How the Oil Drop Experiment WorkedThe original experiment was performed in 1909 by Robert Millikan and Harvey Fletcher by balancing the downward gravitational force and the upward electrical and buoyant forces of charged oil droplets suspended between two metal plates. The mass of the droplets and the density of the oil was known, so the gravitational and buoyant forces could be calculated from the measured radii of the oil drops. Since the electric field was known, the charge on the oil drops could be determined when the drops were held at equilibrium. The value for the charge was calculated for many droplets. The values were multiples of the value of a charge of a single electron. Millikan and Fletcher calculated the charge of an electron to be 1.5924(17)×10−19 C. Their value was within one percent of the currently accepted value for the charge of an electron, which is 1.602176487(40)×10−19 C.
Millikan Oil Drop Experiment ApparatusMillikan's experimental apparatus was based on a pair of parallel horizontal metal plates held apart by a ring of insulating materal. A potential difference was applied across the plates to create a uniform electric field. Holes were cut into the insulating ring to allow for light and a microscope so that the oil drops could be observed.
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 ExperimentInitially, drops fall into the space between the parallel plates with no voltage applied. They fall and achieve terminal velocity. When the voltage is turned on, it is adjusted until some of the drops start to rise. If a drop rises, it indicates the upward electrical force is greater than the downward gravitational force. A drop is selected and allowed to fall. Its terminal velocity in the absence of the electrical field is calculated. The drag on the drop is calculated using Stokes Law:
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