Millikan Oil Drop Experiment

[Battlemage!]

Homework Statement

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.

Homework Equations

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

F_d and F_g

F = ma


F_d = 6πηRv,

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

The Attempt at a Solution

First, because velocity is the terminal velocity,

F_d = F_g

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!

 

[anathema]

Hello,

Your approach is correct, but there are a few minor errors in your calculations. Here is the corrected version:

Starting with Fd = Fg:

6πηRv = mg

Substituting m = ρV:

6πηRv = ρVg

Substituting V = (4/3)πR^3:

6πηRv = (4/3)πR^3ρg

Solving for R:

6πηv = (4/3)πR^2ρg

R^2 = (6πηv)/(4/3)πρg

R = √((6πηv)/(4/3)πρg)

R = √((9ηv)/(8ρg))

Substituting back into the density equation:

m = ρ(4/3)π(R^3)

m = ρ(4/3)π(√((9ηv)/(8ρg))^3)

Simplifying:

m = ρ(4/3)π((9ηv)/(8ρg))^(3/2)

m = ρ(4/3)π(9/8)^(3/2)η^(3/2)v^(3/2)/(ρ^(3/2)g^(3/2))

m = (3/2)ρ(9/8)^(3/2)πη^(3/2)v^(3/2)/g^(3/2)

Finally, substituting in the known value for e/m ≈ 1.76 x 10^11C/kg:

e = (3/2)ρ(9/8)^(3/2)πη^(3/2)v^(3/2)/g^(3/2) * (1.76 x 10^11)

I hope this helps! Let me know if you have any further questions.

 

[thomate1]

Yes, your solution is correct. You have correctly derived the formula to find mass and by proxy, the charge of the particle in the Millikan Oil Drop Experiment. Your use of the density formula and substitution of variables is accurate and shows a clear understanding of the experiment. Well done!