\documentclass[10pt]{article}
\usepackage{graphicx, multicol,wrapfig,exscale,epsfig,fancybox,fullpage}
\pagestyle{empty}

\parindent=0pt
\parskip=.1in

\newcommand\hs{\hspace{6pt}}

\begin{document}

\let\DS=\displaystyle

\newcommand{\bmit}[1]{\mathchoice
        {\mbox{\boldmath$\displaystyle#1$}}
        {\mbox{\boldmath$\textstyle#1$}}
        {\mbox{\boldmath$\scriptstyle#1$}}
        {\mbox{\boldmath$\scriptscriptstyle#1$}}
        }
\newcommand{\vf}[1]{\vec{\bmit#1}}
\newcommand{\VF}[1]{\kern-0.5pt\vec{\kern0.5pt\bmit#1}}
\newcommand{\Hat}[1]{\kern1pt\hat{\kern-1pt\bmit#1}}
\newcommand{\HAT}[1]{\kern-0.25pt\hat{\kern0.25pt\bmit#1}}
\newcommand{\Prime}{{}\kern0.5pt'}

\renewcommand{\AA}{\vf A}
\newcommand{\BB}{\vf B}
\newcommand{\EE}{\vf E}
\newcommand{\FF}{\vf F}
\newcommand{\GG}{\vf G}
\newcommand{\HH}{\vf H}
\newcommand{\II}{\vf I}
\newcommand{\JJ}{\kern1pt\vec{\kern-1pt\bmit J}}
\newcommand{\KK}{\vf K}
\newcommand{\gv}{\VF g}
\newcommand{\rr}{\VF r}
\newcommand{\rrp}{\rr\Prime}
\newcommand{\vv}{\VF v}
\newcommand{\grad}{\vec\nabla}
\newcommand{\zero}{\kern-1pt\vec{\kern1pt\bf 0}}

\newcommand{\ii}{\Hat\imath}
\newcommand{\jj}{\Hat\jmath}
\newcommand{\kk}{\Hat k}
\newcommand{\nn}{\Hat n}
\newcommand{\rhat}{\HAT r}
\newcommand{\xhat}{\Hat x}
\newcommand{\yhat}{\Hat y}
\newcommand{\zhat}{\Hat z}
\newcommand{\that}{\Hat\theta}
\newcommand{\phat}{\Hat\phi}

\newcommand{\LargeMath}[1]{\hbox{\large$#1$}}
\newcommand{\INT}{\LargeMath{\int}}
\newcommand{\OINT}{\LargeMath{\oint}}
\newcommand{\LINT}{\mathop{\INT}\limits_C}

\newcommand{\Int}{\int\limits}
\newcommand{\dint}{\mathchoice{\int\!\!\!\int}{\int\!\!\int}{}{}}
\newcommand{\tint}{\int\!\!\!\int\!\!\!\int}
\newcommand{\DInt}[1]{\int\!\!\!\!\int\limits_{#1~~}}
\newcommand{\TInt}[1]{\int\!\!\!\int\limits_{#1}\!\!\!\int}

\newcommand{\Bint}{\TInt{B}}
\newcommand{\Dint}{\DInt{D}}
\newcommand{\Eint}{\TInt{E}}
\newcommand{\Lint}{\int\limits_C}
\newcommand{\Oint}{\oint\limits_C}
\newcommand{\Rint}{\DInt{R}}
\newcommand{\Sint}{\DInt{S}}

\newcommand{\Partial}[2]{{\partial#1\over\partial#2}}

\newcommand{\Item}{\smallskip\item{$\bullet$}}

\renewcommand{\thesubsection}{\arabic{subsection}}




\section*{Activity 1: Solutions for potential due to 2 point charges}

All solutions will begin the electrical potential due to
point charges
\begin{equation}
V(\rr) = {1\over 4\pi\epsilon_0} \sum_{i=1}^N {q_i\over|\rr-\rr_i|}
\end{equation}
$\rr$ denotes the position in space at which the potential is
measured and $\rr_i$ denotes the position of the charge. In
Cartesian coordinates this becomes
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0} \sum_{i=1}^N {q_i\over\sqrt{(x -
x_i)^2 + (y - y_i)^2 + (z - z_i)^2}}
\end{equation}
Because we are considering only the $x,y$ plane, $z = 0$ and because
the two charges are on the $x$-axis, then $y_i, z_i = 0$, and $N =
2$
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0} \sum_{i=1}^2 {q_i\over\sqrt{(x -
x_i)^2 + y^2}}
\end{equation}

\subsection{$x$-axis}

This section looks at the four cases for the potential on the
$x$-axis. since $y = 0$, then for all four cases on the $x$-axis,

\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0} \sum_{i=1}^2 {q_i\over\sqrt{(x -
x_i)^2}}
\end{equation}


\subsubsection{2 positive charges, $+Q$, one at $D$ and one at $-D$, $|x|<<D$}

With both charges equal to $+Q$, Eq. 4 leads to
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {\left({1\over\sqrt{(x - D)^2}} +
{1\over\sqrt{(x + D)^2}}\right)}
\end{equation}
Because $|x|<<D$, this leads to
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {\left({1\over{D - x}} +
{1\over{D + x}}\right)}
\end{equation}
Factoring out $D$ from the denominator yields
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {1\over D} {\left({1\over{1 -
{x\over D}}} + {1\over{1 + {x\over D}}}\right)}
\end{equation}
Which can be rewritten as
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {1\over D} {\left(\left(1 -
{x\over D}\right)^{-1} + \left(1 + {x\over D}\right)^{-1}\right)}
\end{equation}
Using the power series $(1 + z)^p = 1 + pz + {p(p-1)\over 2!}{z^2} +
...$ results in
\begin{eqnarray}
V(x,y,z) = &  &{Q\over 4\pi\epsilon_0}\, {1\over D}\left( 1 + {x\over D} + {x^2\over D^2} + {x^3\over D^3} + ...\right)\nonumber\\
&+&{Q\over 4\pi\epsilon_0}\, {1\over D}\left( 1 - {x\over D} +
{x^2\over D^2} - {x^3\over D^3} + ...\right)
\end{eqnarray}
The odd powers cancel to produce the expansion
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0}\, {2\over D}\left( 1 + {x^2\over
D^2} + {x^4\over D^4} + ...\right)
\end{equation}

\subsubsection{Opposite charges, $+Q$ at $+D$, $-Q$ at $-D$, $|x|<<D$}
Eq. 4 now leads to the same results for Eq. 5 except for a sign
change, becoming
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {\left({1\over\sqrt{(x - D)^2}} -
{1\over\sqrt{(x + D)^2}}\right)}
\end{equation}
Using the same procedure as in Eq.6 - 9 before, we now have
\begin{eqnarray}
V(x,y,z) = &  &{Q\over 4\pi\epsilon_0}\, {1\over D}\left( 1 + {x\over D} + {x^2\over D^2} + {x^3\over D^3} + ...\right)\nonumber\\
&-&{Q\over 4\pi\epsilon_0}\, {1\over D}\left( 1 - {x\over D} +
{x^2\over D^2} - {x^3\over D^3} + ...\right)
\end{eqnarray}
Now the even powers cancel to become
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0}\, {2\over D}\left({x\over D} +
{x^3\over D^3} + {x^5\over D^5} + ...\right)
\end{equation}
\subsubsection{2 positive charges, $+Q$, one at $D$ and one at $-D$, $x>>D$}
Starting with Eq.5, but now with $x>>D$,
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {\left({1\over{x - D}} +
{1\over{x + D}}\right)}
\end{equation}
Factoring out $x$ from the denominator yields
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0} {1\over x} {\left({1\over{1 -
{D\over x}}} + {1\over{1 + {D\over x}}}\right)}
\end{equation}
Using the Laurent series expansion now results in
\begin{eqnarray}
V(x,y,z) = &  &{Q\over 4\pi\epsilon_0}\, {1\over x}\left( 1 + {D\over x} + {D^2\over x^2} + {D^3\over x^3} + ...\right)\nonumber\\
&+&{Q\over 4\pi\epsilon_0}\, {1\over x}\left( 1 - {D\over x} +
{D^2\over x^2} - {D^3\over x^3} + ...\right)
\end{eqnarray}
The odd powers of the expansion cancel to become
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0}\, {2\over x}\left( 1 + {D^2\over
x^2} + {D^4\over x^4} + ...\right)
\end{equation}
However, it should noted that this is an odd function, and
multiplying through by $1\over x$ results in
\begin{equation}
V(x,y,z) = {2Q\over 4\pi\epsilon_0}\, \left({1\over x} + {D^2\over
x^3} + {D^4\over x^5} + ...\right)
\end{equation}
\subsubsection{Opposite charges, $+Q$ at $+D$, $-Q$ at $-D$, $x>>D$}
Changing the sign in Eq. 14 results in the even powers of the
expansion cancelling and
\begin{equation}
V(x,y,z) = {Q\over 4\pi\epsilon_0}\, {2\over x}\left({D\over x} +
{D^3\over x^3} + {D^5\over x^5} + ...\right)
\end{equation}
Which can be rewritten as
\begin{equation}
V(x,y,z) = {2Q\over 4\pi\epsilon_0}\, \left({D\over x^2} + {D^3\over
x^4} + {D^5\over x^6} + ...\right)
\end{equation}

\subsection{$y$-axis}

This section looks at the four cases for the potential on the
$y$-axis, where we now consider that $x = 0$ and Eq. 3 becomes

\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0} \sum_{i=1}^2 {q_i\over\sqrt{
x_i^2 + y^2}}
\end{equation}
Because $x_i = \pm D$
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0} \sum_{i=1}^2 {q_i\over\sqrt{D^2 +
y^2}}
\end{equation}
\subsubsection{2 positive charges, $+Q$, one at $D$ and one at $-D$, $|y|<<D$}
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0}{2Q\over\sqrt{D^2 + y^2}}
\end{equation}
Factor out $D$ from the denominator yields
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0}\,{1\over D}\,{2Q\over\sqrt{1 +
{{y^2}\over {D^2}}}}
\end{equation}
Which can be rewritten as
\begin{equation}
V(x,y,z) = {2Q\over 4\pi\epsilon_0} {1\over D} {\left(1 +
{{y^2}\over {D^2}}\right)^{{-1}\over 2}}
\end{equation}
Using the power series $(1 + z)^p = 1 + pz + {p(p-1)\over 2!}{z^2} +
...$ results in
\begin{equation}
V(x,y,z) = {2Q\over 4\pi\epsilon_0} {1\over D} {\left(1 - {1\over
2}{{y^2}\over {D^2}} + {3\over 8}{{y^4}\over {D^4}} + ...\right)}
\end{equation}
\subsubsection{Opposite charges, $+Q$ at $+D$, $-Q$ at $-D$, either $|x|<<D$ or $x>>D$}
Either inspection or calculation reveals that the potential is
always zero on the $y$-axis for this case
\begin{equation}
V(x,y,z) = 0
\end{equation}
\subsubsection{2 positive charges, $+Q$, one at $D$ and one at $-D$, $y>>D$}
Beginning with Eq. 24, Factoring out $y$ from the denominator yields
\begin{equation}
V(x,y,z) = {1\over 4\pi\epsilon_0}\,{1\over y}\,{2Q\over\sqrt{1 +
{{D^2}\over {y^2}}}}
\end{equation}
Following the same method as in Eq. 25 and 26 results is the
Laurent series expansion
\begin{equation}
V(x,y,z) = {2Q\over 4\pi\epsilon_0} {1\over y} {\left(1 - {1\over
2}{{D^2}\over {y^2}} + {3\over 8}{{D^4}\over {y^4}} + ...\right)}
\end{equation}

\vfill\eject
\end{document}