%% This document created by Scientific Word (R) Version 2.0 %% %% This lecture is not available in HTML format. This is a LaTex file. %% It is best to download this file on a PC that has Scientific Workplace %% If you want to use a generic latex2dvi driver, erase the lines so marked %% Even so, you may have problems, especially with figures. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \documentclass[12pt]{report} \usepackage{amsfonts} %%%%% \usepackage{sw20jrep} %%ERASE THIS LINE IF NOT USING SW %TCIDATA{TCIstyle=Report/report.lat,jrep,report} \input tcilatex \pagestyle{myheadings} \markright{{\sf Lecture 3 (Ch 1 part 3) Capacitance}} \addtolength{\topmargin}{-0.4in} \addtolength{\textheight}{1in} \addtolength{\oddsidemargin}{-0.2in} \addtolength{\textwidth}{0.8in} \setcounter{chapter}{2} \thispagestyle{empty} \QQQ{Language}{American English} %%ERASE THIS LINE IF NOT USING SW \begin{document} \chapter{Ch.1, part 3} ------------------------ \vspace{1.0in} \textsc{Lecture 3: Capacitance and capacitors}\ \newpage\ \section{A single conductor} If an isolated conductor is raised to the potential $V$, the potential everywhere is $\Phi (\mathbf{x)}=V\phi (\mathbf{x})$, where $\phi (\mathbf{x)% }$ $=1$ on the conductor and vanishes at infinity. The charge on the conductor, as given by Gauss's theorem, is \[ 4\pi Q=-V\int_{\mathcal{S}}\frac{\partial \phi }{\partial n}\,da \] where $\mathcal{S}$ is (conveniently) the surface of the conductor. We see that $Q$ is proportional to $V$. By definition, the capacitance is \begin{equation} C=\frac QV=-\frac 1{4\pi }\int_{\mathcal{S}}\frac{\partial \phi }{\partial n}% \,da \label{C} \end{equation} The energy of the charged conductor is \[ W=\frac 12QV=\frac 12\ CV^2=\frac 12\frac{Q^2}C \] Another expression for $W$ is the usual \[ W=\frac 1{8\pi }\int_{\mathcal{V}}\left( V\text{\thinspace }\mathbf{\nabla }% \phi \right) ^2d^3x \] where $\mathcal{V}$ is the volume external to the conductor. Comparing these two expressions for $W$, we get another formula for $C$% \[ C=\frac 1{4\pi }\int_{\mathcal{V}}(\mathbf{\nabla }\phi )^2d^3x \] (The earlier formula is recovered from this with an integration by parts). We see that $C$ is always positive. This follows also from the fact that $% \phi (\mathbf{x)}$ has its maximum, $\phi =1$, on the surface $\mathcal{S}$, so that $\partial \phi /\partial n$ is negative on $\mathcal{S}$. \noindent ------------------------------ In esu, the unit of $C$ the $\mathrm{cm}$. In the MKSA system, it is the $% \mathrm{farad}$: \[ \mathrm{farad}=\frac{\mathrm{coul}}{\mathrm{volt}}=\frac{3\cdot 10^9\mathrm{% statcoul}}{\left( 1/300\right) \,\mathrm{statvolt}}=9\cdot 10^{11}\mathrm{cm} \] For a sphere of radius $a$, the (esu) capacitance is $C=a$. The farad is a huge capacitance. In common use is the picofarad: \[ \mathrm{pF}=\mu \mu \mathrm{F}=0.9\,\mathrm{cm} \] \noindent --------------------------------- For a simple body, $C$ is of the order of the size of the body. For a sphere of radius $a,$ we have the basic result $C=a$ (in gaussian units, of course; in SI units $C=4\pi \epsilon _0a$). If body A can be inscribed in body B, it can be shown that $C_A>b$): \[ C=\dfrac a{\ln \dfrac{2a}b} \] \end{itemize} \item oblate ellipsoid, axes $a,a,b$, focal distance $c=\sqrt{a^2-b^2}$: \[ C=\frac c{\arctan \dfrac cb} \] \begin{itemize} \item disk ($a>>b$): \[ C=\frac{2a}\pi \] \end{itemize} \item thin toroidal ring, ring radius $a,$ cross-section radius $b:C=\dfrac{% \pi a}{\ln \dfrac{8a}b}$ \item spherical bowl, radius $a,$ subtended angle $2\theta :$% \[ C=\frac a\pi \left( \theta +\sin \theta \right) \] \FRAME{itbpF}{5.0984cm}{5.0786cm}{0cm}{}{}{Plot }{\special{language "Scientific Word";type "MAPLEPLOT";width 5.0984cm;height 5.0786cm;depth 0cm;display "PICT";plot_snapshots TRUE;function \TEXUX{$1$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$.01$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "-2";xmax "2";xviewmin "-2";xviewmax "2";yviewmin "-1.04";yviewmax "1.041";rangeset"X";recompute TRUE;phi 45;theta 45;plottype 8;constrained TRUE;numpoints 49;axesstyle "none";xis \TEXUX{v627};var1name \TEXUX{$\theta $};valid_file "T";tempfilename 'C:/SW20/DOCS/DL8UAU99.wmf';tempfile-properties "XP";}} \smallskip\ \FRAME{% itbpFU}{5.1972cm}{4.9094cm}{0cm}{\Qcb{$C/a$ vs. $x=\theta /\pi $}}{}{Plot }{% \special{language "Scientific Word";type "MAPLEPLOT";width 5.1972cm;height 4.9094cm;depth 0cm;display "PICT";plot_snapshots TRUE;function \TEXUX{$x+\left( 1/\pi \right) \sin \left( \pi x\right) $};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "0";xmax "1";xviewmin "0";xviewmax "1";yviewmin "-0.02";yviewmax "1.02";rangeset"X";recompute TRUE;phi 45;theta 45;plottype 4;numpoints 49;axesstyle "boxed";xis \TEXUX{x};var1name \TEXUX{$x$};valid_file "T";tempfilename 'C:/SW20/DOCS/DL8UAV8A.wmf';tempfile-properties "XP";}} Note that the results for the ellipsoids and for the bowl reduce to $C=a$ in the appropriate limit ( $b=a$ for the ellipsoids, $\theta =\pi $ for the bowl). Also, for small $\theta $ the bowl reduces to a disk of radius $% a\theta $ and its capacitance is $C=2a\theta /\pi $ as expected. \end{itemize} ----------------------------------------------------- \section{Many conductors} For two bodies, the relation between charges and voltages is \label{C2} \begin{eqnarray} Q_1 &=&C_{11}V_1+C_{12}V_2 \label{C2} \\ Q_2 &=&C_{21}V_1+C_{22}V_2 \nonumber \end{eqnarray} with $C_{21}=C_{12}.$ It can be shown that $C_{11}$ and $C_{22}$ are positive, $C_{12}$ is negative, and $C_{11}\geq \left| C_{12}\right| $ as well as $C_{22}\geq \left| C_{12}\right| $. To obtain these results, start from the two basic configurations where $% V_1=1,V_2=0$ and $V_1=0,V_2=1$. Call the corresponding potentials $\phi _1(x) $ and $\phi _2(x)$. In the general configuration the potential is $% \Phi (x)=V_1\phi _1(x)+V_2\phi _2(x)$ and Gauss's theorem gives \[ 4\pi Q_1=-V_1\int_{\mathcal{S}_1}\frac{\partial \phi _1}{\partial n}% \,da-V_2\int_{\mathcal{S}_1}\frac{\partial \phi _2}{\partial n}\,da \] \[ 4\pi Q_2=-V_1\int_{\mathcal{S}_2}\frac{\partial \phi _1}{\partial n}% \,da-V_2\int_{\mathcal{S}_2}\frac{\partial \phi _2}{\partial n}\,da \] This argument clearly can be generalized to any number of conductors. --------------------------- \textsc{Homework Assignment} Analogously to the one-conductor case, give explicit expressions for $C_{ij}$ and use energy arguments, or flux arguments, to show that $C_{ii}$ is positive, $C_{ij}$ negative for $i\neq j$, and $\sum_jC_{ij}$ is non-negative. Use Green's theorem to show that $C_{ij}=C_{ji}$. ------------------------------ \subsection{Capacitors} Of practical interest is the configuration with zero net charge, $Q_1=Q$ and $Q_2=-Q$. \begin{itemize} \item In an ideal \textit{capacitor }$C_{12}=-C_{11}$ (so that no flux escapes): then eq. (\ref{C2}) gives $Q=C_{11}\Delta V$, where $\Delta V=V_1-V_2$ (note that $C_{11}$ is not the same as the capacitance of the isolated conductor 1). \item In the case of a symmetrical capacitor, $V_1=\Delta V/2$ and $% V_2=-\Delta V/2$. \item Quite generally we can solve for $V_1$ and $V_2$ to obtain $\Delta V=Q/C$, with the \textit{internal capacitance }now given by \begin{equation} C=\frac{C_{11}C_{22}-C_{12}C_{21}}{C_{11}+C_{22}+C_{12}+C_{21}} \label{Cm} \end{equation} \end{itemize} --------------------------- \textsc{Homework Assignment} Obtain eq. (\ref{Cm}) and its simpler forms that apply when: \begin{itemize} \item The two conductors are equal \item Conductor 2 surrounds conductor 1 \end{itemize} A spherical capacitor consists of a metal sphere of radius $a$ and a metal shell of inner radius $b_1$, outer radius $b_2$. Find $C$ and all the $% C_{ij} $ elements for this capacitor. When the capacitor is charged ($Q$ on the inner sphere, $-Q$ on the outer spherical shell), what are the potentials $V_1$ and $V_2$? How is the charge distributed on the outer shell (how much is on the inner surface, how much on the outer surface? \FRAME{dtbpF}{7.62cm}{5.08cm}{0pt}{}{}{Plot }{\special{language "Scientific Word";type "MAPLEPLOT";width 7.62cm;height 5.08cm;depth 0pt;display "PICT";maintain-aspect-ratio TRUE;plot_snapshots TRUE;function \TEXUX{$1$};linecolor "red";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$1.5$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";function \TEXUX{$1.55$};linecolor "black";linestyle 1;linethickness 1;pointstyle "point";xmin "-3.1416";xmax "3.1416";xviewmin "-3.142";xviewmax "3.142";yviewmin "-1.612";yviewmax "1.613";rangeset"X";phi 45;theta 45;plottype 8;constrained TRUE;numpoints 49;axesstyle "none";xis \TEXUX{v627};var1name \TEXUX{$\theta $};valid_file "T";tempfilename 'C:/SW20/DOCS/DL8PNGIK.wmf';tempfile-properties "XP";}} --------------------------------------------- Here is a list of $C$ for various capacitors. \begin{itemize} \item Two closely spaced parallel plates of area $A$, separation $d$ ($% A>>d^2$): \[ C=\frac A{4\pi d} \] \item Two concentric cylinders, radii $a$ and $b$ with $a>b$, length $% L>>(a-b)$: \[ C=\frac L{2\ln \dfrac ab} \] \item Two parallel cylinders, both of radius $a$, separation $D>2a$, length $L>>D:$% \[ C=\frac L{4\ln \left( \dfrac D{2a}+\sqrt{\left( \dfrac D{2a}\right) ^2-1}% \right) } \] \begin{itemize} \item for $D>>a$, this reduces to \[ C=\frac L{4\ln \dfrac Da} \] \end{itemize} \item Any two parallel cylinders, radii $a,b$, separation $D,$ length $L:$ \[ C=\frac L{2\ln \left( x+\sqrt{x^2-1}\right) } \] where \[ x=\left| \frac{D^2-a^2-b^2}{2ab}\right| \] This formula applies when the cylinders are external to each other $(D>a+b)$ and $L>>D,$ as well as when they are inside each other $(D<\left| a-b\right| )$ and $L>>\left| a-b\right| .$ The case of two intersecting cylinders is different, since they must be at the same potential. \item Two concentric spheres, radii $a$ and $b$, with $a>b:$% \[ C=\frac 1{\dfrac 1b-\dfrac 1a} \] \end{itemize} \subsection{Forces on charged conductors} Forces can be obtained simply by differentiating at constant $Q_i$ the energy formula \[ W=\frac 12\sum_iV_iQ_i=\frac 12\sum_{ij}\left( \Bbb{C\,}^{-1}\right) _{ij}Q_iQ_j \] where $\Bbb{C\,}^{-1}$ is the inverse of the capacitance matrix. The force in the general direction of the parameter $\xi $ is \[ F_\xi =-\left( \frac{\partial W}{\partial \xi }\right) _Q=-\frac 12\sum_{ij}Q_iQ_j\frac \partial {\partial \xi }\left( \Bbb{C\,}^{-1}\right) _{ij} \] \begin{itemize} \item For a single conductor \[ W=\frac{Q^2}{2C} \] and the force in the general direction of the parameter $\xi $ is \begin{equation} F_\xi =-\left( \frac{\partial W}{\partial \xi }\right) _Q=-\frac{Q^2}2\frac{% \partial (1/C)}{\partial \xi }=\frac{Q^2}{2C^2}\frac{\partial C}{\partial \xi } \label{FQ} \end{equation} For example, a charged sphere of radius $r$ experiences a radial force \[ F_r=-\frac{\partial W}{\partial r}=\frac{Q^2}{2r^2} \] As expected, this force pushes outward, expanding the sphere. \item For a capacitor the force between the plates is also given by eq. (% \ref{FQ}) where $C$ is now the internal capacitance. For example, the mutual force between the plates of a thin capacitor ($C=A/4\pi x)$ is \[ F_x=-Q^2\frac \partial {\partial x}\left( \frac 12\frac{4\pi x}A\right) =-% \frac{2\pi Q^2}A \] As expected, this force is attractive and its magnitude per unit area is $% 2\pi \sigma ^2$, where $\sigma =Q/A$ is the surface charge density. \end{itemize} \noindent --------------------------------- If we want to obtain the forces by differentiating at constant $V$, we must use the formula (with the seemingly ``wrong'' sign) \begin{equation} F_x=\left( \frac{\partial W}{\partial x}\right) _V=\frac \partial {\partial x}\left( \frac 12CV^2\right) =\frac{V^2}2\frac{\partial C}{\partial x} \label{F2} \end{equation} The reason for the ''wrong'' sign is that the virtual work at constant $V$ is $\delta W-V\delta Q$, where the extra term is the work (typically done by batteries) to restore the charge. Since $\delta W=\frac 12V\delta Q$, we find that $\delta W-V\delta Q=-\delta W$. \end{document}