# User:Randy Jay Lafler/Notebook/Physics 307L/Milikan Oil Drop Summary

My partner was Emran.

## Summary

In this lab we used a Millikan Oil Drop Apparatus to measure the charge on individual oil drops. Our goal was to notice that the charges on the drops were an integer multiple of an elementry charge e. First we sprayed oil into the chamber. Then we had to focus on a drop, and measure its fall and rise times. The fall times were the amount of time it took for the drop to fall from one major reticule line to the next without the voltage. The rise times were the time it took for the drop to rise the same distance under the influence of a voltage of 500v. From these times I calculated average rise and fall velocities. Then I solved for the charges on the drops by using a very complicated formula. Here is a link to our notebook.

## Data overview

Particle 2

• average fall time = 12.68s
• standard div. fall time = 0.963s

average fall velocity $v_f=3.94(28)*10^-5m/s\,\!$

• average rise time = 3.63s
• standard div. rise time = 0.128s

average rise velocity $v_r=1.38(5)*10^-4m/s\,\!$ calculating the charge for particle 2

• I calculated the charge q in three parts.
• First I calculated the radius a of the oil drop
$a=\sqrt{(b/2p)^2+9nv_f/2gp_r}{-b/2p}\,\!$
• Then I calculated the mass
$m=4*pi*a^3*p_r\,\!$
• I calculated the total charge
$a=\frac{mg(v_f+v_r)}{(V/d)v_f}\,\!$
$q=4.25*10(40)^-19C\,\!$

Elementry charge

• Knowing that the accepted value for e is $e=1.6*10^-19C\,\!$, it appears that particle 2 has 3e.
• Dividing q by 3 I get:
$e=1.4(1)^-19C\,\!$

Particle 4

• average fall time = 7.80s
• standard div. fall time =

average fall velocity $v_f=6.41*10^-5m/s\,\!$

• average rise time = 1.58s
• standard div. rise time =

average rise velocity $v_r=3.16*10^-4m/s\,\!$ Calculating charge on particle 4

• The total charge
$q=1.22*10^-18C\,\!$

Elementry charge

$e=1.2*10^-19C\,\!$

Particle 5

• average fall time = 18.64s
• standard div. fall time = 3.31s

average fall velocity $v_f=2.68(40)^-5m/s\,\!$

• average rise time = 3.56s
• standard div. rise time = 0.90s

average rise velocity $v_r=1.4(3)*10^-4m/s\,\!$

• The total charge
$q=2(2)*10^-19C\,\!$

Elementry charge

$e=2(2)^-19C\,\!$

## Error

Our error could have been from adjusting the droplet focusing ring rather than the reticule focusing ring while we took measurements. Our error also was due to the fall times of the oil drops. The drops did not fall straight down every time were took measurements and therefore did not seem to fall at a constant rate.

## Citing and acknowledgements

• We followed the Pasco Millikan Manual for our experiment
• Emran for the pictures.
• Professor Koch for telling us after (SJK 05:21, 21 December 2010 (EST)
05:21, 21 December 2010 (EST)
Sorry it was after!
we had taken our measurements that we were not supposed to adjust the droplet focusing ring on the viewing scope and telling us that adjusting it instead of the reticule focusing ring was one reason for our inconsistent measurements.