From e2fed4f3049dd18e785a14d5d42ec00ce85ab72a Mon Sep 17 00:00:00 2001 From: bai2795 <dieter.horns@desy.de> Date: Fri, 3 May 2024 19:52:40 +0200 Subject: [PATCH] added appendix, including formula explicit integration over charged sphere to demonstrate the result obtained with Gauss law. tbd: * make the differentials everywhere non-slanted * add the atmospheric electricity part --- v0.1/Formelsammlung.tex | 2 +- v0.1/ac_circuit.tex | 2 +- v0.1/appendix.tex | 2 + v0.1/chargedshell.tex | 57 +++ v0.1/elektrostatik.tex | 24 +- v0.1/figures/drawing_sphere.png | Bin 0 -> 13086 bytes v0.1/figures/drawing_sphere.svg | 678 ++++++++++++++++++++++++++++++++ v0.1/maxwell_gleichungen.tex | 2 +- v0.1/skript.tex | 6 +- 9 files changed, 766 insertions(+), 7 deletions(-) create mode 100644 v0.1/appendix.tex create mode 100644 v0.1/chargedshell.tex create mode 100644 v0.1/figures/drawing_sphere.png create mode 100644 v0.1/figures/drawing_sphere.svg diff --git a/v0.1/Formelsammlung.tex b/v0.1/Formelsammlung.tex index ee893d8..1003877 100644 --- a/v0.1/Formelsammlung.tex +++ b/v0.1/Formelsammlung.tex @@ -1,4 +1,4 @@ -\chapter*{Formelsammlung} +\chapter{Formelsammlung} \begin{enumerate}[a)] \item \textbf{Ladung/Ströme} \begin{enumerate}[i)] diff --git a/v0.1/ac_circuit.tex b/v0.1/ac_circuit.tex index 41874b4..ca140e1 100644 --- a/v0.1/ac_circuit.tex +++ b/v0.1/ac_circuit.tex @@ -7,7 +7,7 @@ Die Kapazität \index{Kapazität} $C$ (abhängig von der Geometrie, in Einheiten Farad\index{Farad}) gibt das Verhältnis von Ladung $Q$ zu Spannung $U$ an (z.B. Plattenkondensator mit Fläche $A$ und Plattenabstand $d$: $C=\epsilon_0\epsilon_r \frac{A}{d}$). Damit ist die Ladung $Q$ bei - einer Spannung $U$ gegeben als $$Q=CU$$ und die Energie im elektrischen Feld $W_E=\frac{Q}{2C^2}$. + einer Spannung $U$ gegeben als $$Q=CU$$ und die Energie im elektrischen Feld des Kondensators $W_E=\frac{Q}{2C^2}$. \item \textit{Induktivität}:\index{Induktivität} Die Induktivität $L$ (abhängig von der Geometrie, Einheit Henry\index{Henry}) gibt das Verhältnis von magnetischen Fluss $\phi_m$ zur Stromstärke $I$ einer Spule an (z.B. Solenoidspule mit $N$ Windungen, Querschnittsfläche $A$ und der L\"ange $\ell$ hat $L=\mu_0\mu_r\frac{AN^2}{\ell}$). Der magnetische Fluss $\phi_m$ ergibt sich bei einem Strom $I$ zu $$\phi_m = LI$$ und die induzierte Spannung $U_\mathrm{ind}$: $\dot \phi_m=L\dot I=-U_\mathrm{ind}$ diff --git a/v0.1/appendix.tex b/v0.1/appendix.tex new file mode 100644 index 0000000..e1976b2 --- /dev/null +++ b/v0.1/appendix.tex @@ -0,0 +1,2 @@ +\chapter{Elektrisches Feld: Geladene Kugelfläche ohne Gauss'sches Gesetz} +\end{appendices} diff --git a/v0.1/chargedshell.tex b/v0.1/chargedshell.tex new file mode 100644 index 0000000..26ead01 --- /dev/null +++ b/v0.1/chargedshell.tex @@ -0,0 +1,57 @@ +\chapter{Elektrisches Feld einer geladenen Kugelfläche: ohne Gauss'sches Gesetz} +\label{appendix:chargedshell} +Im Abschnitt \ref{section:gauss} haben wir das elektrische Feld in der Nähe einer geladenen Kugelschale +mit dem Gauß'schen Gesetz bestimmt. Hier demonstrieren wir, dass das Ergebnis auch ohne das Gauß'sche Gesetz (allerdings mit mehr Aufwand) gefunden werden kann. + +Wir bestimmen das Feld an einem Punkt $P$, der sich im Abstand $r$ von dem Mittelpunkt $O$ einer Kugelschale mit Radius $R$ befindet. Wir nehmen zunächst an, +dass der Punkt $P$ außerhalb der Kugelschale liegt, also $r>R$ gilt. Wir wählen den Ursprung des Koordinatensystems den Punkt $O$ und legen die $z$-Achse entlang +der Richtung $\bar{OP}$, so dass $\mathbf{r} = r\mathbf{e}_z$. +Auf der dünnen Kugelschale befindet sich die Gesamtladung $Q$ über die gesamte Fläche $4\pi R^2$ gleichförmig verteilt (siehe Abbildung~\ref{fig:spherecharge}). + +Um das elektrische Feld am Punkt $P$ auszuwerten, müssen wir über sämtliche Ladungen auf der Oberfläche integrieren. Aufgrund der Symmetrie ist +das resultierende Feld $\mathbf{E}_P = E_P~\mathbf{e}_z$ (ähnlich wie beim Ring im Beispiel im Abschnitt~\ref{sec:efield}): Die Feldkomponenten in der Richtung senkrecht +zur $z$-Richtung addieren sich zu Null, wenn wir über die Kugelsegmente integrieren. + +\[ + E_P = \frac{1}{4\pi\epsilon_0}~\int \mathrm{d}A \frac{Q}{4\pi R^2} \frac{(\mathbf{r}-R\mathbf{e}_{r'})\cdot \mathbf{e}_z}{|\mathbf{r}-R\mathbf{e}_{r'}|^3}, +\] +wobei das Flächenelement $\mathrm{d}A = R^2\mathrm{d}\phi \sin\vartheta\mathrm{d}\vartheta$ und $\vartheta$ der Winkel zwischen $\mathbf{e}_{r'}$ und $\mathbf{e}_z$ ist. +Wir erhalten nach Integration über $\phi$ und Ausmultiplizieren von Zähler und Nenned: +\[ +E_P = \frac{Q}{8\pi\epsilon_0} \int\limits_0^\pi \mathrm{d}\vartheta \sin\vartheta \frac{r-R\cos\vartheta}{(r^2+R^2-2Rr\cos\vartheta)^{3/2}}. +\] +Wir substituieren $z=r^2+R^2-2Rr\cos\vartheta$, so dass $\mathrm{d}z=2Rr\sin\vartheta~\mathrm{d}\vartheta$ und +$R\cos\vartheta = (r^2+R^2-z)/2r$: +\[ +E_P = \frac{Q}{32\pi\epsilon_0 Rr^2} \int\limits_{(R-r)^2}^{(R+r)^2} \mathrm{d}z \frac{r^2-R^2+z}{z^{3/2}}. +\] +Wir werten die Stammfunktionen an den Grenzen aus: +\[ +E_P = \frac{Q}{32\pi\epsilon_0 Rr^2}\left. \left( 2 z^{1/2} - 2\frac{r^2-R^2}{z^{1/2}} \right) \right|_{(R-r)^2}^{(R+r)^2} +\] +so dass wir mit +\[ + E_P = \frac{Q}{16\epsilon_0 Rr^2} \left(R+r - |R-r| - \left(\frac{(r-R)(r+R)}{R+r} - \frac{(r-R)(r+R)}{|R-r|}\right)\right) +\] +eine Fallunterscheidung machen müssen zwischen $r>R$ und $r<R$. +Für $r>R$ ist $|R-r| = r-R$, so dass +\[ +E_P (r>R) = \frac{Q}{16\pi \epsilon_0 Rr^2} 4R = \frac{Q}{4\pi \epsilon_0 r^2}, +\] +und für $r<R$ mit $|R-r| = R-r$: +\[ +E_P(r<R) = \frac{Q}{16\pi\epsilon_0 Rr^2} (R+r-R+r - (r-R +r+R)) = 0. +\] +Damit ist das Ergebnis aus dem Abschnitt~\ref{sec:anwendunggauss} bestätigt: Außerhalb einer geladenen Kugelschale mit Gesamtladung $Q$ fällt das +elektrische Feld genauso ab, als ob sich im Mittelpunkt eine Punktladung $Q$ befinden würde. Innerhalb der Kugelschale verschwindet das Feld. + +\begin{figure} +\parbox{0.7\linewidth}{ +\includegraphics[width=1.\linewidth]{figures/drawing_sphere.png}} +\parbox{0.28\linewidth}{ +\caption{Zur Bestimmung des elektrischen Feldes im Punkt $P$ wird über die geladene Kugelfläche mit Radius $a$ integriert. +\label{fig:spherecharge}} +} +\end{figure} + + diff --git a/v0.1/elektrostatik.tex b/v0.1/elektrostatik.tex index 3b5e2da..f23c8cf 100644 --- a/v0.1/elektrostatik.tex +++ b/v0.1/elektrostatik.tex @@ -200,6 +200,7 @@ Das Superpositionsprinzip lässt sich in dieser Weise auch durch ein Volumeninte \section{Das elektrische Feld} +\label{sec:efield} Wir haben mit dem Coulomb-Gesetz \ref{eq:coulomb1} ein neues Kraftgesetz gefunden (vgl. Gravitationskraft, s. Physik I). Ein Kraftgesetz erlaubt es uns, die Kräfte zwischen zwei Körpern anzugeben. Wir hatten hierbei durch die Notation $\mathbf{F}_{1,2}$ angedeutet, dass die Kraft auf die Ladung $Q_2$ ursächlich durch @@ -238,8 +239,8 @@ beliebigen Ladungsverteilung \lambda = \frac{Q}{2\pi a}. \] Wir teilen das elektrische Feld am Punkt -$P$ im Abstand $x$ in zwei Komponenten auf $E_x=|\mathbf{E}~\cos\theta$ und $E_\perp$. Wenn wir entlang dem Ring -integrieren bekommen wir für +$P$ im Abstand $x$ in zwei Komponenten auf $E_x=|\mathbf{E}|~\cos\theta$ und $E_\perp$. Wenn wir entlang dem Ring +integrieren, bekommen wir für \[ E_\perp = \frac{1}{4\pi\epsilon_0}\int d\ell \frac{\lambda}{r^2}=0 \] @@ -721,7 +722,7 @@ ist proportional zur Ladung in dem umschlossenen Volumen.} \[ \mathbf{E}(R) = \frac{1}{4\pi\epsilon_0} \frac{Q}{R^2} \mathbf{e}_r. \] - Das Flächenelement in Kugelkoordinaten $d\mathbf{A} = R^2\sin\vartheta d\theta d\phi \mathbf{e}_r$ ist nach Konvention für geschlossene Flächen so ausgerichtet, dass der Normalenvektor + Das Flächenelement in Kugelkoordinaten $d\mathbf{A} = R^2\sin\vartheta d\vartheta d\phi \mathbf{e}_r$ ist nach Konvention für geschlossene Flächen so ausgerichtet, dass der Normalenvektor nach außen zeigt ($\mathbf{n}=\mathbf{e}_r$). Wir können das Integral sofort auswerten: \[ \Phi_E = \oint d\mathbf{A}\cdot \mathbf{E} = R^2 \frac{Q}{4\pi\epsilon_0 R^2} \underbrace{\int\limits_{0}^{2\pi}d\phi\int\limits_{0}^\pi d\vartheta \sin\vartheta}_{=2\pi\cdot 2=4\pi}=\frac{Q}{\epsilon_0} @@ -758,6 +759,7 @@ die wir integrieren. \subsection{Anwendungen Gauß'sches Gesetz} +\label{sec:anwendunggauss} Das Gauß'sche Gesetz ermöglicht es prinzipiell, das elektrische Feld ohne genaue Kenntnis der Ladungsverteilung zu bestimmen. Der elektrische Fluss bzw. die Ladung innerhalb einer geschlossenen Fläche genügt, um in bestimmten Fällen @@ -803,6 +805,22 @@ Insbesondere kann mit einer sinnvoll gewählten Oberfläche \end{example} +\begin{example}[Geladene Kugelschale] + Wir betrachten eine Kugelschale mit Radius $R$, die die Ladung $Q$ trägt. Wir berechnen analog zum vorigen Beispiel + den Fluß und setzen den Fluß gleich mit der eingeschlossenen Ladung: + \[ + \oint \mathrm{d}\mathbf{A}\cdot E = 4\pi r^2 E = \frac{Q}{\epsilon_0}, + \] + für $r>R$. Damit ergibt sich das Feld zu + \[ + E(r) = \frac{Q}{4\pi\epsilon_0 r^2}, + \] + also identisch zu dem Feld einer Punktladung im Mittelpunkt der Kugelschale. Für $r<R$ ist der Fluß null, weil sich innerhalb der Integrationsfläche + keine Ladungen befinden. +Im Anhang \ref{appendix:chargedshell}, bestimmen wir zum Vergleich das elektrische Feld ohne +Anwendung des Gauß'schen Gesetzes. +\end{example} + \begin{example}[Homogen geladene unendlich ausgedehnte Platte] Abbildung~\ref{fig:gauss12} rechts: Wir nehmen an, dass die Platte mit einer konstanten Flächenladungsdichte $\sigma$ aufgeladen ist. 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\usepackage{enumerate} \usepackage{bm} \usepackage{color} @@ -144,6 +145,9 @@ Konsultieren Sie gegebenenfalls auch die empfohlenen Lehrb\"ucher.\\ \input{em_waves} \input{geometrical_optics} \input{wellenoptik} -\input{Formelsammlung} \printindex +\begin{appendices} +\input{Formelsammlung} +\input{chargedshell} +\end{appendices} \end{document} -- GitLab