diff options
Diffstat (limited to 'Master/texmf-dist/doc/generic/pst-electricfield/pst-electricfield-docEN.tex')
-rw-r--r-- | Master/texmf-dist/doc/generic/pst-electricfield/pst-electricfield-docEN.tex | 332 |
1 files changed, 332 insertions, 0 deletions
diff --git a/Master/texmf-dist/doc/generic/pst-electricfield/pst-electricfield-docEN.tex b/Master/texmf-dist/doc/generic/pst-electricfield/pst-electricfield-docEN.tex new file mode 100644 index 00000000000..1aab052e4bf --- /dev/null +++ b/Master/texmf-dist/doc/generic/pst-electricfield/pst-electricfield-docEN.tex @@ -0,0 +1,332 @@ +%% $Id: pst-electricfield-docFR.tex 336 2010-05-29 18:38:59Z herbert $ +\documentclass[11pt,english,english,BCOR10mm,DIV12,bibliography=totoc,parskip=false,smallheadings + headexclude,footexclude,oneside]{pst-doc} +\usepackage[latin1]{inputenc} +\usepackage{pst-electricfield} + +\usepackage{pst-electricfield} +\usepackage{pst-func} +\usepackage{pst-exa}% only when running pst2pdf +\usepackage{esint} + + +\let\pstEFfv\fileversion +\lstset{pos=t,language=PSTricks, + morekeywords={psElectricfield,psEquipotential},basicstyle=\footnotesize\ttfamily} +\newcommand\Cadre[1]{\psframebox[fillstyle=solid,fillcolor=black,linestyle=none,framesep=0]{#1}} +% +\begin{document} + +\title{\texttt{pst-electricfield}} +\subtitle{Electric field lines of charges; v.\pstEFfv} +\author{Juergen Gilg\\ Manuel Luque\\Patrice Megret\\Herbert Vo\ss} +%\docauthor{Juergen Gilg\\Manuel Luque\\Herbert Vo\ss} +\date{\today} +\maketitle + + +\section{Method based on electric flux (Patrice Mégret)} + +Equipotential surfaces and E-field lines can be drawn by using the package \LPack{pst-func} and the command \Lcs{psplotImp}\verb+[options](x1,y1)(x2,y2)+. + +The following explanations describe the theory on which this is based. + + +Gauss theorem states that the electric flux across a closed surface $S$ and defined by: +\begin{equation}\label{pm-eq-a} +\psi = \oiint\limits_S \vec{D} \cdot \vec{u}_n \mathrm{d} S = Q +\end{equation} +is equal to the real charge $Q$ inside $S$. As a consequence, in place where there is no charge ($Q=0$), the electric flux is a conservative quantity. + + +A tube of flux is a tube constructed on D-field lines and without charge, the flux going inside any cross-section of the tube is equal to the flux going outside any cross-section of the tube. This means that, by following a tube of a given flux, we automatically follow a D-field line. By using this technique, it is thus possible to obtain a scalar equation that describes the D-field lines. This equation is an implicit equation and can be derived for systems with simple geometrical properties. + +Here the analysis will be limited to point charges and the D-field lines will thus be identical to the E-field lines as there is no electric polarization. + + +For a point charge $q$, located at the origin of the coordinate system, the electric field and the potential are given by: +\begin{equation}\label{pm-eq-b} +\vec{E} = \frac{1}{4 \pi \varepsilon_0 \varepsilon_r} q \frac{\vec{r}}{|\vec{r}|^3} +\end{equation} +\begin{equation}\label{pm-eq-c} +V = \frac{1}{4 \pi \varepsilon_0 \varepsilon_r} \frac{q}{r} +\end{equation} + +The flux across a portion of a sphere of surface $S$ and with an aperture angle $\theta$, is simply given by: +\begin{equation}\label{pm-eq-d} +\psi = \varepsilon_0 \varepsilon_r E S = \frac{1}{2} q (1 -\cos\theta) +\end{equation} +because $S= 2\pi r^2 (1 - \cos\theta)$ and from (\ref{pm-eq-a}) $4 \pi r^2 \varepsilon_0 \varepsilon_r E =q$. + +\begin{center} +\begin{pspicture}(-3,-3)(3,3) +%\psgrid +\psdot[dotscale=2](0,0) +\uput[-135](0,0){$q$} +\psaxes[labels=none,ticks=none]{->}(0,0)(-2.5,-2.5)(2.5,2.5)[$x$,-90][$y$,0] +\pswedge(0,0){2}{-30}{30} +\psarc{->}(0,0){1}{0}{30} +\rput(1.2,0.2){$\theta$} +\rput(2.2,0.7){$S$} +\end{pspicture} +\end{center} + +To find the implicit expression of the E-field lines, it is sufficient to express the flux invariance: +\begin{equation}\label{pm-eq-e} +\psi(x,y) = \frac{1}{2} q (1 -\cos\theta) = \mathrm{cte} +\end{equation} +This relation simply shows that E-field lines correspond to $\theta=\mathrm{cte}$, so that they are clearly radial lines. + +For the E-field lines in the $xy$ plane, expression (\ref{pm-eq-e}) in Cartesian coordinates is: +\begin{equation}\label{pm-eq-f} +\frac{x}{\sqrt{x^2+y^2}} = \mathrm{cte} +\end{equation} + +For the equipotential surface, relation (\ref{pm-eq-c}) is already in implicit form, therefor $V=\mathrm{cte}$ is the wanted equation: +\begin{equation}\label{pm-eq-g} +\frac{1}{\sqrt{x^2+y^2}} = \mathrm{cte} +\end{equation} + +The following graph shows the field and equipotential for this point charge obtained by implicit plotting of functions (\ref{pm-eq-f}) and (\ref{pm-eq-g}). It is clear that the E-field lines are radial ones and the equipotential surfaces cross the $xy$ plane along circles orthogonal to the E-field lines. +\begin{center} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +% \psElectricfield[Q={[1 0 0]}] +% \psEquipotential[Q={[1 0 0]}](-5,-5)(5,5) +\multido{\r=-1+0.1}{20}{% +\psplotImp[linestyle=solid,linecolor=blue](-6,-6)(6,6){% +x y 2 exp x 2 exp add sqrt div \r \space sub}} +\multido{\r=0.0+0.1}{10}{% +\psplotImp[linestyle=solid,linecolor=red](-6,-6)(6,6){% +x 2 exp y 2 exp add sqrt 1 exch div \r \space sub}} +\end{pspicture*} +\end{center} + + +\begin{verbatim} +%% E-field lines +\multido{\r=-1+0.1}{20}{% +\psplotImp[linestyle=solid,linecolor=blue](-6,-6)(6,6){% +x y 2 exp x 2 exp add sqrt div \r \space sub}} + +%% equipotential +\multido{\r=0.0+0.1}{10}{% +\psplotImp[linestyle=solid,linecolor=red](-6,-6)(6,6){% +x 2 exp y 2 exp add sqrt 1 exch div \r \space sub}} +\end{verbatim} + + + +Let's now generalize to point charges distributed arbitrarily along a \textbf{line}. The charge $i$ is $q_i$ and is placed at $(x_i,0)$. +\begin{center} +\begin{pspicture}(0,-3)(12,3) +%\psgrid +\psset{dotscale=2} +\dotnode(0,0){NA}\nput{-45}{NA}{$q_1$} +\dotnode(2,0){NB}\nput{-90}{NB}{$q_2$} +\dotnode(5,0){NC}\nput{-90}{NC}{$q_n$} +\dotnode[linecolor=red](4,2){ND}\nput{90}{ND}{$P(x,y)$} +\ncline{NA}{ND}\naput{$r_1,\theta_1$} +\ncline{NB}{ND}\nbput{$r_2,\theta_2$} +\ncline{NC}{ND}\nbput{$r_n,\theta_n$} +\psaxes[labels=none,ticks=none]{->}(0,0)(0,-2.5)(11,2.5)[$x$,-90][$y$,0] +\psarc{->}(5,0){0.7}{0}{116.5} +\rput(6,0.5){$\theta_n$} +\dotnode[linecolor=blue](4,-2){NE} +\nccurve[ncurv=2,linecolor=green!40!black]{ND}{NE} +\end{pspicture} +\end{center} + +This problem possesses a cylindrical symmetry: it is thus sufficient to study the field and the potential in $xy$ half-plane and the complete results are obtained by rotation around the $x$-axis. + +By rotation around $x$-axis, the E-field line in $P$ creates a tube of flux. The flux across any surface including $P(x,y)$ and crossing $x$-axis beyond the last charge (the trace of this surface in the $xy$ plane is drawn in green) is obtained from (\ref{pm-eq-d}): +\begin{equation}\label{pm-eq-h} +\psi = \frac{1}{2} \sum_{i=1}^{n} q_i (1 -\cos\theta_i) +\end{equation} + +E-field lines are easily computed by the condition $\psi = \mathrm{cte}$, which is expressed as: +\begin{equation}\label{pm-eq-i} +\sum_{i=1}^{n} q_i \frac{x-x_i}{\sqrt{(x-x_i)^2+y^2}} = \mathrm{cte} +\end{equation} +in Cartesian coordinates. + +For the potential, the solution is trivial: +\begin{equation}\label{pm-eq-j} +\sum_{i=1}^{n} \frac{q_i}{\sqrt{(x-x_i)^2+y^2}} = \mathrm{cte} +\end{equation} + +\begin{center} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 -2 0][-1 2 0]}] +\psEquipotential[Q={[1 -2 0][-1 2 0]},Vmin=-2,Vmax=2,stepV=0.25](-5,-5)(5,5) +\multido{\r=-2+0.2}{20}{% +\psplotImp[linestyle=solid,linecolor=red](-6,-6)(6,6){% +x 2 add dup 2 exp y 2 exp add sqrt div 1 mul +x -2 add dup 2 exp y 2 exp add sqrt div -1 mul add +\r \space sub}} +\multido{\r=-0.5+0.1}{10}{% +\psplotImp[linestyle=solid,linecolor=blue](-6,-6)(6,6){% +x 2 add 2 exp y 2 exp add sqrt 1 exch div 1 mul +x -2 add 2 exp y 2 exp add sqrt 1 exch div -1 mul add +\r \space sub}} +\end{pspicture*} +\end{center} + +\begin{verbatim} +%% E-field lines +\multido{\r=-2+0.2}{20}{% +\psplotImp[linestyle=solid,linecolor=red](-6,-6)(6,6){% +x 2 add dup 2 exp y 2 exp add sqrt div 1 mul +x -2 add dup 2 exp y 2 exp add sqrt div -1 mul add +\r \space sub}} +%% equipotential +\multido{\r=-0.5+0.1}{10}{% +\psplotImp[linestyle=solid,linecolor=blue](-6,-6)(6,6){% +x 2 add 2 exp y 2 exp add sqrt 1 exch div 1 mul +x -2 add 2 exp y 2 exp add sqrt 1 exch div -1 mul add +\r \space sub}} +\end{verbatim} + + +The last example corresponds to one charge $+1$ in $(-2,0)$ and one charge $-1$ in $(2,0)$. Here we have superposed the results obtained by implicit functions and those obtained by the direct integration of the equations. +The superposition is perfect, but the method of implicit function is quite slow. Moreover, this method is limited to problem with cylindrical symmetry. + + +\section{Examples} + +\begin{LTXexample}[pos=t] +\begin{pspicture*}(-6,-6)(6,6) +\psframe*[linecolor=lightgray!50](-6,-6)(6,6) +\psgrid[subgriddiv=0,gridcolor=gray,griddots=10] +\psElectricfield[Q={[-1 -2 2][1 2 2][-1 2 -2][1 -2 -2]},linecolor=red] +\psEquipotential[Q={[-1 -2 2][1 2 2][-1 2 -2][1 -2 -2]},linecolor=blue](-6.1,-6.1)(6.1,6.1) +\psEquipotential[Q={[-1 -2 2][1 2 2][-1 2 -2][1 -2 -2]},linecolor=green,linewidth=2\pslinewidth,Vmax=0,Vmin=0](-6.1,-6.1)(6.1,6.1) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t] +\begin{pspicture*}(-6,-6)(6,6) +\psframe*[linecolor=lightgray!50](-6,-6)(6,6) +\psgrid[subgriddiv=0,gridcolor=gray,griddots=10] +\psElectricfield[Q={[-1 -2 2 false][1 2 2 false][-1 2 -2 false][1 -2 -2 false]},radius=1.5pt,linecolor=red] +\psEquipotential[Q={[-1 -2 2][1 2 2][-1 2 -2][1 -2 -2]},linecolor=blue](-6,-6)(6,6) +\psEquipotential[Q={[-1 -2 2][1 2 2][-1 2 -2][1 -2 -2]},linecolor=green,linewidth=2\pslinewidth,Vmax=0,Vmin=0](-6.1,-6.1)(6.1,6.1) +\end{pspicture*} +\end{LTXexample} + + +\begin{LTXexample}[pos=t] +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=lightgray!40](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[-1 -3 1][1 1 -3][-1 2 2]},N=9,linecolor=red,points=1000,posArrow=0.1,Pas=0.015] +\psEquipotential[Q={[-1 -3 1][1 1 -3][-1 2 2]},linecolor=blue](-6,-6)(6,6) +\psEquipotential[Q={[-1 -3 1][1 1 -3][-1 2 2]},linecolor=green,Vmin=-5,Vmax=-5,linewidth=2\pslinewidth](-6,-6)(6,6) +\end{pspicture*} +\end{LTXexample} + + + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 -2 0][-1 2 0]},linecolor=red] +\psEquipotential[Q={[1 -2 0][-1 2 0]},linecolor=blue](-5,-5)(5,5) +\psEquipotential[Q={[1 -2 0][-1 2 0]},linecolor=green,Vmin=0,Vmax=0](-5,-5)(5,5) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 -2 0][1 2 0]},linecolor=red,N=15,points=500] +\psEquipotential[Q={[1 -2 0][1 2 0]},linecolor=blue,Vmin=0,Vmax=20,stepV=2](-5,-5)(5,5) +\psEquipotential[Q={[1 -2 0][1 2 0]},linecolor=green,Vmin=9,Vmax=9](-5,-5)(5,5) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-10,-5)(6,5) +\psframe*[linecolor=lightgray!40](-10,-5)(6,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[600 -60 0 false][-4 0 0] },N=50,points=500,runit=0.8] +\psEquipotential[Q={[600 -60 0 false][-4 0 0]},linecolor=blue,Vmax=100,Vmin=50,stepV=2](-10,-5)(6,5) +\psframe*(-10,-5)(-9.5,5) +\rput(0,0){\textcolor{white}{\large$-$}} +\multido{\rA=4.75+-0.5}{20}{\rput(-9.75,\rA){\textcolor{white}{\large$+$}}} +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-6,-5)(6,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 -2 -2][1 -2 2][1 2 2][1 2 -2]},linecolor={[HTML]{006633}}] +\psEquipotential[Q={[1 -2 -2][1 -2 2][1 2 2][1 2 -2]},Vmax=15,Vmin=0,stepV=1,linecolor=blue](-6,-6)(6,6) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 2 0][1 1 1.732][1 -1 1.732][1 -2 0][1 -1 -1.732][1 1 -1.732]},linecolor=red] +\psEquipotential[Q={[1 2 0][1 1 1.732 12][1 -1 1.732][1 -2 0][1 -1 -1.732][1 1 -1.732]},linecolor=blue,Vmax=50,Vmin=0,stepV=5](-5,-5)(5,5) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-5,-5)(5,5) +\psframe*[linecolor=green!20](-5,-5)(5,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 2 0][1 1 1.732][1 -1 1.732][1 -2 0][1 -1 -1.732][1 1 -1.732][-1 0 0]},linecolor=red] +\psEquipotential[Q={[1 2 0][1 1 1.732 12][1 -1 1.732][1 -2 0][1 -1 -1.732][1 1 -1.732][-1 0 0]},Vmax=40,Vmin=-10,stepV=5,linecolor=blue](-5,-5)(5,5) +\end{pspicture*} +\end{LTXexample} + +\begin{LTXexample}[pos=t,vsep=5mm] +\psset{unit=0.75cm} +\begin{pspicture*}(-6,-5)(6,5) +\psframe*[linecolor=green!20](-6,-5)(6,5) +\psgrid[subgriddiv=0,gridcolor=lightgray,griddots=10] +\psElectricfield[Q={[1 -4 0][1 -2 0 12][1 0 0][1 2 0][1 4 0]},linecolor=red] +\psEquipotential[Q={[1 -4 0][1 -2 0][1 0 0][1 2 0][1 4 0]},linecolor=blue,Vmax=30,Vmin=0,stepV=2](-7,-5)(7,5) +\end{pspicture*} +\end{LTXexample} + + + + +\clearpage +\section{List of all optional arguments for \texttt{pst-electricfield}} + +\xkvview{family=pst-electricfield,columns={key,type,default}} + +\nocite{*} +\bgroup +\raggedright +\bibliographystyle{plain} +\bibliography{pst-electricfield-doc} +\egroup + + +\printindex + + + + + + + +\end{document} |