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#