\documentclass[12pt]{article} % BB Nov 21 2011.
\usepackage{epsfig}
\textheight 9.0in \textwidth  6.0in \topmargin -0.4in \headsep 0.4in  
\oddsidemargin  0.25in  \evensidemargin 0.25in  

\def\sqr#1#2{{\vcenter{\hrule height.#2pt
   \hbox{\vrule width.#2pt height#1pt \kern#1pt
   \vrule width.#2pt} \hrule height.#2pt}}}
\def\square{{\mathchoice\sqr64\sqr64\sqr{4.2}3\sqr{3.0}3}}
\def\undertext#1{$\underline{\smash{\hbox{#1}}}$}
\pagestyle{empty}

\begin{document}

\begin{center}
{\bf Dirichlet Boundary Condition: Paul Trap} \\%
\end{center}
\bigskip 

A Paul Trap (Physics Nobel Prize 1989) is used to confine charged 
particles. Let us consider a cylindrical symmetric trap, which is 
conveniently described in cylindrical coordinates $(\rho,\theta,z)$.
The trap electrodes are hyperbolas of revolution about the $z$ axis.
The figure depicts a cross section of the trap. 
\smallskip

By rotation about the $z$ axis the two grounded hyperbolas with potential 
$V(t)=0$ form the upper and lower end electrodes, defined by the equation 
$z^2=\rho^2/2+d^2$.
\smallskip

The hyperbolas with potential $V(t)=V_0\,\sin(wt)$ get connected and 
form the ring electrode, defined by the equation $\rho^2/2=z^2+d^2/2$.
\smallskip

Suppose $\omega\ll c/d$ ($c$ is the speed of light). Use the 
quasi-static approximation and Poisson's equation 
$$\nabla^2\Phi=0$$ 
to calculate the electric field inside the trap. 
\smallskip

Hint: Find first the
most general potential $\Phi$ that fulfills the boundary
condition on the end electrodes. Proceed to implement the 
boundary condition of the ring electrode and show that the 
obtained expression is a solution of the Poisson equation.
% $\nabla^2\,(z^2-\rho^2/2-d^2)=0$.

\begin{figure}[tb]
\vspace{5pc}
 \centerline{\hbox{ \psfig{figure=PaulTrap.eps,width=12.5cm} }}
% \caption{Ladder for assignment~14.} \label{fig_lader}
\end{figure}

\end{document}
