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Jonas Hahn
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117 lines
3.4 KiB
Typst
117 lines
3.4 KiB
Typst
// Main VL template
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#import "../preamble.typ": *
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// Fix theorems to be shown the right way in this document
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#import "@preview/ctheorems:1.1.3": *
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#show: thmrules
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// Main settings call
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#show: conf.with(
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// May add more flags here in the future
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num: 6,
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type: 0, // 0 normal, 1 exercise
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date: datetime.today().display(),
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//date: datetime(
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// year: 2025,
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// month: 5,
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// day: 1,
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//).display(),
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)
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= Uebersicht
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== Kondensatoren und Kapazitaet
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Wir betrachten ein Volumen $V$ mit verschiedenen Ladungen $S_(i) $ (Leitern) im Inneren des Volumens.
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Die Laplace Gleichung
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$
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Delta f_(i) (arrow(x)) = 0 space forall arrow(x) in V
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$
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mit den Randbedingungen
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$
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f_(i) (arrow(x)) |_(S_(i)) = 1 \, space forall j != i : f_(i) |_(S_(j)) = 0.
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$
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Linearitaet der Maxwellgleichungen
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$
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Phi (arrow(x)) = sum_(i = 1)^(N + 1) phi_(i) f_(i) (arrow(x)).
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$
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Ladung auf Leiter
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$
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Q_(i) = integral _(S_(i) ) dif S sigma_(i) = epsilon_0 integral _(S_(i) ) dif S arrow(E) * hat(n) = - epsilon_0 integral _(S_(i) ) dif arrow(S) * arrow(nabla) Phi \
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= - epsilon_0 sum _(j = 1) ^(N + 1) phi_(j) integral _(S_(i) ) dif arrow(S)* arrow(nabla) f_(j) \
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= sum _(j = 1) ^(N) C_(i j) phi_(j).
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$
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Mit der Kapazitaetsmatrix
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$
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C_(i j) = underbrace(- epsilon_0 integral _(S_(i) ) dif arrow(S) * arrow(nabla) f_(j), "Definition") = - epsilon_0 integral _(partial V) dif arrow(S)* (arrow(nabla) f_(j)) f_(i) \
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= ^("Gauss") epsilon_0 integral _(V) dif ^3 x arrow(nabla) * (f_(i) arrow(nabla) f_(j) ) = epsilon_0 integral _(V) dif ^3 x (arrow(nabla) f_(i) * arrow(nabla) f_(j) ) + 0
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$
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welche nur von der Geometrie des Problems abhaengt. Diese hat eine Reihe von Eigenschaften
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- Sie ist symmetrisch $C_(i j) = C_(j i) $
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Es gilt
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$ sum _(i) C_(i j) = - epsilon_0 sum _(i) integral _(S_(i) ) dif arrow(S) * arrow(nabla) f_(j) \
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= epsilon_0 integral _(partial V) dif arrow(S) * arrow(nabla) f_(j) \
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= ^("Gauss") epsilon_0 integral _(V) dif^3 x underbrace(arrow(nabla) * (arrow(nabla) f_(j) ), = 0) = 0.
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$
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#example[
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Plattenkondensator.
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Es gilt
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$
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C = mat(
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C_(1 1) , C_(1 2) ;
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C_(2 1) , C_(2 2) ;
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) = mat(
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C_(1 1) , - C_(1 1) ;
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- C_(1 1) , C_(1 1) ;
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).
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$
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Dieser hat also einfach eine Kapazitaet von $C$.
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Konkret gilt
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$
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arrow(E) = (sigma) / (epsilon_0 ) hat(z) \
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V = phi (0) - phi (d) = E d = (sigma d) / (epsilon_0 ) = (Q d) / (A epsilon_0 ) \
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C = (Q) / (V) = (A epsilon_0 ) / (d).
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$
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]
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In einem Kondensator wird Energie gespeichert
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$
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U = (epsilon_0 ) / (2) integral dif^3 x arrow(E)^2 (arrow(x)) = (epsilon_0 ) / (2) A d ((sigma) / (epsilon_0 ) )^2 \
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= 1/(2 epsilon_0 ) Q^2 /A d = 1/2 (Q^2 ) / (C) = 1/2 C V^2.
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$
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Fuer eine allgemeine Konfiguration
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$
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U = 1/2 sum _(i) Q_(i) phi_(i) \
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= 1/2 sum _(i, j) C_(i j) phi_(i) phi_(j).
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$
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= Magnetfelder
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Zum Zeitpunkt $arrow(j)$ ist auch der Zeitpunkt $arrow(B)$. Es gilt $rho = 0 ==> arrow(E) = 0$.
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Maxwellgleichungen
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$
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arrow(nabla) times arrow(B)= mu_0 arrow(j) \
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arrow(nabla) * arrow(B) = 0 space "es gibt keine magnetischen Monopole".
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$
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=== Konsistenzcheck
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Konti
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$
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(diff rho) / (diff t) + arrow(nabla) arrow(j) = 0 \
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"Zeitunabhaengig" ==> arrow(nabla) * arrow(j) = 0 \
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arrow(nabla) (arrow(nabla) times arrow(B)) = 0.
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$
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American Journal of Physics: 92 (2024) 583 \
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J. Franklin, D. Griffiths, D. Schroeter \
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A taxonomy of magnetic field lines
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#remark[
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Magnetische Feldlinien muessen nicht geschlossen sein.
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]
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