Quantum and Nuclear

The Millikan experiment

Teaching Guidance for 14-16 PRACTICAL PHYISCS

The Millikan experiment is hugely important. Between 1910 and 1911 Robert Millikan used some clever ideas and careful experimentation to show that charge is quantized. He then determined a value for the fundamental quantum of charge, known as unit charge.

He could not measure a single quantum of charge. Instead, he measured the charge on a number of oil drops and deduced that they could all be divided by a single factor, which must be the basic unit of charge.

The Millikan experiment is very fiddly and difficult to perform in school. It is more likely that you will use a film clip or simulation of the experiment to show its principles to students. The principles are:

  • An oil drop will fall through air under its own weight. If the drop is given a charge, it can be suspended using an electric field. At this point the electrostatic force balances the weight of the drop. The size of the electrostatic force depends on the charge on the drop. So Millikan could work out the charge as long as he knew the weight.
  • In order to find the weight of the drop, Millikan allowed the drop to fall through air. It quickly reaches its terminal velocity. At this point, the weight is being balanced by the viscous drag of the air. The drag can be calculated from Stokes' Law which allowed Millikan to determine the weight.
  • Millikan repeated the experiment for over 150 oil drops. He selected 58 of his results and found the highest common factor. That is, the single unit of charge which could be multiplied up to give the charge he measured on all his oil drops.

The calculations

1 When the oil drop is in the electric field, there is an electric force, F , acting upwards. This is given by:

F = Eq where q is the charge on the oil drop and E is the field strength.

= Vqd where V is the voltage on the plates and d is their separation.

The drop is being pulled down by its weight, mg. When the drop is suspended and stationary, the net force is zero. So:

  • Vqdmg = 0
  • Vqd = mg
  • so q = mgdV

2 To find the weight of the drop, Millikan let it fall through the air and measured its terminal velocity. At this point, the net force is zero – i.e. weight is balanced by the viscous drag. The viscous drag, D , is given by:

D = 6πη vr where η is the viscosity of air, r is the radius of a spherical drop and v is its speed.

This allowed him to work out the radius of the drop and therefore its weight.

[An experimental alternative is to use the electric field to change the terminal velocity of the oil drop rather than keep it stationary. This is easier to achieve and measure experimentally. It is then possible to derive an expression for the charge which is related to the two velocities. However, for school students, the principle of the stationary drop is probably easier to grasp.]

3 Having measured the charge on a number of oil drops, q1, q2, q3, etc, Millikan reasoned that each charge must be a whole number multiple of the fundamental charge, e. So:

q1 ,=n1 e , q2 ,=n2 e , q3 ,=n3 e and so on, where n is a whole number in each case.

So he found e by finding the highest common factor of all the values of charge that he measured on the oil drops.

is a constituent in our description of Beta Decay
is a type of Lepton
can exhibit Wave-Particle Duality
is a constituent of the Plum Pudding Model
has the quantity Charge
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