The mathematics may at first appear a little complicated, but it is actually a lot simpler than a real game I think. The complication arises from having to understand the gravitation and electrical forces involved for the simulation to make sense and from having to manipulate the equations to get a numerical result.

The experiment is tedious to perform in real life. It takes a long time to collect a little data; and the effort involved does not improve understanding. It involves a lot of squinting, focusing, and luck. I think it’s an experiment ideal for a simulation that would produce realistic data that you can then analyze.

Here’s a quote from an MIT lab manual. “The timing measurements are like a video game in which practice makes perfect. N repetitions of any given measurement will reduce the random error of the mean in proportion to N^1/2.”

 http://web.mit.edu/afs/athena/course/8/8.13/JLExperiments/JLExp_02.pdf

Background

The Millikan oil-drop experiment is a famous experiment done in 1909 that measured the magnitude of the charge on the electron. It works because the oil drop is so small that the change of one charge makes a noticeable difference. The experiment is a typical high-school or freshman physics lab. Millikan got the Nobel Prize in 1923 for this work.

You get an atomizer filled with oil, spray the oil as a mist into a two-level chamber. The drops fall through a little hole into the lower chamber.

Here’s a schematic picture from http://www68.pair.com/willisb/millikan/experiment.html 

Although the figure below does not show it, Millikan, pointed an x-ray source into the lower chamber to charge the drops. The lower chamber is between two capacitor plates. The drops are small enough that you have to use a microscope, and you often illuminate the lower chamber with a bright light (also not shown in the figure,)

This is more or less how Millikan did it. Today, people fill an atomizer with tiny latex balls, often have just one chamber, and skip the x-rays. Instead, you let the drop get its charge from rubbing against the nozzle of the atomizer or from encounters with stray charges left in the air by cosmic rays.

Here’s a photo from http://www.juliantrubin.com/bigten/millikanoildrop.html of Millikan's actual apparatus.

The experiment consists of varying the voltage (hundreds of volts) until you can get a drop to be rising at a constant rate. The distance between the plates varies from about 0.5 to 2 cm. More on the equations later.

There exist some simulations on the net. The URL http://www68.pair.com/willisb/millikan/experiment.html has the nicest one I’ve seen. It’s two-dimensional and a Java applet. That bottom line always reads 0V; I think it should increase as you increase the electric field. Here’s how  it looks.

Here’s a website that describes a version of the experiment using latex balls, one chamber, and no x rays.

 http://www.phy.davidson.edu/StuHome/MiLee/JLab/Ex4/title.htm

 You can buy an apparatus that makes the experiment easier and more reproducible. It even has a connection that allows you to project what the microscope seen onto a screen for an entire audience to see

 http://store.pasco.com/pascostore

The Experiment

Time the trips an oil drop makes over a known distance at various voltages. The falling velocities u_d (in the absence of voltage) are all the same because the oil doesn’t evaporate (early versions in Millikan’s lab used water droplets) and hence the weight is constant. The rising velocities (the drop rises when you turn up the voltage) vary with the charge q (which we know, but the experiment proves is an integral number of electronic charges, ne.)

The Equations

The symbols here all represent positive quantities so that the signs are explicit.

The force on a falling oil drop is

where w is the drop’s weight due to gravity, B is the buoyant force of the air and ku_d is the air resistance, which depends on the velocity. u_d is the downward velocity where the d means down. At terminal downward velocity the force is zero; the buoyant force and the air resistance exactly compensate for the weight. Note that the resistance is opposite to the weight because it is always opposite to the motion, and the drop is falling.

The force on a rising drop is

Note that in this case the weight and the air resistance have the same sign because the drop is rising and the air resistance is opposite to the motion. Also the velocity is a constant upward velocity so the subscript is u.

Replace w – B with ku_d and solve for q and you get

This looks deceptively simple. You set and measure the voltage V and measure the velocities. The electric field E is V/s where s is the distance between the plates. Fringe effects at the edges of the plates are negligible so the complication does not arise from that.

The complication comes from getting k, the Stokes constant. It depends on the radius of the drop and the air’s coefficient of viscosity η,

First of all we don’t know the radius of the drop. We do know it’s small, so small that the  equation above needs a correction because the radii of the drops is of the same order of magnitude of the mean free path of an air molecule which is 2.2 x 10^-6 cm.

High school versions often ignore these complications. You can read how to deal with them in an MIT lab manual at http://web.mit.edu/afs/athena/course/8/8.13/JLExperiments/JLExp_02.pdf