# Oil drop experiment and quantization of charge

How to systematically show that the resulting charges in oil drop experiment are integers multiplied by $e$ in other word how to extract $e$ from the data?

• This question is very vage. I have no clue what you are asking. You don't give any background info. – Bernhard Feb 1 '14 at 12:39
• More on Millikan's oil drop experiment: physics.stackexchange.com/search?q=millikan+oil – Qmechanic Feb 1 '14 at 14:21
• I give a simplified version of this question as a problem in Modern Physics. Students with OCD-like tendency find some pretty decent solutions. That said there is often some confusions about exactly what the data is in Milikan's experiment and you should almost certainly say what you think it is so that everyone is on the same foot. – dmckee Feb 1 '14 at 14:41
• @richard: Isn't this something which is taught in basic freshman physics? Also, the article you linked explains everything in complete detail, so I don't understand the point of the question. – DumpsterDoofus Feb 1 '14 at 15:45
• As far as I know you fiddle around with different values until you find one that gives the best fit. I must admit I don't know of a systematic way to do it. I would be interested in an explanation for some systematic procedure if someone would like to provide an answer. Incidentally, I suspect the downvote and criticism stem from a misunderstanding of what you're asking. The question is how, starting from a list of charges, do you find the quantity that they are all integer multiples of. Especially bearing in mind there will be experimental errors in the charges. – John Rennie Feb 1 '14 at 16:03

## 2 Answers

To address John Rennie's comment in the comment section regarding the existence of a systematic, human-guess-independent algorithm for determining the LCM of a data series in the presence of significant experimental error and without the aid of single-electron-charged droplets to make a human-sensible guess:

a = 12.5654;
L = 400;
list = Table[a (RandomInteger[{6, 35}] + RandomReal[{-0.25, 0.25}]), {k, L}];
f[b_] := Module[{g = Nearest[b Range[L]]}, Sum[Abs[g[list[[k]]][] - list[[k]]],
{k, L}]/b];
ListPlot[list, PlotRange -> All]
Plot[f[x], {x, 6, 15}, PlotRange -> All]  There's no way a human could look at that plot of the noisy raw data and guess the LCM, but a computer can handle it just fine. Note that this is reliably indicating the LCM even though the "measurement" error is on the order of 50%. I used uniformly-distributed error, but it works with Gaussian-distributed errors just as fine.

As an interesting mathematical aside, in the absence of noise the LCM appears as the largest zero of the merit function, which has a sequence of zeros whose density of zeros tends as $(a x)^{-1}$ where $a$ is the LCM and $x$ is the guess. As $x\rightarrow 0$ the there is an oscillatory singularity, and for $x>a$, there are no further zeros. • Is there any chance that you can explain the code also ? Plus, it is Mathematica ? – onurcanbektas Feb 7 '18 at 4:45

If the experiment was done with sufficient accuracy, simply plotting the calculated charge values should give obvious clustering. (Two measurements per particle: mass from free fall velocity, and voltage to achieve zero velocity is how I remember the experiment, but that is from a fifty year-old memory of high-school physics... plot voltage/mass.)

R.J.Doe has a set of directions (with an amusing apocalyptic conclusion) on writing up a somewhat different version of the experiment: http://www.phys.ksu.edu/personal/cocke/classes/phys506/aasamplewriteup.htm using both a downward and upward acceleration to give three velocities per particle. I'm wondering if that might have the advantage that you would not need to depend on a previously measured value for the viscosity of air.

I see that DumpsterDoofus is expressing annoyance at a lack of effort and suggest perhaps the use of http://webphysics.davidson.edu/applets/pqp_preview/contents/pqp_errata/cd_errata_fixes/section4_5.html to generate dome "data" would mollify him. It would be more interesting to see data gathered this way than to look at his generation of data which I suspect is very much unlike what was gathered by Millikan. (I also disagree that we could not have done such data analysis without computers.)

• Sorry, I probably should've been a little more clear in my answer: I wasn't saying that doing such data analysis without computers is impossible, I was saying that for the example I gave, it would be difficult for humans to see the clustering. In actual charged-drop experiments, there is enough low-integer charged droplets that you can look at the data and guess the answer; the point of my answer was to provide a general, human-independent algorithm for estimating LCM's on noisy data with non-obvious clustering and large amounts of charge. – DumpsterDoofus Feb 1 '14 at 20:20
• But I think the Millikan data was presented (at least as I remeber it) as having typically single or doubly or triply charged droplets, so I think your computer generated example was not a realistically constructed example. – DWin Feb 1 '14 at 20:33
• Yes, you remember correctly. When you do the Millikan droplet experiment, the droplets usually have only a couple electrons on them if made properly, so it's not hard to estimate the charge just by looking at the data, which is why it's done in some freshman physics labs. The point of my answer was to answer the more general question of how to attack the problem for similar types of data analysis involving noisy, quantized data that is not amenable to human attack, rather than just Millikan drop data. – DumpsterDoofus Feb 1 '14 at 20:52
• It is very hard to get down to just a few electrons in practice (though that is the idealized version of the experiment usually presented). In Miliken's actual experiment he ran a single drop up and down until a charge is randomly canceled by ambient ionization. Then you compute the difference in the changes before and after as deduced from the voltage--velocity relationship. Repeat with many drops and multi-neutralization events per drop and you can do very well, but it takes a lot of patience. – dmckee Feb 1 '14 at 21:44