Nearly every explanation of oil drop experiment I ever found concludes along the lines of

The charge $q$ on the droplets were thus measured, and were all found to be integral multiple of $1.6 \cdot 10^{-19}\,\rm C$.

How exactly do you get this random number? I mean, I could say, they were all integral multiples of $0.8 \cdot 10^{-19}$, and I'll still be right since $0.8 = 1.6/2$. In fact, I can just divide $1.6$ by any number $n$, and conclude that it is the fundamental unit of charge since the charge on the droplets is a multiple of it.

Until and unless we are certain that a drop gained exactly one electron and measure it, there is no way to conclude the value of $e$, isn't it?

  • 2
    $\begingroup$ Millikan found the highest common factor for the charges on his oil drops which would rule out your value of 1.6/2 which he never found. $\endgroup$ – Farcher Feb 12 '17 at 7:11
  • 1
    $\begingroup$ Why should I believe you that the charge on an electron is the HCF of those on the droplet? That'll require an extra assumption that at least one droplet lost/gained no more than 1 electron $\endgroup$ – Arnav Ramkrishnan Feb 12 '17 at 8:25
  • 3
    $\begingroup$ @Barbaad Because the alternative is to believe that every one of your thousands of drops picked up an even number of electrons, or that they all picked up multiples of $n$ electrons for some $n$. What mechanism would you propose to explain that? $\endgroup$ – David Richerby Dec 24 '17 at 18:36
  • $\begingroup$ It is simpler than you might think. All charges are multiples of that number, so the smallest difference between measured charge values must be that number. Find the smallest difference, in an experiment where very small amounts of charge are used, and you will likely have the charge of a single electron. $\endgroup$ – Sam Gallagher Apr 25 '18 at 13:45

Millikan's oil drop experiment first and foremost serves to establish that electron charge is quantized. However, as you say in and of itself it does not exclusively specify the charge $e$, as it could be $e/2$ as you mention. However, the simplest interpretation of the data is to say that the charges are quantized with charge $e$; if it were $e/2$ instead, you'd have to tell me why should I expect every droplet to have an even number of electrons. Essentially, you'd need some kind of conspiracy going on, some extra physics to observe such phenomena.

However, there are also other ways to measure $e$, such as measuring the shot noise in a current-carrying wire. Such experiments give us even more confidence that the electron charge is $e$. In fact, in the fractional quantum hall effect, if one tries to do the same shot-noise measurement to experimentally verify $e$, one gets a rational fraction such as $e/3$ instead. This phenomenon actually led to the Nobel Prize in 1998 because of the non-trivial physics that was discovered. All in all, because of this and more, the value of $e$ is more or less an established fact (the evidence is more than the oil-drop experiment).


Keep in mind that in doing this experiment one observers drops of many different charges, because the triboelectric effect that causes the drops to be charged in the first place is stochastic.

That is, you see values that would eventually be identified as $1e$, $2e$, $3e$, and on up to as high a value as you are able to measure with the apparatus.

Moreover, in Millikan's real experiment (rather than the simplified version presented in many basic treatments) you watch a drop while for long enough to record one or more instances of the drop's charge being reduced (an effect of cosmic radiation), so you can observe the steps down toward neutral.

In any case, when you have enough data it becomes clear that you have discrete levels, and that the spacing is the same as the lowest level. At that point the least hypothesis is pretty unavoidable.

  • 1
    $\begingroup$ I am not arguing that the charge is not quantized. That is definitely deducible from the results. But I can claim e to be e/2 and I can still explain every single one of Millikan's result. He had no way of knowing how many electrons he was losing or gaining in one go. Maybe he was losing millions, and e is actually e/1000000?? $\endgroup$ – Arnav Ramkrishnan Feb 12 '17 at 7:10
  • 1
    $\begingroup$ That fails the least hypothesis test, because you have to assume that somehow the random process that gives each drop it's own charge is always giving you a multiple of some number and that none-the-less those groups are detachable but that it won't happen in this process. Yes, you can be pedantic about it, but no one is going to be impressed. $\endgroup$ – dmckee Feb 12 '17 at 7:13

Gonna be my first answer on this site, since I had this question right from school till college until I read the actual paper by Millikan from my library archive.

You're absolutely correct. For all we know 1.6 can be the wrong number, and instead the actual value of e is 1.6/n where n is some integer. After all, concluding what Millikan concluded requires the extra assumption that there exists at least one droplet which lost exactly one electron. However, hear me out-

  1. The version presented in books is a very abridged one. Millikan watched the droplets for prolonged period of time, where he would zap the droplets by X-ray, and keep experimenting on it till it lost all charge, then zap it again randomly. And he observed MANY such drops.

  2. The drops always lost charges in simple whole number ratios. First and foremost, this proved that charge is quantized.

  3. Here's the interesting part- The common denominator the droplets were losing charge in, also happened to be the lowest measured charge in all the trials.

  4. Let's say that e is actually e/2. This would require you to explain why, in a random process, are electrons always lost in even numbers. Now let's say e is e/3. Here, electrons can be lost in both even and odd numbers (3, 6, 9 etc), but then, it'd require you to explain why they're lost in triplets, and so on.

  5. The above argument would break down for, say, e= e/million. Because then, the errors in measurement would far exceed our resolution to distinguish between 1 million vs 1 million and 1 electrons. This is a plausible scenario.

  6. But there's a thing in Science called Occam's razor. If two theories explain the same thing perfectly, you choose the less complicated one. For all we know, a neutron is actually made up of two neutrons of half the accepted mass of a neutron. But the thing is, it doesn't matter. If all observable experiments can make do by considering the simplest hypothesis, that hypothesis stands until it breaks down in some other experiment

  7. e/m ratio and mass of electrons can be measured independently. They, and every other experiment, agree on the charge found by Millikan (save for some error). However, we need to be very careful not to include experiments which already have the charge of electron secretly put in there (like those involving an oscilloscope)."First principle is you must not fool yourself".


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.