How can I find the smallest positive integer number without dividing q by e?

In summary, the conversation discusses an experimental method for determining the elementary charge, e, by observing the movement of oil drops or latex spheres in an electric field. The participants discuss how to identify and group drops with the same number of charges, n, and how to use that information to calculate e. It is noted that a larger amount of data is needed for more accurate results, and that students often end up with too much charge on their drops or spheres. A method is suggested for grouping and averaging the data to determine the value of n for each group. However, concerns are raised about the validity of this method in accurately determining e.
  • #1
jenny777
28
0
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
 
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  • #2
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
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
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.
 
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  • #5
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
You probably cannot exactly :frown:, 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
jenny777 said:
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!

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
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
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
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
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
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
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
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
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
n=6 is not a bad idea, after drawing your own conclusions from your data. Just say you compared to the literature.
 
  • #17
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
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
For drop 1 and 3 I get 10.8/6.4=1.6875=16875/10000=27/16.
 
  • #20
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...
 
  • #21
Vanadium 50 said:
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.

Hi,

1) You can consider the literature to find out what the correct value should be, and discuss why you came to another conclusion. At least that is how I did it I am many lab reports at my non-US university.
2) I was not converting N values into N^2 values (or just accidentally), just taking the differences of the smallest charge-multiples with the measured charges. This was similarly also considered by others. Most of these values can be thrown away since their absolute value is too big to be considered an elementary charge.
 
  • #22
To add - most people are adopting the strategy "guess at the number of electrons on the sphere with the least charge, and see how well that agrees with the other measurements". This strategy fails if that sphere is mismeasured, even if all the others are perfect.

A better way, especially if you can use Excel, is to pick a candidate elementary charge, vary it, and find the charge that gives you the minimum total deviation from integer charges of the ensemble. If you do this, the charge you get will not be the book value, but it is a lot more honest.
 
  • #23
I was actually going to start another rant about how the number of usable values by my method was actually not squared and how it is pure rhetorics to claim it was, even if it seems mathematically right. But I consider your method with the total (standard?) deviation to be sophisticated and cool, so maybe you could elaborate a little more so I don't have to start to babble again. :wink:
 
  • #24
I tried to find a charge that when divided into each one of the measured charges gives a result within an error bar or two-three of an integer for each and every charge... but there was none in my 10 data sets

and by the way, after getting those 100 values, how should I plot it? what would my x-axis be?

Thank you :)!
 
  • #25
Jenny, please look at what other people are saying. The "100 values" technique will not help you. More importantly, you need more data. The data you have is not good enough for you to measure anything. You don't quote uncertainties, but you probably have at best about a 5% measurement of the charge of each sphere. That means that the only data points that will help you constrain e are those from spheres with fewer than 10 electrons on them: a 5% uncertainty means you cannot distinguish a 10 electron sphere that was measured high from an 11 electron sphere that was measured low.

DarthMatter, the way the method works is as follows. Assume the smallest charge is q. For each measurement q, find the smallest non-integral part of the number Q/q. If Q=11.5 and q = 2, Q/q = 5.75, so write down 0.25. (6-5.75 is 0.25). That's the minimum error on each measurement, assuming q = e. Now total all those up. For some value of q, this is at a minimum: that's the q for which the values of Q/q look most like integers, and is the best estimator of the fudamental charge.

This can be made more complicated in multiple ways, but the idea is to answer the question "What value of q makes my measured values of Q/q look most like integers?"
 
  • #26
@Vanadium 50:

So I was wondering why the differences of the original measured values should be worse data than the original values itself. The upside seems to be that you can cover more possibilities of sphere charges, by taking away all the charges of one sphere (which you measured) from another one (which you measured), even multiple times. But the answer seems to be: Because subtracting a 5% measurement from another may give a much bigger uncertainty? Would taking the differences be ok if you had made a very precise measurement of the charges?

I like the method with the minimum error best, I have to admit.

@Jenny777:

The graph shows each of your values minus i times your smallest measured charges for each measured data point, the idea was to just sort out the smallest values to get some candidates for the elementary charge. But only the values which have E-19 at the end can be considered as such candidates, since your smallest measured charge already is in this regime (which are actually around twenty). So the x-axis has no real meaning here, 0 on it would be your first data value minus 0 times the smallest charge, 1 on it would be your first data value minus 1 times your smallest charge, an so on, until you get to the next data point at "x=10". But I think that, considering the measurement uncertainity, the method Vanadium 50 described is more accurate.
 
  • #27
DarthMatter said:
Would taking the differences be ok if you had made a very precise measurement of the charges?

The difference between two values is meaningful only if it is large compared to the error in the two values. This is true no matter how precise the measurements of the two values are.
 
  • #28
Reason 1: There is no more information in 100 differences (or even 45) than there is in the original 10. There is, however, more calculational work required. So you cannot gain.

Reason 2: The uncertainty on the differences is, on average, 40% larger than the uncertainties on the original measurements. So you cannot even break even.

But the fundamental problem here is not the analysis technique. It's that there is not enough data, and in particular, there is not enough data with low charges. A sphere with q=27e does not help very much.
 
  • #29
I would (in this experiment, where the guy or girl in the lab assumes to have an elementary charge) not generally agree on reason 1. Let's say you (or me) knew nothing about integer numbers and you wanted to find the smallest positive integer number. Some guy gave you or me 10 integer numbers with the additional information that each sum of difference of these numbers is another integer number. If this data came out as ## \{3,7,45,34,9,14,18,22,26,15 \} ## all the differences could be taken and no smaller positive difference than 1 would be found, so therefore 1 would be the smallest positive integer number. If you had another data set where the '1' doesn't turn up as a difference you might even check the differences of the differences until you find 1 as the smallest positive difference.

If one has continuous data with non-negligible uncertainity, this may be another story. But I think reason 2 is reason :wink: enough to go with the smallest error method.

To add, I am not sure what the point is to ask a student (maybe even, as in my case, with questionable equipment) in the lab to take just 10 values in an experiment which would need many more and afterwards find out how little he or she will find out with this data. :rolleyes:
 
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What is the Millikan Oil Drop Experiment?

The Millikan Oil Drop Experiment was a scientific experiment conducted by physicist Robert Millikan in 1909, in order to measure the charge of an electron.

How does the Millikan Oil Drop Experiment work?

The experiment involved suspending tiny oil droplets in a chamber using an electric field. By measuring the rate at which the droplets fell, Millikan was able to calculate the charge of the droplets and thus the charge of an electron.

What was the significance of the Millikan Oil Drop Experiment?

The Millikan Oil Drop Experiment provided the first accurate measurement of the charge of an electron, which was a crucial step in understanding the structure of the atom and the nature of electricity.

What challenges did Millikan face during the experiment?

One of the biggest challenges Millikan faced was accurately measuring the tiny oil droplets, which required a high level of precision. He also had to account for other factors such as air resistance and contamination from the oil, which could affect the results.

What impact did the Millikan Oil Drop Experiment have on the scientific community?

The experiment was highly influential in the scientific community, as it provided solid evidence for the existence and properties of subatomic particles. It also opened up new areas of study in atomic and nuclear physics.

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