Why so much electric charge? I'm working on the Millikan experiment using the following apparatus:
http://www.unitedsci.com/product-catalog/millikan-oil-drop-apparatus-0
when I compute the value of charge of the oil drops, the vast majority is around 200 times the charge of the electron, but according to the manual and many websites with just 10 times is bad data. So my question is what could I been doing wrong?
At first I thought it was because I was using the mercury lamp all the time during the experiment, but after turning it off the oil drops become neutral again. Also I used the data in the manual to make sure my equations were correct. So if someone has experienced this before let me know.
 A: I'm not sure how this manufactured device looks from the inside. During my undergraduate my friend and I built our own device. But I can give you some possible explanations.
First of all there is a slight chance that the mercury lamp causes this due to the emission of UV. Is it build in? Because it also generates a lot of heat (not responsible for the charge). Second, what source for ionization do you use? Some radiation sources emit a lot of radiation, this causes the oil drops to become charged with a huge amount of electrons. Another possibility is more mathematical. The best statistical analysis is minimizing $\chi^2$ distribution of the measured electron charges. More info on https://www.physics.uci.edu/~advanlab/millikan.pdf
A: If the drops had 200 times the charge on the electron that would mean that the drops moved very fast though the apparatus or you had produced very large oil drops.
I would expect you to be able to observe the motion of the droplets over a number of seconds.  If this was the case then the oil drops that you were producing were too large or you have an error in your calculations.  An error which is often made is to use a density in g/cm$^3$ rather than in kg/m$^3$ or the separation of the plates in centimetres rather than in metres.
I note that the manual for the Pasco version of this apparatus gives a formula for the electric field strength in electrostatic units (esu).  Use the SI units the electric field strength - potential difference between the plates (volt) divided by the separation of the plates (metre).
