# Millikan Oil Drop Experiment

Daniel Luo

## Homework Statement

When an oil drop falls freely, the velocity first increases, but afterwards it quickly reaches a constant speed, since the air resistance becomes equal to the weight of the oil drop. The air resistance is given by:

Fair=6$\pi$r$\eta$v,

where $\eta$ is the viscosity of air with a value of 1.759*10-5 N s/m2, Oil's density is 918.7 kg/m3. In Millikan's experiment, the speed of the oil drop is 5.449*10-4 m/s.

- Find the radius and mass of the oil drop.

Afterwards, electricity was sent through the plates, and the speed of the oil drop increases by 5.746*10-4 m/s. The upward electric force acting on the oil drop is given by:

FE=qE,

where E is the electric field strength: E = 3.178*105 N/C.

Find the electric charge of the oil drop, q.

## Homework Equations

For no. 1:

Weight = Air resistance
mg = Fair=6$\pi$r$\eta$v,
$\rho$Vg=Fair=6$\pi$r$\eta$v,

No. 2:

FE=qE
mg = FE + Fair

## The Attempt at a Solution

All right. What I did in no. 1 was to say that the weight was equal to the air resistance. Assuming the drop to be spherical, I said:

$\rho$(4/3$\pi$r3)g = 6$\pi$r$\eta$v. Then I solved for r:

r = 2.19*10-6 m

And I found m:

m = $\rho$(4/3$\pi$r3) = 4.04*10-14 kg (This seems to be too light even for a raindrop?)

For no. 2: I said that the weight was equal to the sum of air resistance and the electric force:

mg = 6$\pi$r$\eta$v + qE

Solving for q gave: q = -1.31*10-18 C.

However, this answer seems wrong, as Millikan tried to find the charge of the electron e from this experiment, which is 1.602*10-19 C.

Have I done something wrong? In case I did, what was my mistake?

Thanks a lot!

Last edited:

Homework Helper
Well a raindrop is water and may have a radius around 1-2mm.
The oil drop has a radius of around 3 microns ... about 1000x smaller.
Millican needed a microscope to see them.

Technically, Millican found the electron charge-to-mass ratio from this experiment.
To find out how much charge a single electron has, he needed to know how many electrons were responsible for the charge.

How many excess electrons are there in your oil drop?

Homework Helper
Gold Member
Technically, Millican found the electron charge-to-mass ratio from this experiment.
No, I think that was obtained through a much earlier experiment by someone else. The point of Millikan's experiment was to find the absolute charge. (The mass of the electrons would have been immaterial in the oil drop.). The technique was to measure the charge on many oil drops and find the GCD.

Daniel Luo
How do I find out? Do I use the mass, molar mass and Avogadro's number?

Homework Helper
Gold Member
How do I find out? Do I use the mass, molar mass and Avogadro's number?
You have found the charge, what's the problem? If you like, you can estimate the number of excess electrons in the drop.

Daniel Luo
It just doesn't seem right? I get the no. of electron in the drop to be 8 (rounded).

Daniel Luo
Am I not supposed to find the actual electric charge of an electron? Isn't that he point of the experiment?

Homework Helper
No, I think that was obtained through a much earlier experiment by someone else.
Oh you are right - that was J. J. Thomson who did the q/m thingy.
I keep getting them mixed up.

You still can't do it with just one drop.

Daniel Luo said:
It just doesn't seem right? I get the no. of electron in the drop to be 8 (rounded).
1. you were not asked to find the charge of an electron.
2. I doubt you found the total number of electrons in the oil drop ... recall: neutral oil has electrons (-) and atomic nuclei (+) in equal amounts. Since your oil drop is negatively charged ...............

Daniel Luo
So I should just go ahead with my answer which corresponds to approx. the charge of 8 electrons?

That is quite a good result, the charge is not much more than 8 electrons. Usually, the oil drop gains more than one electrons in that experiment. Anyway, when I did it as a student during Laboratory Practice, I got usually half-integer results 