# Millikan Oil Drop Experiment

1. Nov 7, 2008

### Battlemage!

1. The problem statement, all variables and given/known data

If the charge to mass ratio of a particle, e/m, is known derive a formula to find m, and then by proxy e.

Drops of oil are allowed to drop down through a potential difference. With the electric field you can keep a particle suspended for however long you need, droping it and repeating again and again. When the terminal velocity of the particle is reached, it's velocity is found by timing how long the drop takes to pass through some graduated crosshairs of a microscope used to watch the particle fall.

2. Relevant equations

At terminal velocity, the force from drag - the force due to gravitation = 0

Fd and Fg

F = ma

Fd = 6πηRv,

π≈3.14, η = viscocity, R = radius of drop, v = velocity​

3. The attempt at a solution

First, because velocity is the terminal velocity,

Fd = Fg

6πηRv = mg

Now, obviously, mass is unknown, and you really can't measure it. So, using the density formula:

ρ = m/V, V is volume
==>

m = ρV​

Now, because the drop is approximately a sphere, V is given by:

V = (4/3)πR3

So, substituting this into the equation relating drag and gravitation:

6πηRv = ρ(4/3)πR3g

Now, because v, η and ρ are measureable, R can be derived. Solving for R:

6πηv = ρ(4/3)πR2g

R2= 6πηv/(ρ(4/3)πg)

R = √(6πηv/(ρ(4/3)πg))

R = √(18ηv/(ρ4g))

R = √(9ηv/(2ρg))​

From there, substitue R back into the density equation:

m = ρ(4/3)π(√(9ηv/(2ρg)))3

And that should give mass. From there, you can find e by using the already determined ration of e/m ≈ 1.76 x 10 11C/kg.

Is this correct?

Thanks!