Turning Points Tutorial 2 – Millikan’s Experiment



Millikan's Experiment

Stokes' Law

Quantisation of Charge


Now physicists knew that electrons were orbiting the nucleus.  But the nature of the electron remained a mystery.  They were not the neat little spheres that are found in many a textbook (or, for that matter, these notes).  They are quantum beings.  The closer you get to the little brutes, the harder they are to catch.



Millikan's Experiment

Robert Millikan used a simple but famous experiment that served to confirm the unit electronic charge as 1.6 × 10-19 C.    He sprayed oil drops into a space between two charged plates.  Each tiny oil droplet was charged up with a negative charge by friction as it left the sprayer.  The theory was simple; the attractive electrostatic force between the negatively charged droplet and the positively charged plate would balance out the weight of the droplet.  His apparatus was like this:



He would select a particular oil drop and hold it stationary by altering the voltage between the two plates. 


Question 1

What are the forces acting on the plates?  What is the resultant force?



  The forces on the stationary drop are like this:



We know that:


These two combine to give:



It doesn't take a genius to see that:



Question 2

How might you find the weight of the drop?




Stokes' Law

This is not a very satisfactory way because the uncertainty is too great.  So another method was used.  Millikan turned off the plates and watched the oil drop.  Very quickly the oil drop reached terminal speed.


Question 3

What forces are acting on an object at terminal speed?  What is the resultant force?



So we can write:

mg = drag force


The drag force can be worked out indirectly using Stokes' Law, which describes the force acting on a sphere falling at terminal speed through a viscous fluid.  Normally we think of viscous fluids as thick gooey oils;  for a tiny oil droplet, air is pretty viscous.   The equation that describes Stokes' Law is:


[r - radius of the sphere, v - terminal speed (m s-1 ).]


The strange looking symbol, h, is "eta", a Greek lower case letter long 'ē', the Physics Code for the coefficient of the viscosity of a fluid. 


The units for h are N s m-2. For air, h = 1.8 × 10-5 N s m-2.



Question 4

A small metal sphere of radius 0.50 mm has mass 1.0 × 10-3 kg is dropped into oil of which the viscosity is 0.36 N s m-2.  What is the terminal velocity at which it falls?



[Note:  this model only works for objects falling at low speed.  At higher speed, turbulence has an effect.]


In question 2 we looked at the intuitive way of finding the weight by:


Weight = density × volume × g


We can write this as:


So we can bring in the Stokes' Law equation in by writing:



Rearranging and cancelling out gives us:



So by observing the terminal speed, we can work out the radius.  From that, we can work out the volume, hence the mass and weight.  Although it seems long-winded, this method produces much less uncertainty than attempting a direct measurement of the radius.



Worked example

The data below are from an experiment similar to Millikan’s experiment.

  • Density of oil = 900 kg m-3

  • Pd across the plates = 613 V

  • Plate separation = 0.010 m

  • Viscosity of air = 1.8 × 10-5 N s m-2

  When the voltage between the plates is turned off, the droplet falls steadily a distance of 2.50 × 10-3 m in a time of 22 s.  What is the charge?  Take g as 9.8 m s-2.


Work out the speed = distance ÷ time =  2.50 × 10-3 m ÷ 22 s = 1.14 × 10-4 m s-1.


Now work out the radius using the equation:


Substitute in the numbers:


 Remember to take the square root.

 r = 1.02 × 10-6 m



Mass = volume × density:




We know that weight = force from the electric field



Rearranging gives us:



q = 6.4 × 10-19 C




Quantisation of Charge

The answer above is a charge of magnitude 4e (where e = -1.602 × 10-19 C).


The important finding from this experiment was that the charge was always a whole number multiple of 1.6 × 10-19 C.  Therefore the electron charge came in definite amounts (quanta); therefore it is said to be quantised.  Electrons are indivisible, i.e. fundamental particles of matter.



Question 5

In an experiment to determine the charge on a charged oil droplet, the droplet was held stationary in a vertical electric field of strength 57 kV m-1.  After the field was switched off, the droplet fell at a steady speed, taking 18.3 s to fall through a vertical distance of 2.0 mm


Viscosity of air = 1.80 × 10-5 N s m-2

Density of oil = 970 kg m-3

g = 9.8 m s-2


(a)    Calculate the speed of the droplet as it falls.

(b)   Show that the droplet’s radius is 9.7 × 10-7 m

(c)    Calculate the charge of the droplet.

(d)   Compare this to the electronic charge.  What does it suggest?



Once we know the charge on an electron, a simple calculation will tell us the mass of the electron, 9.11 × 10-31 kg.



Robert Andrews Millikan (1868 - 1953) was an American physicist who won the Nobel Prize for this work and other work he did on the photoelectric effect.  However, his first degree was not in Physics, but in Classics.  He was a promising scholar in Latin and Greek.  During his final year his Greek professor asked him to take the foundation Physics class, to which he, not unnaturally, objected on the grounds that he knew no physics.  The professor told him, "If you can do well in Greek with me, you can teach physics."  After warning the professor that he would not be responsible for the outcome, Millikan bought a text book, and kept three pages ahead of his class, discovering that he really enjoyed it.  And that rubbed off onto his class.


Gaining a doctorate in 1895, Millikan soon became professor at Chicago University.  In 1908 he started work on his oil drop experiment.


Millikan enjoyed the teaching aspect of his work.  He wrote a number of text books that were well ahead of their time, that encouraged students to "think in physics", rather than to apply formulae in a mechanical manner.



This experiment won Millikan the Nobel Prize in 1923.