PlanetPhysics/Electric Field of a Charged Disk

Electric Field of a Charged Disk

The Electric Field of a charged disk can teach us important [[../PreciseIdea/|concepts]] that you will see over and over in physics: superposition, cylindrical coordinates and non-constant basis [[../Vectors/|vectors]]. To get a glimpse of the [[../Power/|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 [[../ElectricField/|Electric Field]] at a point P above the center of a charged disk with radius of R and a uniform surface [[../Charge/|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/|formula]] for a surface charge

$𝐄=\frac{1}{4\pi {ϵ}_0}\int \frac{\sigma (r^{\prime })(𝐫-{𝐫}^{\prime })d{a}^{\prime }}{{|𝐫-{𝐫}^{\prime }|}^3}$

choose a coordinate [[../SimilarityAndAnalogousSystemsDynamicAdjointnessAndTopologicalEquivalence/|system]]. A disk clearly lends itself to cylindrical coordinates. As a refresher, the next figure shows the infinitesimal displacement, where we have the infinitesmal area $d{a}^{\prime }$

cartesian coordinates:

$d{a}^{\prime }=d{x}^{\prime }d{y}^{\prime }$

cylindrical coordinates:

$d{a}^{\prime }=s^{\prime }d{s}^{\prime }d{\varphi }^{\prime }$

\begin{figure} \includegraphics[scale=.5]{InfDis.eps} \vspace{20 pt} \end{figure}

The vectors to the source and [[../CosmologicalConstant2/|field]] points that are needed for the integration in cylindrical coordinates

$𝐫=z\hat{z}$ ${𝐫}^{\prime }=s^{\prime }{\hat{\varphi }}^{\prime }$

therefore

$𝐫-{𝐫}^{\prime }=z\hat{z}-s^{\prime }{\hat{\varphi }}^{\prime }$ $|𝐫-{𝐫}^{\prime }|=\sqrt{s{\text{'}}^2+z^2}$

substituting these relationships into (1) gives us

$𝐄=\frac{\sigma }{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}\frac{s^{\prime }d{s}^{\prime }d{\varphi }^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}(z\hat{z}-s^{\prime }{\hat{\varphi }}^{\prime })$

As usual break up the integration into the $z$ and $\varphi$ components

{\mathbf z} component:

${𝐄}_z=\frac{\sigma }{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}\frac{z\hat{z}{s}^{\prime }d{s}^{\prime }d{\varphi }^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}$

Since $\hat{z}$ is always in the same direction and has the same [[../AbsoluteMagnitude/|magnitude]] ([[../PureState/|unit vector]]), it is constant and can be brought out of the integration. Integrating the ds them

${𝐄}_z=\frac{\sigma z\hat{z}}{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}d{\varphi }^{\prime }\underset{0}{\overset{R}{\int }}\frac{s^{\prime }d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}$

using u substitution

$u=s{\text{'}}^2+z^2$ $du=2s^{\prime }ds$ $ds=\frac{du}{2s}$

with the limits of integration becoming

$u(s^{\prime }=0)=z^2$ $u(s^{\prime }=R)=R^2+z^2$

trasnforming the integral to

${𝐄}_z=\frac{\sigma z\hat{z}}{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}d{\varphi }^{\prime }\underset{z^2}{\overset{R^2+z^2}{\int }}\frac{u^{-3/2}du}{2}$

integrating

${𝐄}_z=\frac{\sigma z\hat{z}}{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}d{\varphi }^{\prime }{|}_{z^2}^{R^2+z^2}-u^{-1/2}du$

evaluating the limits

${𝐄}_z=\frac{\sigma z\hat{z}}{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}})d{\varphi }^{\prime }$

integrating again simply gives

${𝐄}_z=\frac{\sigma z\hat{z}}{2{ϵ}_0}(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}})$

$\mathit{\varphi }$ component:

${𝐄}_{\varphi }=\frac{\sigma }{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}-\frac{s{\text{'}}^2{\hat{\varphi }}^{\prime }d{s}^{\prime }d{\varphi }^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}$

If you cannot simply see how the $\varphi$ 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 $\varphi$ component, is to realize that $\hat{\varphi }$ 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 $\hat{\varphi }$. An important result from cylindrical coordinates is the [[../Bijective/|relation]] between its unit vectros and those of cartesian coordinates.

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

Plugging in the ${\hat{\varphi }}^{\prime }$ into our integral

${𝐄}_{\varphi }=\frac{\sigma }{4\pi {ϵ}_0}\underset{0}{\overset{2\pi }{\int }}\underset{0}{\overset{R}{\int }}-\frac{s{\text{'}}^2(-\sin {\varphi }^{\prime }\hat{x}+\cos {\varphi }^{\prime }\hat{y})d{s}^{\prime }d{\varphi }^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}$

$𝐱$ component:

To make our job easier, let us first integrate $d{\varphi }^{\prime }$

${𝐄}_{\varphi }^x=\frac{\sigma \hat{x}}{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^{2}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}\underset{0}{\overset{2\pi }{\int }}\sin {\varphi }^{\prime }d{\varphi }^{\prime }$

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

${𝐄}_{\varphi }^x=\frac{\sigma \hat{x}}{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^{2}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}{|}_0^{2\pi }-\cos {\varphi }^{\prime }$

Evaluating the limits, gives us the result we expected.

${𝐄}_{\varphi }^x=\frac{\sigma }{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^2\hat{x}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}(-1-(-1))=0$

$𝐲$ component:

${𝐄}_{\varphi }^y=\frac{\sigma \hat{y}}{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^{2}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}\underset{0}{\overset{2\pi }{\int }}-\cos {\varphi }^{\prime }d{\varphi }^{\prime }$

integrating

${𝐄}_{\varphi }^y=\frac{\sigma \hat{y}}{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^{2}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}{|}_0^{2\pi }-\sin {\varphi }^{\prime }$

which once again yeilds a zero.

${𝐄}_{\varphi }^y=\frac{\sigma \hat{y}}{4\pi {ϵ}_0}\underset{0}{\overset{R}{\int }}\frac{s{\text{'}}^{2}d{s}^{\prime }}{{(s{\text{'}}^2+z^2)}^{3/2}}(0-0)=0$

Since the x and y components are zero

${𝐄}_{\varphi }=0$

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

$𝐄=\frac{\sigma z\hat{z}}{2{ϵ}_0}(\frac{1}{z}-\frac{1}{\sqrt{R^2+z^2}})$

and rearranging

$𝐄=\frac{\sigma }{2{ϵ}_0}(1-\frac{z}{\sqrt{R^2+z^2}})\hat{z}$

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.