Millikan’s oil-drop experiment

Millikan’s experiment determines the charge on a small oil droplet by observing its motion between two oppositely charged parallel plates. The droplet’s weight acts downward, while the electric field between the plates exerts a force on the charged droplet. The direction of that force depends on the direction of the field and the sign of the droplet’s charge.

Let $$Q$$ be the magnitude of the droplet’s charge. If the potential difference between plates separated by distance $$d$$ is $$V$$, the uniform electric field strength is $$E=V/d$$. The magnitude of the electric force is therefore $$QE=QV/d$$.

A stationary oil droplet

When the electric force acts upward and the droplet is stationary, the upward electric force balances its weight. For a droplet of mass $$m$$, this gives:

Here $$g$$ is the gravitational field strength. Once the droplet’s mass is known, its charge magnitude can be calculated as $$Q=mgd/V$$.

Finding the droplet’s radius and mass

With the electric field switched off, the droplet falls under its weight. As it speeds up, the upward viscous drag from the air increases. It reaches a terminal speed $$v_0$$ when the drag balances its weight.

Stokes’ law gives the drag magnitude as $$F=6\pi\eta rv$$, where $$\eta$$ is the viscosity of air, $$r$$ is the droplet’s radius and $$v$$ is its speed. Neglecting air buoyancy, the force balance at terminal speed is:

The droplet’s mass is $$m=\frac{4}{3}\pi r^3\rho$$, where $$\rho$$ is the oil density. Substituting this into the force balance gives the radius:

The mass can then be found from $$m=\frac{4}{3}\pi r^3\rho$$.

Finding charge from terminal speeds

The charge can also be found when the droplet continues to fall with the field switched on. Suppose the electric force acts upward and the droplet falls at terminal speed $$v_E$$. Both the electric force and the drag then act upward, balancing the weight:

Using the field-off balance $$mg=6\pi\eta r v_0$$ and rearranging gives:

The measured field-off speed gives $$r$$, while the difference between the terminal speeds gives the electric force. With an upward electric force, the droplet falls more slowly than it does with the field off.

If the electric force acts downward instead, it adds to the weight. For a falling droplet at terminal speed $$v_E$$, the upward drag balances both downward forces: $$6\pi\eta r v_E=mg+QV/d$$. The droplet then falls faster than it does with the field off.

Quantisation of electric charge

Electric charge is quantised: the net charge on an oil droplet can take only certain discrete values, rather than any value in a continuous range. Each allowed value is a whole-number multiple of a charge unit called the elementary charge.

Here, $$Q$$ is the droplet’s net charge, $$n$$ is an integer, and $$e \approx 1.60 \times 10^{-19},\mathrm{C}$$ is the magnitude of an electron’s charge. An electron has charge $$-e$$. The integer $$n$$ may be positive, negative or zero, so the droplet’s net charge may be positive, negative or zero.

For example, a droplet with $$n=-3$$ has charge $$Q=-3e$$, or approximately $$-4.80 \times 10^{-19},\mathrm{C}$$. A charge such as $$-2.5e$$ is not an allowed net charge for the droplet.

Millikan’s evidence for quantisation

Millikan measured the charges on oil droplets and found that the results were consistent with whole-number multiples of a common smallest charge unit. This pattern provided evidence that charge is not transferred in arbitrary continuous amounts. By finding the size of the common unit, Millikan could determine the elementary charge.