The charges on the oil droplets are expected to be some integral multiple of a unknown fundamental unit of charge e. When the data on the charges in the oil droplets are sorted, we expect to see a relation of the form q_n = e.n.
If we plot q_n Vs n we expect to see a straight line whose slope would be this fundamental unit of charge (e). Since Millikan's data may have errors it may not be a perfect line but one can make a least-squares linear fit of the data.
Remember that all entries in the sequence {q_n} may not be present. In fact it need not even start with q_1. The sequence interval \Deltaq_n = q_n-q_{n-1} = e, gives a rough estimate of e that is used mainly to figure out n.
Millikan's own data is given below. The third column shows that the sequence interval, a rough estimate of e. It is clear that the first data (19.66) is roughly 4 times the sequence interval (4.9). So it is obvious that q_1, q_2 and q_3 are missing. Similarly if you look at the 16^{th} row you see that sequence interval suddenly jumps showing a missing q_15
Millikan's own data is reproduced below.
n\qquad\qquad q_n\qquad\qquad\Deltaq_n
\quad\quad\times10^{-10}\times10^{-10}
\qquad\qquad( esu )\qquad(esu)
01 \qquad..........\qquad..........
02\qquad.......... \qquad..........
03\qquad.......... \qquad..........
04\qquad19.66\qquad\quad..........
05\qquad24.60\qquad\quad4.94
06\qquad29.62\qquad\quad5.02
07 \qquad34.47\qquad\quad4.91
08\qquad39.38\qquad\quad4.91
09\qquad44.42\qquad\quad5.04
10\qquad49.41\qquad\quad4.99
11\qquad53.91\qquad\quad4.50
12\qquad59.12\qquad\quad5.21
13\qquad63.68\qquad\quad4.56
14\qquad68.65\qquad\quad4.97
15\qquad..........\qquad..........
16\qquad78.34\qquad\quad9.69
17\qquad83.22\qquad\quad4.88
Millikan's Experiement Data Source:
http://physics.nyu.edu/~physlab/Classical%20and%20Quantum%20Wave%20Lab/Millikan%20oil%20drop%2002-06-2009.pdf
Least Squares Fit: When we have measured data (x_i, y_i) which we want to make a linear fit of the form, y=ax, the least squares method tries to minimise the square of the deviation from the fit value:
S^2 = \sum_i(y_i-ax_i)^2=\sum_i(y_i^2+a^2x_i^2-2ax_iy_i)
To find the a that minimises S, differentiate S with respect to a, set it to zero and solve for a.
\frac{\delS^2}{\dela} = \sum_i2ax_i^2-2x_iy_i = 0
\bara=\frac{\sum_ix_iy_i}{\sum_ix_i^2}
Applying this to Millikan's data:
\bare = \frac{\sum_{n=4}^{n=17} nq_n}{\sum_{n=4}^{n=17} n^2} = (7589\times10^{-10}\quad esu)/1546
\quad = 4.909\times10^{-10}\quad esu
Therefore the least-square fit value of the fundamental unit of charge for Millikan's data, written to 4 significant digits is:
\bare_{esu} = 4.909\times10^{-10}\quad esu
Converting from esu to "coulomb"
\bare_{C} = \bare_{esu}\times\sqrt{4\pi\epsilon_0}\times10^{-6} = 1.637\times10^{-19}\quad C
