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\rightline{(revised 1/23/02)}
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\begin{center}
\bf MILLIKAN OIL-DROP EXPERIMENT
\end{center}
\vspace{1cm}
\begin{center}
Advanced Laboratory, Physics 407\\
University of Wisconsin\\
Madison, Wisconsin 53706
\end{center}
\vspace{.5in}
\begin{center}
\underline{\bf Abstract}
\end{center}
\bigskip
The charge of the electron is measured using the classic technique of Millikan.
Measurements are made of the rise and fall times of oil drops illuminated by
light from a Helium--Neon laser. A radioactive source is used to enhance the
probability that a given drop will change its charge during observation. The
emphasis of the experiment is to make an accurate measurement with a full
analysis of statistical and systematic errors.
\newpage
\pagestyle{plain}
\setcounter{page}{2}
\begin{flushleft}
\underline{\bf Introduction}
\end{flushleft}
\bigskip
Robert A.\ Millikan performed a set of experiments which gave two important 
results:

\begin{itemize}
\item[(1)] Electric charge is quantized.  All electric charges are integral 
multiples of a unique elementary charge $e$.

\item[(2)] The elementary charge was measured and found to have the value $e = 
1.60 \times 10^{-19}$ Coulombs.
\end{itemize}

Of these two results, the first is the most significant since it makes an 
absolute assertion about the nature of matter.  We now recognize $e$ as the 
elementary charge carried by the electron and other elementary particles.  More 
precise measurements have given the value
\[
 e = (1.60217733\pm 0.00000049) \times 10^{-19}\  \hbox{Coulombs} \]

The electric charge carried by a particle may be calculated by measuring the 
force experienced by the particle in an electric field of known strength.  
Although it is relatively easy to produce a known electric field, the force 
exerted by such a field on a particle carrying only one or several excess 
electrons is very small.  For example, a field of 1000 volts per cm would exert 
a force of only $1.6\times 10^{-14}$ N on a particle bearing one excess 
electron.  This is a force comparable to the weight of $10^{-12}$ grams.

The success of the Millikan Oil-Drop experiment depends on the ability to 
measure small forces. The behavior of small charged droplets of oil,  
weighing only $10^{-12}$ gram or less, is observed in a gravitational and 
electric field.  Measuring the velocity of fall of the drop in air enables, 
with the use of Stokes' Law, the calculation of the mass of the drop.  The 
observation of the velocity of the drop rising in an electric field then 
permits a calculation of the force on, and hence the charge, carried by the oil 
drop.

Although this experiment will allow one to measure the total charge $q$ on a 
drop, it is only through an analysis of the data obtained and a certain degree 
of experimental skill that the charges can be shown to be quantized.  By 
selecting droplets which rise and fall slowly, one can be certain that the drop 
has a fairly small charge.  A number of such drops should be observed and their 
respective charges $q$ calculated.  If the charges $q$ 
on these drops are integral 
multiples of a certain smallest charge $e$, then this is an observation that 
charge is quantized.

\def\ll{\leftline} \def\ub{\underbar}

\smallskip{\samepage
\begin{flushleft}
\underline{Question}
\end{flushleft}\nopagebreak
Some students measure a medium size charge $q$ and then divide it by whatever 
large integer, $n$, which will give 
\[
 {q \over n} = 1.60 \times 10^{-19}\ \hbox{coulombs!} \]
\noindent
What is wrong with this?}

Since a different droplet has been used for measuring each charge, there 
remains the question as to the effect of the drop itself on the charge.  This 
uncertainty can be eliminated by changing the charge on a single drop while the 
drop is under observation.  An ionization source placed near the drop will 
accomplish this.  In fact, it is possible to change the charge on the same 
drop several times.  If the results of measurements on the same drop then yield 
charges $q$ which are integral multiples of some smallest charge $e$, then this 
strongly suggests that charge is quantized.

The measurement of the charge of the electron also permits the calculation of 
Avogadro's number.  The charge F 
required to electrodeposit one gram equivalent 
of an element on an electrode (the Faraday) is equal to the charge of the 
electron multiplied by the number of molecules in a mole.
Through electrolysis experiments, the Faraday has been found to be
F = $9.625 \times10^7$ coulombs per kilogram equivalent weight.
\begin{eqnarray*}
\arraycolsep=2pt
\mbox{Hence Avogadro's Number }N = F/e &=& \frac{9.625 \times 10^7\ 
{\rm coulombs/kg\ equiv.\ wt.}}{1.60 \times 10^{-19}\ 
{\rm coulombs/molecule}}\\
& = & 6.02 \times 10^{26}\mbox{ molecules/kg equiv.\  wt.}
\end{eqnarray*}

\bigskip
\begin{flushleft}
\underline{\bf Equations for calculating the charge on a drop}
\end{flushleft}

An analysis of the forces acting on the oil drop
let will yield the equations 
for the determination of the charge carried by the droplet.

Figure 1a shows the forces acting on the drop when it is falling in air and has 
reached its terminal velocity (terminal velocity is reached in a few 
milliseconds for the droplets used in this experiment).  In Fig. 1a, $v_{\rm f}$ 
is the velocity of fall, $k$ is the coefficient of friction between the air and 
the drop, $m_1$ is the mass of the drop, $m_2$ is the mass of air displaced by 
the drop and g is the acceleration due to gravity.

The downward force due to gravity is $m_1g - m_2g.$ The viscous retarding force 
is $kv_f$.  Then 
\[ m_1g - m_2g = kv_f\quad.\]
Let $m = (m_1 - m_2)$, then 
\begin{equation}
 mg = kv_f\quad. 
\end{equation}
\begin{figure}
\begin{center}
\includegraphics[width=3.0in]{fig1_2.pdf}
\caption{Fig. 1a and Fig. 1b}
\end{center}
\end{figure}

Figure 1b shows the forces acting on the drop when it is rising under the 
influence of an electric field.  In Fig. 1b, $E$ 
is the electric intensity, $q$ is 
the charge carried by the drop and $v_r$ is the velocity of rise.  Adding the 
forces vectorially yields:
\begin{equation}                       
Eq = mg + kv_r\quad. 
\end{equation}

Eliminating $k$ from equations (1) and (2) and solving for q yields:
\begin{equation}
q = {mg(v_f + v_r) \over Ev_f}\quad.
\end{equation}

To eliminate $m$ from equation (3), one uses the expression for the volume of a 
sphere:
\begin{eqnarray}
m &=& (m_1 - m_2)\nonumber\\
m &=& (4/3)\pi a^3(\sigma_1 - \sigma_2)
\end{eqnarray}
where 
$a$ is the radius of the droplet, $\sigma_1$ is the density of the oil and 
$\sigma_2$ is the density of the air.  Equations (3) and (4) are combined:
\begin{equation}
q = {4 \pi \over 3} {g \over E} (\sigma_1 - \sigma_2)(1 + {v_r \over 
v_f})a^3\quad. 
\end{equation}


To calculate $a$, one employs Stokes' Law, relating the radius a of any spherical 
body to its velocity of fall in a viscous medium (with the coefficient of 
viscosity, $\eta$):
\[
{\rm velocity~falling} = {2 \over 9}~{ga^2 \over \eta}(\sigma_1 - \sigma_2)
\]

In the theoretical derivation of Stokes' Law the following five assumptions are 
made:
\begin{itemize}
\item[(1)] that the inhomogeneities in the medium are small in comparison with 
the size of the sphere;

\item[(2)] that the sphere falls as it would in a medium of unlimited extent;

\item[(3)] that the sphere is smooth and rigid;

\item[(4)] that there is no slipping of the medium over the surface of the 
sphere;

\item[(5)] that the velocity with which the sphere is moving is so small that 
the resistance to the motion is all due to the viscosity of the medium and not 
at all due to the inertia of such portion of the medium as is being pushed 
forward by the motion of the sphere.
\end{itemize}

In the case of our small drops, the assumptions (2), (3), (4) and (5) are 
valid. However, the assumption (1) is not completely valid since the drop radii 
are about 1 or 2 microns and not much greater than the mean free path of the 
air molecules.

The drop will tend to fall more quickly in the ``holes" between the air 
molecules.  Kinetic theory indicates that a correction must be made to the 
formula for the falling velocity
\[
v_f = {2 \over 9}~{ga^2 \over \eta} (\sigma_1 - \sigma_2)\left(1 + A{\ell \over 
a}\right)\quad,\]
where $\ell$ is the mean free path of the air molecules and $A$ 
is a dimensionless 
correction factor.

The mean free path $\ell$ is dependent upon the air pressure $P$ and so we use a 
more convenient form
\begin{equation}
v_f = {2 \over 9}~{ga^2 \over \eta} (\sigma_1 - \sigma_2)\left(1 + {b \over 
Pa}\right)\end{equation}
where $b$ is a correction factor, and $P$ is the pressure.

To calculate the radius $a$ we must solve this equation.  First rearrange.
\[
{9 \eta v_f \over 2g(\sigma_1 - \sigma_2)} = a^2 + {b \over P}a\quad.
\]
Let
\begin{equation}
\theta = b/2P\quad, 
\end{equation}
and 
\begin{equation}
\phi = {9 \eta v_f \over 2g(\sigma_1 - \sigma_2)}\quad. 
\end{equation}
Then $a^2 + 2\theta a - \phi = 0$.  The solution must be the positive root
\begin{equation}
a = -\theta + \sqrt{\theta^2 + \phi}\quad. 
\end{equation}

The electric intensity is given by $E = V/d$, where $V$ 
is the potential difference 
across the parallel plates separated by a distance $d$.  $E$, $V$ and $d$ 
are all 
expressed in the mks system of units and so $E$ is measured in volts/meter.

The charge $q$ may be obtained  by calculating $\theta$ and $\phi$ with equations 
(7) and (8) then calculating the radius $a$ with equation (9) and 
finally $q$ with 
equation (5).

\newpage

\begin{flushleft}
\underline{\bf Calculations}
\end{flushleft}
\smallskip

It is suggested that you 
lay out the results of your calculations for each drop in the form of a 
table so that errors may be found more easily.  Use MKS units and the following 
columns:

\begin{center}
\begin{tabular}{|c|c|c|c|c|c|c|c|c|c|}\hline
\parbox{1.5cm}{\centering \strut average\\ fall\\ time\\ $t_f$\strut}
&\parbox{1.5cm}{\centering \strut average\\ rise\\ time\\ $t_r$\strut}
&\parbox{2.5cm}{\centering $\displaystyle{\left(1+\frac{t_f}{t_r}\right)}$\\
$\displaystyle{=\left(1+\frac{v_r}{v_f}\right)}$\strut}
&\parbox{2cm}{\centering fall\\ velocity\\ $v_f$\strut}
&$\phi$ & $(\theta^2+\phi)$ & $\sqrt{\theta^2+\phi}$
& $a$ & $a^3$ & $q$ \\
\hline
\end{tabular}
\end{center}

\begin{itemize}
\item[(1)] The fall velocity $v_f = s/t_f$ where $s$ is distance between the
images of the graticule lines. You will have to determine $s$ experimentally.

\item[(2)] From $v_f$, calculate $\phi$ by multiplying by a factor 
9$\eta/2g(\sigma_1 - \sigma_2$) where:
\begin{itemize}
\item[] $\eta$ = ?  $\rm N\;sec\;m^2$ (Use graph in Fig. 2: 1 
$\rm N\;sec\;m^2 = 1\ kg\;m^{-1}\,sec^{-1}$).

\item[] g = 9.81 meters/sec$^2$.

\item[] $\sigma_1$ = ? kg/meters$^3$ (density of oil to be measured).

\item[] $\sigma_2$ = 1.192 kg/meters$^3$ (density of air at 1 atmosphere and 
22$^\circ$C).
\end{itemize}
\item[(3)] Calculate $\theta$ from the barometric pressure $P$ 
measured in cm of 
mercury and use $b = 6.17 \times 10^{-6}$ (cm of Hg-meter).

\item[(4)] Add $\theta^2 + \phi$. The $\theta^2$ will be small.

\item[(5)] Take the square root $\sqrt{\theta^2 + \phi}$.

\item[(6)] Subtract $\theta$ to obtain $a$.

\item[(7)] At this stage, check that the radius  $a$ is reasonable.

\item[(8)] Obtain a$^3$.
\item[(9)] Compute a multiplier ${4\pi \over 3}{gd(\sigma_1 - \sigma_2) \over 
V}$. (You will have to measure $d$.)

\item[(10)] Combine the multiplier, $(1+\frac{v_r}{v_f})$ and $a^3$ to obtain 
$q$.
\end{itemize}
\begin{figure}
\begin{center}
\includegraphics[width=4.0in]{millikan_1.pdf}
\caption{Viscosity($\eta$) of dry air as a function of temperature 
($^{\circ}$C).
Vertical scale is in units of $10^{-5}\ {\rm kg\ m^{-1}\ sec^{-1}}$.}
\end{center}
\end{figure}

\begin{flushleft}
\underline{\bf Apparatus}
\end{flushleft}
\smallskip
\noindent
The experiment uses:
\begin{enumerate}
\item
A PASCO SCIENTIFIC CORP apparatus.

\item
A 4 milliwatt He-Ne laser for illumination of the drops.  The laser 
is fastened to the supporting plate so that its light cannot enter your eyes 
accidentally.  However, be careful, and do not thoughtlessly unfasten the laser or 
introduce shiny or reflecting objects into the laser beam.

\item
The high voltage is generated and stabilized within the PASCO 
apparatus and is measured
precisely with a Data Precision Model 1450 multimeter. A correction for the
finite input impedance of the meter may be required.

\item
The time is measured by a small counter controlled by a microswitch 
and a 100 kHz crystal oscillator.  The count is visible on LEDs and is 
precise to the least count of 0.01 seconds.  The counter is zeroed when the 
microswitch is pressed and the count starts.  The count stops when the switch 
is released.

\item
The apparatus and laser are mounted on one plate (for safety) which 
can be adjusted so that the electric field $E$ is vertical.

\item
An atomizer contains a non-volatile oil of known density.
\end{enumerate}
The apparatus controls are:

PLATE CHARGING CONTROL SWITCH - When the three way lever plate control 
switch is in the OFF position the condenser plates are disconnected from the 
high voltage supply and grounded.  When the switch is in the TOP PLATE +,
position the top plate is positive with respect to the bottom plate.  When the 
switch is in the TOP PLATE --, position  the top plate is negative with respect 
to the bottom plate.

RADIATION SOURCE - when the lever is at the OUT position the radiation 
source is shielded on all sides by plastic, so that virtually no radiation 
enters the area of the drops.  At the IN position the plastic shielding is 
removed and the drop area is exposed to the radiation source.   NOTE: move 
the radiation source lever gently to avoid jarring the condenser assembly and 
knocking the droplet from the viewing area.

The radiation source, initial strength and date of initial strength are all 
specified on the radiation tag to the left of the radiation source lever.
\smallskip
The power Supply specifications are:
\begin{itemize}
\item[]Power Input: 110/130 VAC, 50/60 cps.
\item[]Range: 300--400 VDC, continuously variable.
\item[]Regulation: 1\% for 10\% line variation.
\item[]Ripple: less than 0.1 volt.
\item[]Stability: within 1\% after warm up.
\end{itemize}
\newpage

\begin{flushleft}
\underline{\bf Procedure}
\end{flushleft}

Two people are advisable since one person must follow the drops while the other 
records the times.
\begin{enumerate}
\item Darken the room.  A small desk lamp can be used to illuminate your data 
book.

\item Turn the POWER switch ON, the PLATE CONTROL switch to OFF, the RADIATION 
SOURCE LEVER to IN.

\item The RETICULE ILLUMINATION control should be set so that the reticule 
lines are just bright enough to be easily visible.  Excessive illumination of 
these lines may make it difficult to observe very small droplets.

\item Introduce some oil drops into the condenser by placing the nozzle of the 
atomizer into the hole of the condenser housing cover.  A few quick ``squirts" 
of oil will fill the upper chamber of the condenser with drops and begin to 
force some drops into the viewing area.  If no drops are seen, squeeze the 
atomizer bulb gently until drops appear in the viewing area.  If repeated 
``squirts" of the atomizer fail to produce any drops in the viewing area, but 
rather a cloudy brightening of the field, the hole in the top plate is probably 
clogged, and should be cleaned.

The exact technique of introducing drops will have to be developed by the 
experimenter.  The object is to get a small number of drops, not a large, 
bright cloud, from which a single drop can be chosen.  It is important to 
remember that the drops are being forced into the viewing area by the pressure 
of the atomizer.  Therefore, excessive use of the atomizer can cause too many 
drops to be forced into the viewing area and, more important, into the area 
between the condenser wall and the focal point of the scope.  Drops in this 
area prevent observation of drops at the focal point of the scope.

NOTE: If the entire viewing area becomes filled with drops, so that no 
one drop can be isolated, either wait three or four minutes until the drops 
settle out of view, or disassemble the condenser, thus removing the drops.  
When the amount of oil on the condenser parts becomes excessive, clean the 
assembly.

\item Select drops which have a mass which can be accurately measured. If the 
mass is too large the drop will fall quickly and your percentage error in 
timing will be poor.

If the mass is too small, the drop will bounce around due to random collisions 
with air molecules (Brownian motion) and it will be difficult to estimate when 
it crosses a line.  A very small drop may cross a line 10 or 20 times!  The 
laser shows the Brownian motion quite clearly.

We recommend using drops which fall with times between 10 and 40 seconds.

\item Now move the RADIATION SOURCE lever to the OUT position (so that the 
charges will be unlikely to change.)

\item Immediately select those drops with measurable charges.  If the drop 
rises quickly then it has a large charge $q$ and so will not be much use in 
finding the elementary charge.

Look for drops which take 10 to 40 seconds to rise.

\item If you still have many drops in sight, repeat 5 and 7 to concentrate on 
a useful drop.

\item Take about 10 measurements of the fall time and of the rise time of a 
particular drop.  Do not jar the apparatus or you may lose your drop.

\item Plot the fall time and rise time of this drop {\bf before} measuring the 
next drop.  Write the drop number beside its point.

\begin{center}
\includegraphics[width=4.0in]{millikan_2.pdf}
\end{center}

\item Calculate the $q$ for the first drop {\bf before} you make any further 
measurements.  If the drop is outside the range of 1 to $10\times10^{-19}$ 
coulombs then find your procedural or arithmetic error before continuing.

\item Repeat 4 through 10 for about {\bf 30 drops}.  Choose the fall and rise 
times so that you have a fairly uniform scatter within the 10--40 sec 
square on your fall time, rise time plot.  This will prevent you from selecting 
too many identical charges.  Include a few points outside the square if you 
wish.

\item Try a variation on one or two drops by changing the charge.  Drops are 
again introduced into the viewing area and a new drop is selected.  After about 
20 measurements on this drop have been made, the drop is brought to the top of 
the field of view and allowed to fall with the RADIATION SOURCE lever at the IN 
position. A few seconds later the plates should be charged, and, if the rising 
velocity has changed, the RADIATION SOURCE is moved to the OUT position and a 
new series of measurements taken.  If, however, the charge has not changed, 
then turn the Plate Control Switch to OFF and allow the drop to continue 
falling.  After a few seconds, again check for a change in the rising velocity. 
 Continue this procedure until the drop has captured an ion.

If the drop captures an ion such that the drop moves rapidly downward, then 
reverse the polarity of the plates so that the drop can be made to rise.

Make about 20 measurements of the rising and falling velocity of the drop, and, 
if possible, change the charge again and repeat the measurement procedure.

\item Record the barometric pressure and the voltage on the 
condenser for each measurement set.

\item At this time (before doing the calculations) decide which of your 
observations may be in error.  It is very important that you {\bf not} reject 
any data after you see the result of the calculations.  Why?

\item Now calculate $q$ for each of your drops.

\item Make a histogram of your charges $q$.  Use a bin width of 
$0.2\times 10^{-19}$ 
coulombs.  Label each square in the histogram with the drop number.

\item Does your evidence indicate that all charges are integral multiples of 
an elementary charge?

\item Choose a charge $Q$ such that all charges less than $Q$, fall in groups 
which are obvious multiples of an elementary charge.  (The errors of the 
charges larger than $Q$ are too large for clean grouping).

\item Tabulate {\bf all} charges $q$ less than $Q$ 
(it is now too late to reject a 
charge), divide each $q$ 
by the integer appropriate for its group. The final step is an average charge
determined from the average charge of each of the integer subgroups. The
final result should include an error resulting from correct error propagation
of individual error contributions. Be sure to separate systematic and
statistical errors.

\end{enumerate}
\end{document}













