# Homework Help: Millikans Oil Drop Experiment

1. Jan 21, 2004

### Inquiring_Mike

We recently did a lab where we were given an animation of the oil drop experiment, and we were supposed to find the velocity of the droplet when Fg and FQ were in the same direction (v1) and when Fg and FQ were in opposite directions (v2)... Then we found (v1 - v2) and were told to analyze and prove the existence of a smallest unit of electric charge...
Here is the work I've done so far...

v1 (is directly proportional to) Fg + FQ
v2 (is directly proportional to) Fg - FQ

Therefore,
(v1 - v2) (is directly proportional to) (Fg + FQ) - (Fg - FQ)
(v1 - v2) (is directly proportional to) Fg -Fg + 2FQ
(v1 - v2) (is directly proportional to) 2FQ

Am I completely off track, can the fact that (v1-v2) is directly proportional to 2FQ be used to prove the existence of the smallest unit of electric charge?

2. Jan 21, 2004

### Njorl

The definate answer is ... maybe.

You need to do this many times. Each time, the charge, Q, will be different. If you accurately determine Q, you will eventually see that it is always an integral number times a smaller charge, q. If you are lucky, that smaller charge q=e (the charge on an electron). If you are not lucky, it might equal 2e, or 3e. As you do more trials, the chance of being unlucky becomes vanishingly small.

I always thought this was one of the cleverest experiments ever.

Njorl

edited to add- the nasty part is to determine the mass of each droplet.

Last edited: Jan 21, 2004
3. Jan 21, 2004

### Inquiring_Mike

When I did the experiment, I found that (v1 - v2) was constant... Well pretty much, it fluctuated a little bit...