Calculating Charge of an Electron w/ the Millikan Oil Drop Experiment

In summary, In order to calculate the charge of an electron, you must account for the effective viscosity of air. This number is multiplied by a corrective factor to determine the effective velocity of the oil particle. To get a charge on the order of ##10^-19##, you need a velocity on the order of ##10^-5 to 10^-6## m/s.
  • #1
guyvsdcsniper
264
37
Homework Statement
Find the charge of an electron
Relevant Equations
q=ne
I am doing the Millikan Oil Drop experiment to determine the charge of a single electron. I have been following the lab manual provided by the manufacturer, https://hepweb.ucsd.edu/2dl/pasco/Millikans Oil Drop Manual (AP-8210).pdf.

The manual defines a simple method to calculate for the charge of the electron, which accounts for the effective viscosity of air. Page 12 and 13 of the linked lab manual defines all constants and list the method as described above.

I am using 500V from a DC power supply and have an effective viscosity of 1.809

Determining the charge essentially comes down to the velocity of the oil particle during free fall and its rising velocity under the influence of the electric field when the top capacitor plate is positive.

After analyzing over 15 particles, both respective velocities come out to be on the order of ##10^-5 to 10^-6## m/s.

Velocities on this order give a charge on the order of ##10^-10 to 10^-12## C. This is obviously too large.

I have checked the dimensional analysis of the formula provided by the manual and when done with S.I. Units, everything checks out.

After playing with numbers, I would need a velocity on the order of nm/s to get a charge on the order of ##10^-19##.

That velocity seems very excessive, given the other lab reports I have seen. Am I doing something wrong here or missing out a conversion factor?
 

Attachments

  • Screenshot 2023-07-14 at 3.16.33 PM.png
    Screenshot 2023-07-14 at 3.16.33 PM.png
    18.2 KB · Views: 108
  • Screenshot 2023-07-16 at 6.00.32 PM.png
    Screenshot 2023-07-16 at 6.00.32 PM.png
    22.2 KB · Views: 75
Physics news on Phys.org
  • #2
guyvsdcsniper said:
Homework Statement: Find the charge of an electron
Relevant Equations: q=ne

have an effective viscosity of 1.809
What does this mean?
 
  • Like
Likes TSny
  • #3
hutchphd said:
What does this mean?

Sorry I should have been more clear. Its Viscosity of Dry Air as a Function of Temperature. To calculate the radius of the oil drop, stokes law is used. But according to the lab manual

"Stokes’ Law, however, becomes incorrect when the velocity of fall of the droplets is less than 0.1 cm/s. (Droplets having this and smaller velocities have radii, on the order of 2 microns, comparable to the mean free path of air molecules, a condition which violates one of the assumptions made in deriving Stokes’ Law.)"

So the viscosity is multiplied by a corrective factor to give the effective velocity, ##n_{eff}##.
 
  • #4
Sorry, please refer to page 9 of the lab manual in the link for the formula used to calculate q.
 
  • #5
guyvsdcsniper said:
This is obviously too large.
Why is it so obvious that it is too large?
 
  • #6
kuruman said:
Why is it so obvious that it is too large?
Because in S.I. units, 1 electron corresponds to ##1.6 x 10^{-19}## Couloumbs, and what I am calculating is on the order of ##10^{-10-12}## Couloumbs, making my value larger.
 
  • #7
What were the typical times (rise and fall) for your drops? Were they like 10s?
 
  • #8
So recheck your calculations. If you want us to check them, you have to post them here, preferably using LaTeX.
 
  • Like
Likes MatinSAR and hutchphd
  • #9
hutchphd said:
What were the typical times (rise and fall) for your drops? Were they like 10s?
Falling was about a minute and rise was around 5 seconds. In hindsight, id use a lower voltage to reduce the rise speed to get a more accurate velocity.
 
  • Like
Likes hutchphd
  • #10
guyvsdcsniper said:
an effective viscosity of 1.80
What are the units for this number? Are you missing a factor of 10-5?
See viscosity of air.
 
  • Like
Likes MatinSAR, hutchphd and guyvsdcsniper
  • #11
TSny said:
What are the units for this number? Are you missing a factor of 10-5?
See viscosity of air.
Or even ##10^{-10}## if the adjustment was made the wrong way.
The notation on the Y axis at the Pasco link says "##Nsm^{-2}\times 10^{-5}##". Taken literally, that would mean that if the value marked on the axis is 1.8 then 1.8 is the ##Nsm^{-2}## value after multiplying by ##10^{-5}##, i.e. the value is ##1.8\times 10^5##. But what it intends is the reverse.
 
  • Like
Likes MatinSAR and guyvsdcsniper
  • #12
TSny said:
What are the units for this number? Are you missing a factor of 10-5?
See viscosity of air.
haruspex said:
Or even ##10^{-10}## if the adjustment was made the wrong way.
The notation on the Y axis at the Pasco link says "##Nsm^{-2}\times 10^{-5}##". Taken literally, that would mean that if the value marked on the axis is 1.8 then 1.8 is the ##Nsm^{-2}## value after multiplying by ##10^{-5}##, i.e. the value is ##1.8\times 10^5##. But what it intends is the reverse.

TSny was correct, I completely missed the factor of ##10^{-5}##. I am on the right order of magnitude now. I completely missed the units and the ##10^{-5}## on the lab manual, thank you for pointing that out haruspex.

Just a bonehead mistake and reading way too fast. You can close this thread now.
 
  • #13
guyvsdcsniper said:
Just a bonehead mistake and reading way too fast. You can close this thread now.
But it will forever remain a warning beacon marking the shoals of carelessness :warning:
 
  • Like
  • Haha
Likes TSny, Steve4Physics and guyvsdcsniper

FAQ: Calculating Charge of an Electron w/ the Millikan Oil Drop Experiment

What is the Millikan Oil Drop Experiment?

The Millikan Oil Drop Experiment is a famous scientific experiment conducted by Robert A. Millikan in 1909 to measure the charge of a single electron. The experiment involved observing tiny charged oil droplets between two metal plates and balancing the gravitational force with the electric force to determine the charge on the droplets.

How does the Millikan Oil Drop Experiment work?

In the Millikan Oil Drop Experiment, small oil droplets are sprayed into a chamber and allowed to fall between two electrically charged plates. By adjusting the voltage across the plates, the electric force can be balanced against the gravitational force acting on the droplets. When the droplet is suspended in equilibrium, the charge on the droplet can be calculated using the known values of the electric field, the droplet's mass, and the gravitational constant.

What is the significance of the Millikan Oil Drop Experiment?

The significance of the Millikan Oil Drop Experiment lies in its precise measurement of the elementary charge, which is the charge of a single electron. This experiment provided strong evidence for the quantization of electric charge and was crucial in the development of atomic theory and our understanding of fundamental particles.

What equipment is used in the Millikan Oil Drop Experiment?

The equipment used in the Millikan Oil Drop Experiment includes an atomizer to spray tiny oil droplets, a chamber with two parallel metal plates to create an electric field, a light source to illuminate the droplets, a microscope to observe the droplets, and a power supply to adjust the voltage across the plates. Additionally, a stopwatch or timing device is used to measure the fall and rise times of the droplets.

How do you calculate the charge of an electron using the Millikan Oil Drop Experiment?

To calculate the charge of an electron using the Millikan Oil Drop Experiment, you first measure the radius of the oil droplet by observing its fall under gravity. Then, by balancing the gravitational force and the electric force, you determine the charge on the droplet. The formula used is \( q = mgd/V \), where \( q \) is the charge, \( m \) is the mass of the droplet, \( g \) is the acceleration due to gravity, \( d \) is the distance between the plates, and \( V \) is the voltage across the plates. By analyzing multiple droplets, you can determine the smallest common charge, which corresponds to the charge of a single electron.

Back
Top