What error could occur if only a few oil drops were used, or if all the oil drops contained an even number of charges?
[Physics] Millikan’s Experiment- Oil drop containing an even number of charges
charge
Related Solutions
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]]][[1]] - 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.
The Stokes law equation for the drag on the oil droplet is:
$$ F_d = 6 \pi \eta r v $$
wher $\eta$ is the viscosity of the air, $r$ is the radius of the oil drop and $v$ is the velocity of the oil drop. The trouble is that when the oil drop is very small its radius is comparable to the mean free path of the air molecules. That means the air no longer behaves as a completely homogeneous fluid, and while this is a very small effect it is detectable in the oil drop experiment.
The correction that you refer, which is to due to Cunningham, is to write the drag as:
$$ F_d = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$
where the quantity in brackets can be treated as a corrected viscosity. I assume this is what your document refers to, so the $\eta_\text{corr}$ is given by:
$$ \eta_\text{corr} = \frac{\eta}{1 + \frac{b}{P\,r}} $$
In this equation $P$ is the pressure and $b$ is an empirical constant. Some Googling finds that the constant $b$ is approximately $0.00822$ Pa m.
Response to comment:
If we observe an oil drop falling at constant velocity $v$ we know that the drag force is equal to the gravitational force $F_d = mg$, and the mass is just the volume of the drop multiplied by the density of the oil. If we feed this into the Cunningham equation above we get:
$$ \tfrac{4}{3}\pi r^3 \rho g = 6 \pi \left(\frac{\eta}{1 + \frac{b}{p\,r}}\right) r v $$
We know everything in this equation except for $r$, so we can solve the equation to get $r$. It's going to turn into a quadratic equation in $r$, which is easy to solve.
The error due to the Cunningham correction isn't that large. For a one micron sized oil drop it's about 10%, so as a first attempt I would just ignore it and use the uncorrected air vicosity.
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Best Answer
If all the drops contained an even number of electrons (e.g. 2,6,22, and 100), you might look at the charges you measured and determine that the fundamental unit of charge was $2e$ (rather than $e$ which would be the correct answer).
If you only did a few drops, you might have bad luck and get ones that are all divisible by numbers larger than $e$. (maybe you got charges of roughly $7e$, $49e$, and $98e$, all of which are divisible by $7e$).