# 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 $q$? – innisfree Apr 17 '15 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 Apr 17 '15 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. – ACuriousMind Apr 17 '15 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. – innisfree Apr 17 '15 at 12:23

• 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 May 26 '15 at 9:38