experiment.
Theresa Knott

Millikan's oil drop experiment measured the charge of the electron.

How the Oil Drop Experiment Worked


Millikan Oil Drop Experiment Apparatus

The experiment was performed by spraying a mist of oil droplets into a chamber above the metal plates. The choice of oil was important because most oils would evaporate under the heat of the light source, causing the drop to change mass throughout the experiment. Oil for vacuum applications was a good choice because it had a very low vapor pressure. Oil droplets could become electrically charged through friction as they were sprayed through the nozzle or they could be charged by exposing them to ionizing radiation. Charged droplets would enter the space between the parallel plates. Controlling the electric potential across the plates would cause the droplets to rise or fall.

Performing the Millikan Oil Drop Experiment

Fd = 6πrηv1

where r is the drop radius, η is the viscosity of air and v1 is the terminal velocity of the drop.

The weight W of the oil drop is the volume V multiplied by the density ρ and the acceleration due to gravity g.

The apparent weight of the drop in air is the true weight minus the upthrust (equal to the weight of air displaced by the oil drop). If the drop is assumed to be perfectly spherical then the apparent weight can be calculated:

W = 4/3 πr3g (ρ - ρair)

The drop is not accelerating at terminal velocity so the total force acting on it must be zero such that F = W. Under this condition:

r2 = 9ηv1 / 2g(ρ - ρair)

r is calculated so W can be solved. When the voltage is turned on the electric force on the drop is:

FE = qE

where q is the charge on the oil drop and E is the electric potential across the plates. For parallel plates:

E = V/d

where V is the voltage and d is the distance between the plates.

The charge on the drop is determined by increasing the voltage slightly so that the oil drop rises with velocity v2:

qE - W = 6πrηv2

qE - W = Wv2/v1