Millikan's experiment -- data confusion

[Taylor_1989]

Homework Statement

Hi guy I am having a real issue trying to find the fundamental charge from my data.

So here is the background.

Basically I carried out and experiment where we measured an oil droplet the was floating a specific voltage by taking the measurement of 12 oil droplets and calculating the velocity from the distance and time , we could calculate the fall velocity from there, we then use the two equations which I have listed below to calculate the charge and the radius.

Here is my data for the charge and radius using the formula (1), (2)

For now I am not concerning myself with the errors etc as I will add them in etc when I get the correct way I analysing the data.

So the scatter graph of the data is like so:

Now my lowest point is $$2.4*10^{-19}C$$

So this too me is the charge on the smallest oil drop but this dose not tell me the actual charge on a electron, so we have been given a correction factor which I have outlined below but my issue is who do I plot this correction factor?

My thoughts at this moment are that if I use (4) this should give me the actual charge of an electron.

So if I rearrange the equation, I get.

$$q^{-2/3}=e^{-2/3}(1-\frac{a}{r})(5)$$

But I am not really sure how to plot this? Is this equation correct? I am thinking maybe plotting $y^{-2/3}$ vs $1/r$ and then put in a linear trend line and compare the given equation with the (5). Could someone please give some advice on this matter. Any help would be much appreciated.

Homework Equations

$$r=(\frac{9*n_{air}*v_{fall}}{2(\rho-\rho_{air})*g})^{1/2}(1)$$

$$\frac{18\pi*d}{V}*(\frac{n^3*v_{fall}^3}{2(\rho-\rho_{air})*g})^{1/2}(2)$$

correction factor

$$r_{c}=r(1-\frac{a}{r})^{1/2} (3)$$

$$q_{c}=q(1-\frac{a}{r})^{3/2} (4)$$n=viscosity of air[/B]

The Attempt at a Solution

 

[mfb]

What is this correction factor and where does it come from?

With the Millikan experiment you can never be "100% sure" that you get the elementary charge (you could be unlucky and have even multiples of the charge on all drops, for example), but if you do the experiment carefully and with many drops you'll see the values cluster around integer multiples of some value - the elementary charge. Your uncertainty on the individual measurements has to be smaller than the elementary charge, of course, otherwise you don't see anything.
If your values in column C are in Coulomb then the uncertainty is too large to find the elementary charge.

 

[Charles Link]

We did this experiment in 1975 at the University of Illinois at Urbana-Champaign, and I still have the write-up they gave us to go with it. $\\$ Your second equation is for the charge on the oil drop. It should give a reasonably accurate result without any additional correction factors. $\\$ I would suggest you check your computer program carefully for any numerical error before trying to do any additional refinements. $\\$ Your equations are basically the same ones we used except that we had a $v_E$ that was not equal to zero, whereas I believe you adjusted the voltage $V$ to get the drop to stand still. $\\$ For us, they did some additional refinements for increased accuracy: The students in the class submitted the data, and they processed it for us. I'm not sure how much their refinement increased the accuracy, but the average fundamental charge I got was 1.59 E-19 +/- .01 E-19 with about a dozen different drops. The average charge I had on the drops was about 4e. A couple times I had Q=7e or 8e, one time I had 1e,three times I had 2e,twice I had 3e, etc... $\\$ Additional detail is the correction factor you show is, according to the write-up they gave us, a correction to the viscosity $n$ that will also result in a correction to $r$. The primary effect that it has on the computed $Q$ is in the correction to the viscosity $n$, but you need to first get a reasonably accurate result without this refinement. With the refinement, it will likely improve your accuracy, but it would only be worthwhile if you can get e to within about +/-10% or better without it. $\\$ Additional detail : $n=\frac{n_o}{1+\frac{a}{r}}$ is a correction they give us for the viscosity $n$ in your equation (2) for the charge $Q$ where $a \approx 1.0 \, E-7$ meters. This $a$ is essentially the same $a$ in your equation that accounts for this correction. The correction for $Q$ is basically to use this refined $n$ instead of $n_o$. $\\$ Also, the number $n_o$ that we used for the viscosity of air was $n_o=1.832 \, E-5$, so your number of $n_o=1.79 \, E-5$ looks good. $\\$ Looking over a couple of drops you had that were nearly identical in size: 1,2, and 5, it is difficult to see from the charges that you computed how these could be multiples of 1.6 E-19. What was the density of the oil that you used? Also, what did you use for the density of the air?