Millikan's Oil Drop (Finding modified charge)

[fbelzile]

Homework Statement

Hi everyone,

I am working on Millikan's oil drop experiment. The first 2 columns were given to us, while we figured out the rest (chart/formulas below). I now have to complete another column beside "Mass" with the caption: "Excess Electrons" My teacher said that he modified the charge of electrons so we can't use the 1.602x10E-19 constant, instead, we have to find our own. He talked about getting a lowest common multiple of the numbers in the Mass column. How do I do that? He said there was an easy way that would take 5 minutes, yet he never told us how... Once I know the charge of the electrons, I can then find the excess number easily.

Here are the constants given to us:

[tex]\rho[/tex]oil = 930.000 kg/m3                d between plates = 2.000 x 10–2 m
[tex]\rho[/tex]air = 1.230 kg/m3                  d between lens lines = 1.300 x 10-3 m
[tex]\eta[/tex]air = 2.016 x 10-5 Ns/m2           g = 9.807 m/s2

Here is my table of values:

Potential 
Difference (V)	  Time (s)	Velocity (m/s)	    Radius (m)	         Mass (kg)
				
222.656	           21.29	  6.10615E-05	     7.79854E-07	1.84761E-15
-347.614	   18.35	 7.08447E-05 	     8.40007E-07	2.30898E-15
78.688	            31.13	  4.17604E-05	      6.44928E-07	 1.04498E-15
-804.663	   6.61	          0.000196672	     1.39959E-06	1.068E-14
494.85	            23.02	  5.64726E-05	      7.49978E-07	 1.6433E-15
-231.743	   18.35	 7.08447E-05	     8.40007E-07	2.30898E-15
-109.453	   24.98	 5.20416E-05	     7.19954E-07	1.45374E-15
107.376	           30.65	 4.24144E-05	     6.49959E-07	1.06962E-15
-275.638	   16.35	 7.95107E-05	     8.89902E-07	2.74535E-15
-559.107	   33.69	 3.85871E-05	     6.19941E-07	9.28161E-16
-357.002	   21.84 	 5.95238E-05	     7.69972E-07	1.77826E-15
103.731	           49.79 	 2.61097E-05	     5.09953E-07	5.16608E-16
-231.743	   18.35 	 7.08447E-05	     8.40007E-07	2.30898E-15
108.614	           39.86 	 3.26141E-05	     5.69944E-07	7.21222E-16
-153.001	   21.84 	 5.95238E-05	     7.69972E-07	1.77826E-15
-237.659	   23.65   	 5.49683E-05	     7.39921E-07	1.57808E-15
360.178	           17.92	 7.25446E-05	     8.50026E-07	 2.39259E-15
-204.993	   31.62	 4.11132E-05	     6.39912E-07	1.02078E-15
183.994	           17.67	  7.3571E-05	      8.56018E-07	 2.44354E-15
-78.062	           42.81	  3.03667E-05	      5.49957E-07	 6.47973E-16
480.237	           17.92	  7.25446E-05	      8.50026E-07	 2.39259E-15
125.673	           27.6	          4.71014E-05	      6.8493E-07	  1.25173E-15
146.65	           32.63	  3.98406E-05	      6.2993E-07	  9.73753E-16
-220.688	  17.11	         7.5979E-05	      8.69913E-07	 2.56448E-15
-670.694	  18.8	          6.91489E-05	      8.29893E-07	 2.22658E-15
282.012	           28.88	  4.50139E-05	      6.6958E-07	 1.16944E-15
-168.613	  29.73	         4.37269E-05	     6.59939E-07	1.11965E-15
96.35	            29.73	   4.37269E-05	       6.59939E-07	  1.11965E-15
-152.575	  38.5	          3.37662E-05	      5.79923E-07	 7.59772E-16
185.546	           21.29	  6.10615E-05	     7.79854E-07	 1.84761E-15
-323.371	  19.26     	  6.74974E-05	     8.19923E-07	 2.14729E-15
256.295	           14.17	  9.17431E-05	      9.55908E-07	 3.40267E-15
335.137	           12.95	  0.000100386	      9.99922E-07	 3.89466E-15
-746.08	           17.51	  7.42433E-05	      8.5992E-07	  2.47711E-15
-231.743	   18.35	  7.08447E-05	      8.40007E-07	 2.30898E-15
256.889	           27.2	           4.77941E-05	       6.89948E-07	  1.27944E-15
307.489	           31.62	  4.11132E-05	      6.39912E-07	  1.02078E-15
-240.118	   17.92	  7.25446E-05	      8.50026E-07	 2.39259E-15
139.941	           25.69	  5.06033E-05	      7.09936E-07	 1.39389E-15
122.996	           31.62	  4.11132E-05	      6.39912E-07	 1.02078E-15

Homework Equations

velocity = [U]d[tex]_{}lens[/tex][/U]        radius = [U] [tex]\sqrt{}(9/8)(n)(Vo)[/tex][/U]    (square root of the whole fraction)
             t                      g(Pa-Po)


mass = (4/3)(pi)(r^3)(Po)              q = [u](m)(g)(Dp)[/u]
                                             Volts

q = (N)(e)    
     ^
     |
     -----(Where N is the excess electrons, and e is the charge)

The Attempt at a Solution

I tried taking the smallest number and dividing it by all the masses... This got me 40 other numbers. I need one number to be the charge of the electron.



Thanks for any help!

 

[catkin]

Try sorting the numbers by size, replacing any very similar ones with their average, removing duplicates, calculating differences ...

 

[fbelzile]

Been there done that... Tried every possible combination of numbers and operations. I tried using the 1.602x10E-19 constant anyways, but it didn't work (numbers were not like: 32.00000234)

Any real pointers?

 

[catkin]

Let's see the steps of your sorting, group averaging and difference calculations.

 

[fbelzile]

Let's see you say: "I don't know... sorry" Like I said, the values are given to us and should work out without averaging (he told us).

I also handed in the lab (without that part answered) today. It's OK for now, but I'm still curious on how I'd go about solving this.

 

[catkin]

... Tried every possible combination of numbers and operations ...

Methinks you do exaggerate.

 

[fbelzile]

I asked a friend how he did it. He said he subtracted the 2 smallest numbers and then divided that number by ten "a bunch of times" until it worked...

Any reasoning why that worked?

And catkin: "An exaggeration is a truth that has lost its temper." ~Kahlil Gibran

 

[catkin]

Any real pointers?

Yes - my previous post.

Averaging is required iff [sic] ordering the data produces values that are the same except for experimental error as will be the case with typical Milikan experiemnts.

 

[catkin]

Why didn't you calculate the column for q?

My algorithm was for using on the q fiqures and was designed to make charge e visible from "noisy" and unordered q values -- which is what an experiment would produce.

I asked a friend how he did it. He said he subtracted the 2 smallest numbers and then divided that number by ten "a bunch of times" until it worked...

Any reasoning why that worked?

And catkin: "An exaggeration is a truth that has lost its temper." ~Kahlil Gibran

I guess you mean he subtracted the smallest number from the next smallest number. If he was working on q values and they happened to be one e different then he would get a rough value of e. Rough because he was not using averging to reduce the effect of experimental inaccuracies in the data.

It wouldn't make any sense to divide by 10 any number of times except to convert charge from Colulombs to microCoulombs or whatever.