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+\documentclass{article}
+\usepackage{tabularx,amsmath,boxedminipage,epsfig}
+\begin{document}
+\title{Millikan ``Oil Drop'' Experiment}
+\author{} \date{} 
+\maketitle
+	\section*{Introduction and Theory}
+Consider Fig. \ref{moplates}
+\begin{figure}[h]
+\centerline{\epsfig{width=3in, file=moplates.eps}}
+\caption{\label{moplates}Very schematic Millikan Oil Drop System}
+\end{figure}				
+It turns out that very small droplets fall very slowly. Clouds, for
+example, are very small water droplets, trying to fall, but held aloft 
+by very slight air currents.
+
+
+An electric field can take the place of the air current and 
+even cause the oil drop
+to rise. Thus, for a rising oil drop:
+
+\begin{equation}\label{rise}
+Eq=mg+kv_r
+\end{equation}
+where $E$ is the electric field, $q$ is the charge on the droplet, $m$
+is the mass of the droplet, $g$ is the acceleration due to gravity, $v_r$
+is the velocity rising, and $k$ is a drag coefficient which will be 
+related to the viscosity of air and the radius of the droplet.
+
+If the field is off and the droplet is just falling, then:
+\begin{equation}\label{fall}
+mg=kv_f
+\end{equation}
+Combining Eqs. \ref{rise} and \ref{fall} we can find the charge $q$:
+\begin{equation}\label{q}
+q=\frac{mg(v_f+v_r)}{Ev_f}
+\end{equation}
+
+
+The drag coefficient, $k$, can be determined from the viscosity, $\eta$, and
+the radius of the droplet, $a$:
+ using Stokes law: 
+\begin{equation}
+k=6\pi a\eta
+\end{equation}
+
+Since 
+\begin{equation}\label{m}
+m=\frac{4}{3}\pi a^3 \rho 
+\end{equation}
+one may solve for $a$:
+\begin{equation}\label{simple}
+a=\sqrt{\frac{9\eta v_f}{2g\rho}}
+\end{equation}
+Here $rho=.886\cdot 10^3 kg/m^3$ is the density of the oil. (We ignore the
+density of air, which is roughly 1/1000 less.) 
+
+There is a 
+small correction because  the oil drop radius is not so different 
+from the mean free path of air. This leads to an effective viscosity:
+\begin{equation}\label{etaeff}
+\eta_{eff}=\eta\frac{1}{1+\frac{b}{pa}}
+\end{equation}
+where $b\approx 8.20 \times 10^{-3}$ (Pa m) and $p$ is atmospheric
+pressure (1.01 $10^5$ Pa). The idea here is that the effect should be 
+related to the
+ratio of the mean free path to the drop radius. This is the form 
+here since the mean free path is inversely proportional to pressure.
+The particular numerical constant can be obtained experimentally if
+the experiment were performed at several different pressures. A feature
+Milikan's apparatus had, but ours does not.
+
+There are two approaches at this point that one could take. 
+\begin{enumerate}
+\item  One could use Eq. \ref{simple} to determine $a$ using an uncorrected
+$\eta$, then use this to determine $\eta_{eff}$ then use this viscosity
+in Eq. \ref{simple} again to find a somewhat better $a$, and then proceed
+around the loop again until convergence is achieved. If the correction
+is large, this can get tedious.
+
+\item Put $\eta_{eff}$ of Eq. \ref{etaeff}  into Eq. \ref{simple} and
+then solve this more complex equation for $a$. 
+\end{enumerate}
+This second approach leads to:
+\begin{equation}\label{complex}
+a=\sqrt{\frac{9\eta v_f}{2g\rho}+\left(\frac{b}{2p}\right)^2}-\frac{b}{2p}
+\end{equation}
+
+Having found $a$ one can then find $m$ using Eq. \ref{m} and then find
+$q$ from Eq. \ref{q}. I would use this approach rather than the Pasco one.
+Try to find drops that do not rise too quickly for they will likely have
+a large number of electrons on them and, further, it will be difficult to
+determine the $V_r$.  If you can't find slow risers, then lower the voltage
+so as to get better precision. Be sure to measure the space thickness in
+order to determine the field from the voltage.
+
+
+
+\newpage\
+ 
+
+
+
+\subsection*{Required for your report}:
+
+Make a table of your measurements. Identify the drop and charge.
+Determine  the charge in each case. Make a table of charge differences.
+Determine the smallest charge for which all the charges could be multiples
+of this smallest charge.  Estimate the error in your determination of e.
+
+ Answer these questions somewhere in your report:
+
+\begin{enumerate}
+\item You will notice that some drops travel upward and others downward
+   in the applied field. Why is this so? Why do some drops travel
+   very fast, and others slow?
+\item Is the particle motion in a straight line? Or, do you notice that
+   the particle "dances" around ever so slightly? This is due to
+   Brownian motion: the random motion of a small particle in a gas or
+   fluid.
+
+\item Do you notice distinct steps in the terminal velocity in applied
+   field? That is, do the terminal velocities appear to clump around
+   similar values? What does this say about the discrete nature of
+   charge?
+
+\item We made three assumptions in determining the charge from Equation 1
+   above. What are they? Hint: They are related to Stoke's Law.
+
+\item How does the average particle diameter you extracted from the
+   terminal velocity without the field on compare to the value given
+   on the bottle? Try to explain any discrepancies.
+
+\item Would you, like Millikan, spend 10 years on this experiment?
+
+\end{enumerate}
+
+Extra credit:  Millikan and his contemporaries were only able to
+measure integer values of electron charge (as you are). Has anyone
+measured free charges of other than integer multiples of e? 
+
+
+
+
+\end{document}
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