Talk:PlanetPhysics/Electric Field of a Charged Disk

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%%% This file is part of PlanetPhysics snapshot of 2011-09-01 %%% Primary Title: Electric Field of a Charged Disk %%% Primary Category Code: 41.20.Cv %%% Filename: ElectricFieldOfAChargedDisk.tex %%% Version: 8 %%% Owner: bloftin %%% Author(s): bloftin %%% PlanetPhysics is released under the GNU Free Documentation License. %%% You should have received a file called fdl.txt along with this file. %%% If not, please write to gnu@gnu.org. \documentclass[12pt]{article} \pagestyle{empty} \setlength{\paperwidth}{8.5in} \setlength{\paperheight}{11in}

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\begin{document}

\section{Electric Field of a Charged Disk}

The Electric Field of a charged disk can teach us important \htmladdnormallink{concepts}{http://planetphysics.us/encyclopedia/PreciseIdea.html} that you will see over and over in physics: superposition, cylindrical coordinates and non-constant basis \htmladdnormallink{vectors}{http://planetphysics.us/encyclopedia/Vectors.html}. To get a glimpse of the \htmladdnormallink{power}{http://planetphysics.us/encyclopedia/Power.html} 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 \htmladdnormallink{Electric Field}{http://planetphysics.us/encyclopedia/ElectricField.html} at a point P above the \emph{center} of a charged disk with radius of R and a uniform surface \htmladdnormallink{charge}{http://planetphysics.us/encyclopedia/Charge.html} density of $\sigma$ as shown in below figure.

\begin{figure} \includegraphics[scale=.8]{EfieldDisk.eps} \vspace{20 pt} \end{figure}

Starting with the general \htmladdnormallink{formula}{http://planetphysics.us/encyclopedia/Formula.html} 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 \htmladdnormallink{system}{http://planetphysics.us/encyclopedia/SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence.html}. A disk clearly lends itself to cylindrical coordinates. As a refresher, the next figure shows the infinitesimal displacement, where we have the infinitesmal 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 \htmladdnormallink{field}{http://planetphysics.us/encyclopedia/CosmologicalConstant2.html} 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:

$${\bf E}_z = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R \frac{z \hat{z}s'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

Since $\hat{z}$ is always in the same direction and has the same \htmladdnormallink{magnitude}{http://planetphysics.us/encyclopedia/AbsoluteMagnitude.html} (\htmladdnormallink{unit vector}{http://planetphysics.us/encyclopedia/PureState.html}), it is constant and can be brought out of the integration. Integrating the ds them

$${\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_0^R \frac{s'ds'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

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

$${\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \int_{z^2}^{R^2 + z^2} \frac{u^{-3/2}du}{2} $$

integrating

$${\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} d\phi' \left. \right |_{z^2}^{R^2 + z^2} -u^{-1/2}du$$

evaluating the limits

$${\bf E}_z = \frac{\sigma z \hat{z}}{4 \pi \epsilon_0} \int_0^{2\pi} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) d\phi' $$

integrating again simply gives

$${\bf E}_z = \frac{\sigma z \hat{z}}{2 \epsilon_0} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) $$

${\bf \phi}$ component:

$${\bf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R -\frac{s'^2\hat{\phi}'ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

If you cannot simply see how the $\phi$ component is zero through symmetry, then carry out the integration. The key thing to learn here, and why it is not good to just skip over the $\phi$ component, is to realize that $\hat{\phi}$ is not constant throughout the integration. Therefore, one cannot bring it out of the integration. What needs to be done is to substitute in for $\hat{\phi}$. An important result from cylindrical coordinates is the \htmladdnormallink{relation}{http://planetphysics.us/encyclopedia/Bijective.html} between its unit vectros and those of cartesian coordinates.

$$ \hat{s} = \cos \phi \hat{x} + \sin \phi \hat{y} $$ $$ \hat{\phi} = -\sin \phi \hat{x} + \cos \phi \hat{y} $$ $$ \hat{z} = \hat{z} $$

Plugging in the $ \hat{\phi}'$ into our integral

$${\bf E}_{\phi} = \frac{\sigma}{4 \pi \epsilon_0} \int_0^{2\pi} \int_0^R -\frac{s'^2 \left ( -\sin \phi' \hat{x} + \cos \phi' \hat{y} \right )ds'd\phi'}{\left ( s'^2 + z^2 \right )^{3/2}} $$

${\bf x}$ component:

To make our job easier, let us first integrate $d\phi'$

$${\bf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} \sin \phi' d\phi'$$

Note how $\hat{x}$ can be taken out of integral, so we get

$${\bf E}_{\phi}^x = \frac{\sigma \hat{x}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left. \right |_0^{2\pi} - \cos \phi' $$

Evaluating the limits, gives us the result we expected.

$${\bf E}_{\phi}^x = \frac{\sigma}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 \hat{x} ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left ( -1 - (-1) \right ) = 0 $$

${\bf y}$ component:

$${\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \int_0^{2\pi} -\cos \phi' d\phi'$$

integrating

$${\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left. \right |_0^{2\pi} -\sin \phi' $$

which once again yeilds a zero.

$${\bf E}_{\phi}^y = \frac{\sigma \hat{y}}{4 \pi \epsilon_0} \int_0^R \frac{s'^2 ds'}{\left ( s'^2 + z^2 \right )^{3/2}} \left ( 0 - 0 \right ) = 0 $$

Since the x and y components are zero

$$ {\bf E}_{\phi} = 0$$

Therefore, for a charged disk at a point above the center, we have

$${\bf E} = \frac{\sigma z \hat{z}}{2 \epsilon_0} \left ( \frac{1}{z} - \frac{1}{\sqrt{R^2 + z^2}} \right ) $$

and rearranging

$${\bf E} = \frac{\sigma }{2 \epsilon_0} \left ( 1 - \frac{z}{\sqrt{R^2 + z^2}} \right ) \hat{z} $$

\subsection{Superposition}

Before we can apply superposition to this problem, we need to calculate the electric field of a charged ring. This entry is coming soon.

\end{document}