ElectricFieldOfAChargedDisk 2006-09-22 15:42:46 2006-09-22 19:03:39 Example Electric Field % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) \usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \section{Electric Field of a Charged Disk} The Electric Field of a charged disk can teach us important concepts that you will see over and over in physics: superposition, cylindrical coordinates and non-constant basis vectors. To get a glimpse of the power of superposition, we will solve this problem the hard way first and then see how superposition can be a powerfull tool. Let us calculate the electric field at a point P above the \emph{center} of a charged disk with radius of R and a uniform surface charge density of $\sigma$ as shown in below figure. \begin{figure} \includegraphics[scale=.8]{EfieldDisk.eps} \vspace{20 pt} \end{figure} Starting with the general formula for a surface charge \begin{equation} {\bf E} = \frac{1}{4 \pi \epsilon_0} \int \frac{\sigma(r')({\bf r} - {\bf r'}) da'}{\left |{\bf r} - {\bf r'} \right |^3} \end{equation} choose a coordinate system, a disk clearly lends itself to cylindrical coordinates. As a refresher the next figure shows the ininitesimal displacement, where we have the ininitesmal area $da'$ cartesian coordinates: $$da' = dx'dy'$$ cylindrical coordinates: $$da' = s'ds'd\phi'$$ \begin{figure} \includegraphics[scale=.5]{InfDis.eps} \vspace{20 pt} \end{figure} The vectors to the source and field points that are needed for the integration in cylindrical coordinates $${\bf r} = z \hat{z}$$ $${\bf r'} = s' \hat{\phi}'$$ therefore $${\bf r} - {\bf r'} = z \hat{z} - s' \hat{\phi}'$$ $$ \left |{\bf r} - {\bf r'} \right | = \sqrt{s'^2 + z^2} $$ substituting these relationships into (1) gives us \begin{equation} {\bf E} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}}\left ( z \hat{z} - s' \hat{\phi}' \right ) \end{equation} As usual break up the integration into the $z$ and $\phi$ components {\bf z} component: $${\bf E} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{z \hat{z}s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$ Since $\hat{z}$ is always in the same direction and has the same magnitude (unit vector), it is constant and can be brought out of the integration. Integrating the ds them $${\bf E} = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_0^R \frac{s'ds'}{\left ( s'^2 + z^2 \right )^{3/2}} $$ using u substitution $$u = s'^2 + z^2$$ $$du = 2s' ds$$ $$ds = \frac{du}{2s}$$ with the limits of integration becoming $$u(s'=0) = z^2$$ $$u(s'=R) = R^2 + z^2$$ trasnforming the integral to $${\bf E} = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_{z^2}^{R^2 + z^2} \frac{u^{-3/2}du}{2} $$ integrating $${\bf E} = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \left. \right |_{z^2}^{R^2 + z^2} -\frac{u^{-1/2}du}{4} $$ evaluating the limits $${\bf E} = \frac{\sigma z \hat{z}}{16 \pi \epsilon_0} \int_0^{2\pi} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) d\phi' $$