# What is the charge of the oil drop in Millikan's experiment?

• Daniel Luo
In summary, the conversation discusses an experiment done by Millikan to determine the charge of an electron. The experiment involved measuring the velocity of an oil drop falling freely and then increasing its speed by sending electricity through plates. The summary includes equations for finding the radius and mass of the oil drop, and the electric charge of the oil drop. It also mentions that this experiment was used to find the charge-to-mass ratio of an electron, and that a similar experiment was done by J.J. Thomson to determine the charge of an electron. The conversation also mentions that in order to determine the actual charge of an electron, the experiment needs to be repeated with many different oil drops.
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:
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?

Simon Bridge said:
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.

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

Daniel Luo said:
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.

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

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

haruspex said:
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 ...

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

According to what you wrote in post #1, you have only been asked to determine the radius, mass, and charge, for the whole drop. That's all. Nothing about determining anything for the electron nor anything about finding out how many electrons there are on the drop.

In order to discover the charge of an electron by Millican's method, you need to repeat the experiment for a great many different oil drops.
http://webphysics.davidson.edu/appl...ts/pqp_errata/cd_errata_fixes/section4_5.html

Daniel Luo said:
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

ehild

## What is the Millikan Oil Drop Experiment?

The Millikan Oil Drop Experiment, also known as the Millikan's experiment or the oil-drop experiment, was a pioneering experiment conducted by American physicist Robert A. Millikan in 1909. It determined the charge of an electron by measuring the electric force on tiny charged droplets of oil suspended between two metal electrodes.

## Why is the Millikan Oil Drop Experiment important?

The Millikan Oil Drop Experiment is important because it provided the first direct and accurate measurement of the charge of an electron, which was a crucial piece of information in the development of modern physics. It also helped validate the theory of quantized electric charge, which states that electric charge exists in discrete units rather than being continuous.

## What was the procedure of the Millikan Oil Drop Experiment?

The procedure of the Millikan Oil Drop Experiment involved suspending tiny charged oil droplets between two charged metal plates. By measuring the force required to keep the droplets suspended, Millikan was able to calculate the charge on each droplet. He repeated the experiment multiple times to obtain accurate and reproducible results.

## What were the results of the Millikan Oil Drop Experiment?

The results of the Millikan Oil Drop Experiment showed that the charge on the droplets was always a multiple of a single, fundamental unit of charge. It also provided an accurate value for the charge of an electron, which was found to be 1.602 x 10^-19 coulombs.

## What impact did the Millikan Oil Drop Experiment have on the scientific community?

The Millikan Oil Drop Experiment had a significant impact on the scientific community as it provided strong evidence for the existence of individual electrons and their quantized charge. It also helped pave the way for future discoveries in the field of quantum mechanics and solidified the concept of electric charge as a fundamental property of matter.

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