\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 2: Solution for electric potential due to a ring} \subsection*{Find the electrostatic potential in all space due to a ring with total charge $Q$ and radius $R$} \begin{equation} V(\rr) = {1\over 4\pi\epsilon_0} \sum_{i=1}^N {q_i\over|\rr-\rr_i|} \end{equation} For a ring of charge this becomes \begin{eqnarray} V(\rr) = \int\limits_{\hbox{ring}} {1\over 4\pi\epsilon_0} {\lambda(\rrp)\,|d\rrp|\over|\rr-\rrp|} \end{eqnarray} where $\rr$ denotes the position in space at which the potential is measured and $\rrp$ denotes the position of the charge. In cylindrical coordinates, $|d\rrp|=R\,d\phi'$, where $R$ is the radius of the ring. Thus, \begin{eqnarray} V(\rr) = {1\over 4\pi\epsilon_0} \int\limits_0^{2\pi} {\lambda(\rrp)\,R\,d\phi'\over|\rr-\rrp|} \end{eqnarray} Assuming constant linear charge density for a ring with charge Q and radius R, $\lambda(\rrp) = {Q\over{2\pi R}}$ Thus, \begin{eqnarray} V(\rr) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi} {d\phi'\over|\rr-\rrp|} \end{eqnarray} Since $\rr$ and $\rrp$ are not necessarily in the same direction, we cannot simply leave $|\rr-\rrp|$ in curvilinear coordinates and integrate directly. One solution to this problem is to rewrite $|\rr-\rrp|$ in cartesian coordinates $\medskip$ \begin{eqnarray} |\rr-\rrp| &=& \sqrt{(x - x\Prime)^2 +(y - y\Prime)^2 + (z - z\Prime)^2} \end{eqnarray} Setting the ring in the $x,y$ plane with the center at the origin and then rewriting in cylindrical coordinates results in \begin{eqnarray} |\rr-\rrp| = \sqrt{(r\cos\phi - R\cos\phi\Prime)^2 +(r\sin\phi - R\sin\phi\Prime)^2 + (z - 0)^2} \end{eqnarray} Which simplifies to \begin{equation} |\rr-\rrp| = \sqrt{r^2 - 2rR\cos(\phi - \phi\Prime) + R^2 + z^2} \end{equation} Substituting into Eq. 4 results in the elliptic integral \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi} {d\phi'\over\sqrt{r^2 - 2rR\cos(\phi - \phi\Prime) + R^2 + z^2}} \end{eqnarray} \subsection{The $z$ axis} For points on the $z$ axis, $r = 0$ and the integral simplifies to \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}{d\phi'\over\sqrt{R^2 + z^2}} \end{eqnarray} And thus \begin{eqnarray} V(r,\phi,z) = {Q\over 4\pi\epsilon_0}{1\over\sqrt{R^2 + z^2}} \end{eqnarray} \subsubsection{Power series expansions for $z$ axis} To create the power series expansion for $|z|<>R$ is \begin{equation} V(x,y,z) = {Q\over 4\pi\epsilon_0} {1\over z} {\left(1 - {1\over 2}{{R^2}\over {z^2}} + {3\over 8}{{R^4}\over {z^4}} + ...\right)} \end{equation} \subsection{The $x$ axis} For points on the $x$ axis, $z = 0$ and $\phi = 0$, so the integral simplifies to \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi} {d\phi'\over\sqrt{r^2 - 2rR\cos\phi' + R^2}} \end{eqnarray} Which can be rewritten as \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi} {(r^2 - 2rR\cos\phi' + R^2)^{-1/2}{d\phi'}} \end{eqnarray} In this case the power series expansion can be done before integration and then the power series can be integrated. For $x>>R$, factor out an $1/r$ to obtain \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi} \int\limits_0^{2\pi}{1\over r} {\left(1 - {2R\over r}\cos\phi' + {R^2\over r^2}\right)^{-1/2}{d\phi'}} \end{eqnarray} Let $\epsilon = -{2R\over r}\cos\phi' + {R^2\over r^2}$ \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi}}{1\over r} \int\limits_0^{2\pi} {(1 + \epsilon)^{-1/2}{d\phi'} \end{eqnarray} The power series expansion now yields \begin{eqnarray} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi}}{1\over r} \int\limits_0^{2\pi} {\left(1 - {1\over 2}\epsilon + {3\over 8}\epsilon^2 - {15\over 48}\epsilon^3 + ...\right){d\phi'} \end{eqnarray} Substituting $-{2R\over r}\cos\phi' + {R^2\over r^2}$ for $\epsilon$ results in the integrand \begin{eqnarray} 1 + \left(-{1\over 2}\right)\left(-{2R\over r}\cos\phi'+{R^2\over r^2}\right) + \left({3\over 8}\right)\left({4R^2\over r^2}{\cos^2\phi'}-{4R^3\over r^3}{\cos\phi'} +{R^4\over r^4} \right)\\+ \left(-{15\over 48}\right)\left(-{8R^3\over r^3}{\cos^3\phi'}+ {8R^4\over r^4} \cos^2\phi' - {4R^5 \over r^5}\cos\phi' +{R^6 \over r^6} \right)+... \end{eqnarray} Adding like terms and getting rid of any powers greater than third-order in $r$ yields \begin{eqnarray} 1 + {R\over r}\cos\phi'-{R^2\over 2r^2} +{3R^2\over 2r^2}{\cos^2\phi'}-{3R^3\over 2r^3}{\cos\phi'} + {5R^3\over 2r^3}{\cos^3\phi'}+... \end{eqnarray} Using this power series and performing the integral results in the first two non-zero terms for the potential \begin{equation} V(r,\phi,z) = {1\over 4\pi\epsilon_0} {Q\over 2\pi}{1\over r}{\left({2\pi} + {\pi\over 2}{R^2\over r^2} + ...\right)} \end{equation} Which can be simplified to \begin{equation} V(r,\phi,z) = {Q\over 4\pi\epsilon_0} {1\over r}{\left(1 + {1\over 4}{R^2\over r^2} + ...\right)} \end{equation} \vfill \leftline{\it by Corinne Manogue} \leftline{\copyright DATE Corinne A. Manogue} \end{document}