How do I determent the radius and the viscosity of air on an oil drop falling? (Millikan's Oil Drop Experiment) I bought this article with the intentions of learning how to do the calculations in Millikan's oil drop experiment. 
http://iopscience.iop.org/article/10.1088/0143-0807/33/5/1227/meta;jsessionid=BF5A7842586196D302D35310494ED087.ip-10-40-1-98
But in chapter 3, they have this equation:
$Q = \frac{6\pi d}{U}\sqrt{\frac{9\eta^{3}_{corr}}{2(\rho_{oil}-\rho_{air})g}}(v_{fall}+v_{raise})\sqrt{(v_{fall})}$
You are supposed to apply a voltage, U, and measure the speed it rise, $v_{raise}$. Than let it fall and measure the speed it fall, $v_{fall}$. 
d is the distance between the capacitor plates and $\rho$ is the density of air & oil.
The problem is this, it stands:

If the motion of one droplet is measured
  in that way, we can determine its radius and then we can correct the viscosity η of air for that
  case (commonly known as the Cunningham or slip correction factor).

The problem is that I do not know how to do this and the journal do not say how I should do it.
I have found this: 
http://www.tandfonline.com/doi/pdf/10.1080/02786828508959055
But as far I can tell it does not eater say how to do it. Can some one help me sort out all of this?
 A: 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.
