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

[guyvsdcsniper]

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?

 

[hutchphd]

Homework Statement: Find the charge of an electron
Relevant Equations: q=ne

have an effective viscosity of 1.809

What does this mean?

 

[guyvsdcsniper]

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, $n_{eff}$.

 

[guyvsdcsniper]

Sorry, please refer to page 9 of the lab manual in the link for the formula used to calculate q.

 

[kuruman]

This is obviously too large.

Why is it so obvious that it is too large?

 

[guyvsdcsniper]

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.

 

[hutchphd]

What were the typical times (rise and fall) for your drops? Were they like 10s?

 

[kuruman]

So recheck your calculations. If you want us to check them, you have to post them here, preferably using LaTeX.

 

[guyvsdcsniper]

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.

 

[TSny]

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.

 

[haruspex]

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.

 

[guyvsdcsniper]

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.

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.

 

[hutchphd]

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