# Question #bb3ef

Dec 4, 2017

Millikan had data for the charges on the oil droplet which when sorted was expected to show a linear trend ${q}_{n} = e . n$. Slope of this straight line ($e$) is the fundamental unit of charge which is found using the least-square method.

#### Explanation:

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 $\left\{{q}_{n}\right\}$ may not be present. In fact it need not even start with ${q}_{1}$. The sequence interval $\setminus \Delta {q}_{n} = {q}_{n} - {q}_{n - 1} = e$, 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 ${16}^{t h}$ row you see that sequence interval suddenly jumps showing a missing ${q}_{15}$

Millikan's own data is reproduced below.
$n \setminus q \quad \setminus q \quad {q}_{n} \setminus q \quad \setminus q \quad \setminus \Delta {q}_{n}$
$\setminus \quad \setminus \quad \setminus \times {10}^{- 10} \setminus \times {10}^{- 10}$
$\setminus q \quad \setminus q \quad \left(e s u\right) \setminus q \quad \left(e s u\right)$
$01 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$02 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$03 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$04 \setminus q \quad 19.66 \setminus q \quad \setminus \quad \ldots \ldots \ldots .$
$05 \setminus q \quad 24.60 \setminus q \quad \setminus \quad 4.94$
$06 \setminus q \quad 29.62 \setminus q \quad \setminus \quad 5.02$
$07 \setminus q \quad 34.47 \setminus q \quad \setminus \quad 4.91$
$08 \setminus q \quad 39.38 \setminus q \quad \setminus \quad 4.91$
$09 \setminus q \quad 44.42 \setminus q \quad \setminus \quad 5.04$
$10 \setminus q \quad 49.41 \setminus q \quad \setminus \quad 4.99$
$11 \setminus q \quad 53.91 \setminus q \quad \setminus \quad 4.50$
$12 \setminus q \quad 59.12 \setminus q \quad \setminus \quad 5.21$
$13 \setminus q \quad 63.68 \setminus q \quad \setminus \quad 4.56$
$14 \setminus q \quad 68.65 \setminus q \quad \setminus \quad 4.97$
$15 \setminus q \quad \ldots \ldots \ldots . \setminus q \quad \ldots \ldots \ldots .$
$16 \setminus q \quad 78.34 \setminus q \quad \setminus \quad 9.69$
$17 \setminus q \quad 83.22 \setminus q \quad \setminus \quad 4.88$

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 $\left({x}_{i} , {y}_{i}\right)$ which we want to make a linear fit of the form, $y = a x$, the least squares method tries to minimise the square of the deviation from the fit value:
${S}^{2} = \setminus {\sum}_{i} {\left({y}_{i} - a {x}_{i}\right)}^{2} = \setminus {\sum}_{i} \left({y}_{i}^{2} + {a}^{2} {x}_{i}^{2} - 2 a {x}_{i} {y}_{i}\right)$
To find the $a$ that minimises $S$, differentiate $S$ with respect to $a$, set it to zero and solve for $a$.

$\setminus \frac{\setminus \partial {S}^{2}}{\setminus \partial a} = \setminus {\sum}_{i} 2 a {x}_{i}^{2} - 2 {x}_{i} {y}_{i} = 0$
$\setminus \overline{a} = \setminus \frac{\setminus {\sum}_{i} {x}_{i} {y}_{i}}{\setminus {\sum}_{i} {x}_{i}^{2}}$

Applying this to Millikan's data:

$\setminus \overline{e} = \setminus \frac{\setminus {\sum}_{n = 4}^{n = 17} n {q}_{n}}{\setminus {\sum}_{n = 4}^{n = 17} {n}^{2}} = \frac{7589 \setminus \times {10}^{- 10} \setminus \quad e s u}{1546}$
$\setminus \quad = 4.909 \setminus \times {10}^{- 10} \setminus \quad e s u$

Therefore the least-square fit value of the fundamental unit of charge for Millikan's data, written to 4 significant digits is:
$\setminus {\overline{e}}_{e s u} = 4.909 \setminus \times {10}^{- 10} \setminus \quad e s u$

Converting from $e s u$ to $\text{coulomb}$
$\setminus {\overline{e}}_{C} = \setminus {\overline{e}}_{e s u} \setminus \times \setminus \sqrt{4 \setminus \pi \setminus {\epsilon}_{0}} \setminus \times {10}^{- 6} = 1.637 \setminus \times {10}^{- 19} \setminus \quad C$