# Millikan Oil drop experiment

1. Feb 12, 2014

### jenny777

I've got
mg=kvf, when the e-field is zero, (taking downwards direction as positive), k is some constant and vf is the terminal velocity of an oil drop.
Then when the e-field is on, mg+kve=Eq, where Eq is the force from the electric field, and k is the same constant and ve is the drift velocity of an oil drop.
When I isolated q (charge), i got

q=[(ve+vf)/vf]*dmg/V

and q=ne, where n is the number of charge and e is an elementary charge (q is of course the number of charge in an oil drop)

I got something like 8*10^-18 for q, and I'm trying to find n, so that I can plot q vs. n to find the slope of the line (which is e)

but in order for me to find the number of charge (n), don't I have to divide the q by e?
I'm a little confused here because I thought the whole point of doing this experiment is to determine e. but by dividing q by e to obtain n, aren't I misinterpreting the whole the experiment?

How can i find n without dividing q by e?

thank you

2. Feb 12, 2014

### Staff: Mentor

Usually, when people do this experiment, they do a lot of oil drops or latex spheres or whatever, and (hopefully) observe that the values of q bunch together in evenly-spaced clusters. Then they can take the spacing between clusters to be the unit charge.

It's kind of hard to conclude much from a single drop or sphere without assuming something like the value of the charge unit, or the number of charges.

3. Feb 12, 2014

### jenny777

I have 10 data sets for experimentally determined q values,

8.02757E-18
1.56403E-18
2.36235E-18
2.45932E-18
3.57009E-18
4.34722E-18
4.39376E-18
2.38032E-18
9.48661E-19
7.47934E-18

what would be the procedure to determine e??

Thank you!

4. Feb 12, 2014

### DarthMatter

Are these coulomb? I would find out how big the measurement error is and identify drops as having the same charge which only differ by a quantity which is smaller than the measurement error. 4.347E-18 C and 4.393E-18 C are candidates for having the same amount of unit charges, for example. You also probably know that the unit charge must be smaller than the smallest measured value. But 10 values are not many in my opinion.

You can also try taking the differences of integer multiples of the smallest measured value and the other values and taking the smallest such determined difference as the unit charge.

Last edited: Feb 12, 2014
5. Feb 12, 2014

### jenny777

I understand that I should probably treat 4.347E-18 and 4.393E-18 as the charges that have same n, but that still doesn't quite answer my question. I need to find what the value for that n is in order to determine e- since q=ne.

For instance,
I think that 2.36235E-18 2.38032E-18 and 2.45932E-18 probably share the same n value,
4.34722E-18 and 4.39376E-18 probably share the same n value,

But how do I know what that n is? how can i determine the value of n mathematically?
Thank you

6. Feb 12, 2014

### DarthMatter

You probably cannot exactly , because you have not enough data points. But let's say we have only two measured charges, $q_1=n_1e$ and $q_2=n_2e$. Now you can take the difference: $q1-q2=(n_1-n_2)e$. If the absolute value of $(n_1-n_2)e$ is smaller than any other of the measured charges, it is probable that $n_1-n_2$ equals 1 or -1. If this doesn't work so well, you can also multiply one of your data values with integer numbers before taking the difference, since if $q_1=n_1e$ also $mq_1=(m\cdot n_1)e$. If you are lucky you get a pair for which $mq_1-q_2$ is smaller than any of the other differences.

The tricky thing about this that you have to know which charges have to be identified because of the measurement error. But maybe this is not considered so important here.

7. Feb 12, 2014

### MikeGomez

The difference between two data points will either be e or some multiple of e so that is what you are concerned with. For example 2.45932 - 2.36235 = 0.09697, and that value is either a candidate for e 2e or 3e ... or 107e etc. Keep looking at those deltas and try to find the lcd.

You can search for this visually by plotting the values you get, because you should see them clump together at approximate discrete levels, but you need much more data.

8. Feb 13, 2014

### dauto

You find n by looking at many drops and noticing that they form bunches. Each bunch correspond to a different value of n. So you just count the bunches starting from n=0 for those drops that were unaffected by the electric field.

9. Feb 13, 2014

Staff Emeritus
I looked at this data; you need much, much more. One problem is that you have drops with a great many electrons on them. Perhaps as many as 50 or so. So 10 data points means that every drop has a unique number of electrons. (Actually, your #3 and #4 might have the same number)

10. Feb 13, 2014

### Staff: Mentor

Students doing this experiment for the first time almost always end up with too much charge (too many electrons) on their drops or spheres. I had to do it two or three times before I developed enough patience to look for drops that move slowly enough in the applied electric field, while ignoring ones that move faster.

11. Feb 13, 2014

### jenny777

yeah my smallest charge is 9.48661E-19, which is almost 6 times greater than the charge of an electron.
I'm not sure if I should cheat a little bit, or just pretend like 9.48661E-19 is n=1.

Since we already know that e=1.602*10^-19, I can just divide my smallest value (9.48661E-19) by e,and find out that n=6.

I separated my 10 sets of data into groups of 6.

group 1: 9.48661E-19
group 2: 1.56403E-18
group 3: 2.36235E-18, 2.45932E-18, 2.38032E-18
group 4: 3.57009E-18
group 5: 4.39376E-18, 4.34722E-18, 4.39376E-18
group 6: 7.47934E-18, 8.02757E-18

Then I averaged the values in each group, and picked the smallest value (group 1) and divided the rest of 5 groups by that smallest value.
Since we know that the group 1 is n=6, I multiplied the rest of the groups by 6 (instead of going from n=1,2,3...)

But I'm not too sure if I should use that method since the whole point of doing this experiment is to determine e, and by dividing my charges by e to yield n seems a bit redundant to me....

But then again, I can't use my smallest value at n=1, because that would give me like 9000% error...

What should I do!!!!

Thanks

12. Feb 13, 2014

### jenny777

At this point I don't even understand why they wanted me to do this lab. They specifically asked for 10 measurements.... If they really wanted me to find the magnitude of the fundamental charge, shouldn't they have asked for way more data sets?

also, the first part of this lab asks if the drift velocity (under the influence of electric field) has a linear relationship with the Voltage (across capacitors).

v_E= drift velocity
v_f= terminal velocity (free fall)
mg=gravitational force
q=charge
V=potential difference across the capacitor
d=separation of capacitors

v_E=[(q*v_f)/dmg]*V-v_f is the equation

we did this part by taking the v_f measurement without the electric field once. and v_E measurements for voltages 100, 200, 300. what would be the purpose of this part of the lab?

Thank you

13. Feb 13, 2014

### DarthMatter

Lets ask ourselves if your smallest charge $q_1=9.48..E-19$ can be e, given the groups you divided your data into:

group 1: 9.48661E-19
group 2: 1.56403E-18
group 3: 2.36235E-18, 2.45932E-18, 2.38032E-18
group 4: 3.57009E-18
group 5: 4.39376E-18, 4.34722E-18, 4.39376E-18
group 6: 7.47934E-18, 8.02757E-18

I get: $q_2\approx 1.65q_1$
$q_3\approx 2.49 q_1$
$q_4\approx 3.76 q_1$
$q_5\approx 4.6 q_2$

Your group 6 is too wide IMHO, but nevertheless:

$q_6 \approx 7.88 q_1$ or $q_6 \approx 8.47 q_1$.

You could now say that you get no decimal numbers which are even nearly integer multiples of your smallest charge $q_1$, and therefore conclude that e must be smaller. Maybe you can find some charge differences which make better elementary charge units.

I made a plot of your measured data values MINUS i times your smallest charge $q_1$, the number i running from 0 to 9, and found that you can find differences around 1.0E-19 C in this graph. But you would have to explain why you think that one of these numbers makes a better unit charge.

14. Feb 13, 2014

### jenny777

Can you maybe post the plot for me please? I don't quite understand the method (measured data - i times the smallest charge part...)
And what does that plot represent?

So do you think it's a bad idea to find n=6 ?

Thank you :)

15. Feb 13, 2014

### DarthMatter

Lets say you would have measured the charges $q_1=101e$ and $q_2=102e$. Then both the charges would be much bigger than e and no multiples of each other, but the charge *difference* $q_2-q_1$ would be exactly e! Of course, in your measurement this does not have to be the case, but nevertheless each difference is (in the ideal case) also a multiple of e, maybe a smaller one than the original charges.

I seem not to be able to attach pictures from my computer (the forum asks for the URL), but I guess I can just post the data:

The plot simply starts with the first value (in the original post where you gave the values), then simply subtracts your smallest charge from it 9 times, and then goes to the next measured value, so you get 100 values in total.

0 8.02757e-18
1 7.07891e-18
2 6.13025e-18
3 5.18159e-18
4 4.23293e-18
5 3.28426e-18
6 2.3356e-18
7 1.38694e-18
8 4.38282e-19
9 -5.10379e-19
10 1.56403e-18
11 6.15369e-19
12 -3.33292e-19
13 -1.28195e-18
14 -2.23061e-18
15 -3.17927e-18
16 -4.12794e-18
17 -5.0766e-18
18 -6.02526e-18
19 -6.97392e-18
20 2.36235e-18
21 1.41369e-18
22 4.65028e-19
23 -4.83633e-19
24 -1.43229e-18
25 -2.38095e-18
26 -3.32962e-18
27 -4.27828e-18
28 -5.22694e-18
29 -6.1756e-18
30 2.45932e-18
31 1.51066e-18
32 5.61998e-19
33 -3.86663e-19
34 -1.33532e-18
35 -2.28398e-18
36 -3.23265e-18
37 -4.18131e-18
38 -5.12997e-18
39 -6.07863e-18
40 3.57009e-18
41 2.62143e-18
42 1.67277e-18
43 7.24107e-19
44 -2.24554e-19
45 -1.17321e-18
46 -2.12188e-18
47 -3.07054e-18
48 -4.0192e-18
49 -4.96786e-18
50 4.34722e-18
51 3.39856e-18
52 2.4499e-18
53 1.50124e-18
54 5.52576e-19
55 -3.96085e-19
56 -1.34475e-18
57 -2.29341e-18
58 -3.24207e-18
59 -4.19073e-18
60 4.39376e-18
61 3.4451e-18
62 2.49644e-18
63 1.54778e-18
64 5.99116e-19
65 -3.49545e-19
66 -1.29821e-18
67 -2.24687e-18
68 -3.19553e-18
69 -4.14419e-18
70 2.38032e-18
71 1.43166e-18
72 4.82998e-19
73 -4.65663e-19
74 -1.41432e-18
75 -2.36299e-18
76 -3.31165e-18
77 -4.26031e-18
78 -5.20897e-18
79 -6.15763e-18
80 9.48661e-19
81 0
82 -9.48661e-19
83 -1.89732e-18
84 -2.84598e-18
85 -3.79464e-18
86 -4.7433e-18
87 -5.69197e-18
88 -6.64063e-18
89 -7.58929e-18
90 7.47934e-18
91 6.53068e-18
92 5.58202e-18
93 4.63336e-18
94 3.6847e-18
95 2.73604e-18
96 1.78737e-18
97 8.38713e-19
98 -1.09948e-19
99 -1.05861e-18

16. Feb 13, 2014

### DarthMatter

n=6 is not a bad idea, after drawing your own conclusions from your data. Just say you compared to the literature.

17. Feb 13, 2014

### jenny777

so i found this section from the website, (from grade 12 physics textbook surprisingly)

They did a sample calculation for oil drop 1 and 2. and got 45/16=n2/n1
But when I use the method they described for oil drop 1 and 3 instead, I get 3.3750, which can be written as 3375/1000=27/8, which makes the fundamental charge smaller than the one calculated in the sample calculation. but they are saying that it should be the same?

Can you quickly go over the pictures I posted along with this thread?

Thank you :)

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18. Feb 13, 2014

Staff Emeritus
DarthMatter's advice is stunningly bad. If you were to follow it in a class that I was teaching, I would flunk you.

1) The point of a lab is not to see if you can find the "correct" value in a book somewhere and write it down.
2) Converting N measured values into N^2 pairs of values does not add information. It just adds complexity to the analysis.

19. Feb 13, 2014

### DarthMatter

For drop 1 and 3 I get 10.8/6.4=1.6875=16875/10000=27/16.

20. Feb 13, 2014

### jenny777

but that value doesn't make too much sense to me since, drop 2 has less charge than drop 3, but has 45 charges..... I have no idea why the author published such answers/methods on a textbook......