Millikan oildrop experiment: archimedes' principle

[Looh]

Hello!

My physics teacher gave me an assignment to work out (theoretically) how Millikan's oildrop experiment works.

The simple principle of the experiment (as far as I know):
$E = \frac{F}{Q}$
$F = mg$
$mg = qE \rightarrow q = \frac{mg}{E}$

However, after reading a bit on Wikipedia it seems as if there is more involved in order to receive an acceptable value of the particle.

Archimedes' principle:
$F = pVg$

Volume of a sphere (the oildrop):
$V = \frac{4πr^3}{3}$

The weight of the drop:
$w = \frac{4πr^3}{3}(p - p_{air})g$

I don't understand why p is being subtracted by p_air, what density does p refer to?


I'm sorry if I'm ambiguous or outright wrong, I'm not particularly good at physics.
You can read more about the method I'm trying to understand here: http://en.wikipedia.org/wiki/Oil_drop_experiment#Method

Thanks in advance!

 

[Borek]

It is ρ (Greek letter rho) not p.

ρ is a density of the oil.

 

[HallsofIvy]

In other words, $\frac{4\pi r^3}{3}\rho$, the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. $\frac{4\pi r^3}{3}\rho_{air}$ is that same volume times the density of oil. Their difference is the "net weight", the actual weight of the oil minus the weight of the air it displaces.

 

[Looh]

It is ρ (Greek letter rho) not p.

ρ is a density of the oil.

In other words, $\frac{4\pi r^3}{3}\rho$, the volume of the oil drop times the density of oil, is the actual gravitational force on the oil drop. $\frac{4\pi r^3}{3}\rho_{air}$ is that same volume times the density of oil. Their difference is the "net weight", the actual weight of the oil minus the weight of the air it displaces.

Ah, yes, I understand now! Many thanks to both of you!

 

[Nickie]

Hi there!

First of all, great job on trying to understand the Millikan oildrop experiment! It is a very important experiment in the field of physics and it is great that you are taking the time to understand it.

In this experiment, Robert Millikan used the principles of both electrostatics and fluid mechanics to determine the charge of an electron. The experiment involved suspending tiny oil droplets in a chamber and using an electric field to measure the charge on the droplets.

Now, let's talk about Archimedes' principle. This principle states that the buoyant force on an object in a fluid is equal to the weight of the fluid that the object displaces. In simpler terms, it means that the weight of an object in a fluid will be reduced by the weight of the fluid it displaces. In the case of the oildrop experiment, the oil droplet is suspended in air, which can be considered a fluid for this experiment. So, the weight of the droplet will be reduced by the weight of the air it displaces.

Now, let's look at the equation you provided for Archimedes' principle:
F = pVg

Here, p refers to the density of the fluid, which in this case is air. The density of air is much lower than that of the oil droplet, so it is being subtracted from the density of the droplet (p-p_air). This is to account for the weight of the air that the droplet displaces.

I hope this helps you understand the concept better. Keep up the good work and don't be afraid to ask questions if you need further clarification!