# Physics307L:People/Meyers/Millikan Oil Drop Lab Summary II

MILLIKAN OIL DROP LAB (again)

The Apparatus Setup
Other Equipment
Voltage Supply
SJK 17:58, 27 November 2010 (EST)
17:58, 27 November 2010 (EST)
For the time being, see comments on your formal report for this lab

## Purpose

The purpose of this lab was to determine the value of the fundamental charge. A cursory goal was to understand more about the experiment and physics fundamental charge.

## Procedure

The procedure for this lab is posted here

Procedure Notes

Three important things that we forgot about on the second day of lab.

• Clean the equipment fully before each use.
• Make sure the focus on the microscope is sharp.
• If there are no droplets after the above two notes are completed, you can remove the cap off the top capacitor plate to aid in production of drops.

## Data

• Particle 3:

Mean fall velocity: 3.918 * 10 − 5m / s

Mean rise velocity: 8.66 * 10 − 5m / s

• Particle 4:

Mean fall velocity: 3.036 * 10 − 5m / s

Mean rise velocity: 1.134 * 10 − 4m / s

• Particle 5:

Mean fall velocity: 3.269 * 10 − 5m / s

Mean rise velocity: 1.026 * 10 − 4m / s

• Particle 7:

Mean fall velocity: 3.595 * 10 − 5m / s

Mean rise velocity: 1.271 * 10 − 4m / s

• Particle 8:

Mean fall velocity: 3.150 * 10 − 5m / s

Mean rise velocity: 3.85 * 10 − 5m / s

• Particle Thorium:

Mean fall velocity: 3.6250 * 10 − 5m / s

Mean rise velocity: 4.2860 * 10 − 4m / s

## Calculation

Using the rise and fall velocities we can calculate the fundamental charge q

$q={4/3 \pi \rho g}{\Bigg[}{\sqrt{\bigg({\frac{b}{2p}}\bigg)^2+\frac{9ηv_f}{2g\rho}}-\frac{b}{2p}}{\Bigg]^3}\frac{v_f+v_r}{Ev_f}\,\!$

I used this equation to calculate the pressure.

Using this equation and a MATLAB program I wrote I calculated $q\,\!$ to be:

Particle 3: $q=1.566*10^{-19}\pm 1.1*10^{-20}C\,\!$

Particle 4: $q=1.539*10^{-19}\pm 5.1*10^{-21}C\,\!$

Particle 5: $q=1.525*10^{-19}\pm 6.7*10^{-21}C\,\!$

Particle 7: $q=1.954*10^{-19}\pm 4.5*10^{-21}C\,\!$

Particle 8: $q=1.534*10^{-19}\pm 6.9*10^{-21}C\,\!$

Thorium Particle: $q=1.5977*10^{-19}\pm 3.1*10^{-20}C\,\!$

Averaged together we get a q of:

$q=1.643*10^{-19}\pm 6.8*10^{-21}C\,\!$

## Error

With the accepted value of q as:

$q=1.602176487(40)*10^{-19}C\,\!$

This is inside my first standard deviation of mean of my calculated q.

Also the percent error is:

$%error=\frac{(1.643*10^{-19}C)-(1.602*10^{-19}C)}{1.602*10^{-19}C}x100%=2.56%\,\!$

This is very good.

## Conclusion

Compared to the first time we did this lab we have done much better. We found a $q\,\!$ and a standard deviation that contains the accepted value of $q\,\!$. From last time, we changed one fundamental thing, time. We took the time to take more data and make sure that it is good data.

## Citation

1)Pressure versus altitude equation here

2)Altitude of Albuquerque here

3)Wiki math help here

4)Accepted value of q here

## Thanks

1) Nathan for lab help along with data input and help with google docs.

2)Steve Koch for help in trouble shooting the lab.