# Understanding Millikan's oil-drop experiment

This is quoted from A.P. French's Newtonian Mechanicsabout Millikan's oil-drop experiment:

The droplets randomly produced in a mist of oil vapor are of various sizes. The ones that Millikan found most suitable were the smallest. But these droplets were so tiny that even through a medium-power microscope, they appeared against a dark background merely as points of light; no direct measurement of size could be made. However, he used a clever trick of exploiting the law of viscous resistance by applying it to the fall of the droplet under the gravitational force alone. Under these conditions:$$F = \dfrac{4\pi\rho r^3 g}{3}.$$ The terminal velocity of the droplet under gravitational field is then given by: $$v_g = \dfrac{4\pi}{3} \cdot \dfrac{\rho g}{c_1} r^2.$$ Putting in the approximate numerical values, we find, $$v_g \approx 10^8 r^2.$$ Putting $$r \approx 1 \mu = 10^{-6}$$, we have $$v \approx 10^{-4} ~\text{m/s}.$$ Such a droplet would take over 1 min to fall 1 cm in air under its own weight, thus allowing precision measurements of its speed.

It is worth noting the dynamical stability of this system, and indeed of any situation involving a constant driving force that increases monotically with speed. If by chance the droplet should slow down a little, there is a net force that will speed it up. Conversely, if it should speed up, it is subjected to a net retarding force.

[. . .] Millikan was able to follow the motion of a given droplet for many hours on end, using its electric charge as a handle by which to pull it up or down at will. In the course of such protracted observations, the charge on the drop would often change spontaneously, and several different values of terminal velocities would be obtained. The crucial observation was that in such experiment, with a given value of the voltage, the terminal speed was limited to a set of sharp and distinct values, implying that the electric charge comes in discrete units.

1. If he already knew the radius, what advantage did he get by measuring the terminal velocity in the absence of electric field?

2. What thing does speed up the drop when it slows down?

3. Why should the charge on a given droplet change as mentioned in the last para?

• How is $r=1\mu$ found? and is it important to know $r$ to calculate the charge $q$? Commented Apr 17, 2015 at 11:51
• @innisfree: Sir, I'm not understanding it. It is written by the author. And that's what I want to know what was the purpose behind all that if he knew everything
– user36790
Commented Apr 17, 2015 at 12:10
• The quote is clear on the reason: "no direct measurement of size could be made. However, he used a clever trick of exploiting the law of viscous resistance by applying it to the fall of the droplet under the gravitational force alone." - this is done to get the size of the droplet. Commented Apr 17, 2015 at 12:17
• $r=1\mu m$ is a result found from the freefall experiment rather than prior information - he didn't know it before he began his experiment. @ACuriousMind to be fair, i don't think the text is particularly clear on that point. Commented Apr 17, 2015 at 12:23

1. He knows the relationship between radius and terminal velocity, but the drops are too small to measure their radius with any accuracy (1 µm is tiny - looking at such a drop from a distance makes it no more than a speck of light, even with a "micro telescope"). Meauring the terminal velocity, he can deduce the size accurately.
2. If you slow the drop down, it no longer is at terminal velocity - gravity is unchanged, but the drag becomes less. The difference is a net force that accelerates the drop. This is why they stay at the terminal velocity (seems a convoluted way to explain why "at terminal velocity" is a stable state... but that is basically what they are saying)
3. The charge on the droplet is a net charge. Interactions with ions in the air, with ionizing (background) radiation etc can cause an electron to jump on or off (net charge on drop is almost never exactly zero... and if it is non zero, you can expect it not to be absolutely constant). These changes in charge turned out to occur in discrete steps - calculated changes in the electrical force on the droplet corresponded to charge adding or subtracting in discrete units.

It was a very clever experiment - so much to learn from the methods used as well as from the results obtained.

• Again thanking you for this answer, I want to ask you again question no:1. Why did Mr. Millikan performed the experiment in absence of electric field? He wanted to know the radius, right? But as French wrote " Putting in the approximate numerical values, we find, $v_g \approx 10^8 r^2$. Putting $r \approx 1μ = 10^{−6}$...", it seems that he knew before experiment that the radius is one micron. What is it all about then?
– user36790
Commented May 26, 2015 at 9:38
• He knew the radius is approximately one micron - but could only determine it with sufficient accuracy by doing the free fall (no field) observation. Commented May 26, 2015 at 10:24
• "Such a droplet would take over 1 min to fall 1 cm in air under its own weight, thus allowing precision measurements of its speed." So, he measured the velocity first & when it matched, he concluded that the radius was one micron, right?
– user36790
Commented May 26, 2015 at 18:13
• Yes and no. "One micron" is the order of magnitude, not the exact size. Bigger drops fell too fast and literally "dropped out" of the experiment. The experiment selected for very small but observable drops, whose exact size was measured from highly precise terminal velocity measurement. Commented May 26, 2015 at 18:15