Liceo “Francesco Severi” - Milano

Millikan's experiment (electronic charge)

The discovery of electron

In 1897 J.J. Thomson measured the ratio of charge to mass (e/m) for what he called “cathode corpuscles”, small particles that emerged from an electrode called cathode.J.J. called these negative charged particles ‘electrons’.

Mass and charge of electrons

Although the measurement of e/m indicated the identity of electrons, another measurement was required before the charge ‘e’ and the mass ‘m’ could be known separately.

This was first made with precision in 1909 by R.A. Millikan.

Both the charge ‘e’ and the mass ‘m’ of an electron are incredibly small quantities. The mass of any body can be determined from the measurement of the force acting on it when it is accelerated. Even if a single electron could be isolated for study, no instrument could measure its mass directly.

Similarly, the charge on a body can be determined by measuring the force it experiences in an electric field. This method does not require the isolation of a single electron and, since very intense electric fields can be created, a measurable force can be produced.

An experiment to measure ‘e’ must be carried out with a body having so few charges that the change of one charge makes a noticeable difference. Since the experiment must be done with very little charge, the force the body experiences will be small even though a large electric field is utilized.

Mass and charge of electrons

The Millikan’s apparatus

As we can see in the figure, Millikan used a drop of oil as his test body. It was selected from a mist produced by an ordinary atomizer. The drop was so small that it could not be measured optically, but with a microscope it could be seen as a bright spot because it scattered light from an intense beam, like a minute dust particle in bright sunlight.

Drop of oil

where W is the weight, B is the buoyant force (please, remember Archimede!) and R=kv is the resisting force of the air.

When such a drop falls under the influence of gravity, it is hindered by the air it passes through. The way in which the fall of a small spherical body is hindered by air had been described by Stokes, who found that at low velocities such a body experienced a resisting force proportional to its velocity.

So, the resultant force F acting on the drop is:

F =W − B − kv

Electric field off

Initially, the velocity of the drop is zero and the resisting force too; so the resulting force is downward.As its downward velocity increases, the resisting force increases and reaches a value such that the resultant force F is null.

where vG is the terminal velocity.

In this condition the drop falls with a constant velocity called “terminal velocity” (please, remember your last test on electromagnetism!)

W − B = kvG

Electric field off

Now, we can express the proportionally constant ‘k’ in terms of known or measurable quantities: coefficient of viscosity and the density of air, density of oil, gravity acceleration and terminal velocity (if you are interested, ask your teacher for mathematical details…).

Mo’ arivo!!

Ok prof.

But what about the charge?

Electric field off

Electronic charge

With the constant ‘k’ known, we can use the Millikan’s apparatus to determine the electronic charge.

The droplets get their small charge ‘q’ (both positive or negative) from rubbing against the nozzle of the atomizer, or from encounters with stray charges in the air (from cosmic rays, x-ray source, radioactive material nearby).

Electronic charge

In the electric field E a drop will experience a force qE, which can always be directed upward or downward by the proper choice of the direction of E.

We can turn E off and E on so that the drop falls by gravity or rises because of a dominating electric force in the region between the two plates.By timing the trips of the drop over a known distance, the terminal velocities of the drop are found.

Electric field on

When a drop rises under the influence of an electric field E, the resultant force is:

When the terminal velocity vE is reached, F is null and we have:

qE =W − B + kvE

F = qE + B −W − kv

Electronic charge

But, if we remember that W-B=kvG, we finally obtain the charge on the drop:

In the oil-drop experiment, the value of vG is determined for a particular drop with the electric field off, and a whole series of vE’s from the same drop is observed with the field on.

q = kE(vG + vE )

Electricity is “quantized”!!

If we knew that the electronic charge was unique and that there was only one charge on the drop, then the last equation would give the value of this charge at once. But, since the nature of the electronic charge was not known, Millikan repeated the experiment with many different charges on the drop.

This provided a set of q values which he found to be integral multiples of one particular charge which he took to be the ultimate unit of charge, the elementary charge ‘e’.He concluded that electricity must be atomic in character.