Electrostatics – surface charges

Required math: vectors, calculus

Required physics: basics

Reference: Griffiths, David J. (2007) Introduction to Electrodynamics, 3rd Edition; Prentice Hall – Problems 2.6 – 2.8.

Now a few examples of surface charge.

Example 1

First, we consider a circular disk of radius ${R}$ with surface charge density ${\sigma}$ lying in the ${xy}$ plane and centred at the origin. Find the electric field at a point on the ${z}$ axis.

To solve this we can make use of the solution to the circular loop. In this case we’re considering a circular ring of circumference ${2\pi r}$ and thickness ${dr}$ , so the amount of charge in the ring is ${2\pi r\sigma dr}$ and from the earlier solution, the field due to this ring is

$\displaystyle E_{z}=\frac{1}{4\pi\epsilon_{0}}\frac{2\pi r\sigma z}{(z^{2}+r^{2})^{3/2}}dr$

To get the total field from the disk, we integrate over ${r}$ :

$\displaystyle E$	$\displaystyle =$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma z\int_{0}^{R}\frac{r}{(z^{2}+r^{2})^{3/2}}dr$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma\frac{\left(\sqrt{z^{2}+R^{2}}-z\right)}{\sqrt{z^{2}+R^{2}}}$

To get the limiting behaviours we can Taylor-expand the result. For ${z\gg R}$ , we expand about ${R=0}$ and find the leading non-zero term is in ${R^{2}}$ :

$\displaystyle E$	$\displaystyle \rightarrow$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma\frac{1}{2z^{2}}R^{2}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{\pi\sigma R^{2}}{z^{2}}$

This is correct since the total charge on the disk is ${\sigma\pi R^{2}}$ so the field is that due to a point charge of that amount.

If we let ${R\rightarrow\infty}$ we get

$\displaystyle E$	$\displaystyle \rightarrow$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{\sigma}{2\epsilon_{0}}$

This is the field due to an infinite plane of charge. Note that the field is independent of ${z}$ so is the same no matter how far away from the plane we are.

Example 2

We have a spherical shell of charge with radius ${R}$ and surface density ${\sigma}$ , centred at the origin. Again, we seek the field at a point on the ${z}$ axis.

Using spherical coordinates, a point on the sphere has coordinates ${(R,\theta,\phi)}$ where ${\theta}$ is the angle from the positive ${z}$ axis, and ${\phi}$ is the azimuthal angle. We can use the cosine law to write the distance between a point on the sphere and the field point:

$\displaystyle |\mathbf{r}-\mathbf{r}'|=\sqrt{z^{2}+R^{2}-2zR\cos\theta}$

By symmetry, the field will again be in the ${z}$ direction, so we need the ${z}$ component of ${\mathbf{r}-\mathbf{r}'}$ . To get this, we need the angle ${\alpha}$ between ${\mathbf{r}-\mathbf{r}'}$ and the ${z}$ axis. To get this, project the point on the sphere onto the ${z}$ axis; this gives a point with ${z}$ coordinate ${R\cos\theta}$ . The remaining distance along the ${z}$ axis to the field point is therefore ${z-R\cos\theta}$ , but this distance is the projection of ${\mathbf{r}-\mathbf{r}'}$ onto the ${z}$ axis. The cosine of the angle between ${\mathbf{r}-\mathbf{r}'}$ and the ${z}$ axis is this projection divided by ${|\mathbf{r}-\mathbf{r}'|}$ , so we get

$\displaystyle \cos\alpha$	$\displaystyle =$	$\displaystyle \frac{z-R\cos\theta}{\|\mathbf{r}-\mathbf{r}'\|}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{z-R\cos\theta}{\sqrt{z^{2}+R^{2}-2zR\cos\theta}}$

Now for a given value of ${\theta}$ , we have a ring of charge with radius ${R\sin\theta}$ and thickness ${Rd\theta}$ at ${z}$ distance ${|\mathbf{r}-\mathbf{r}'|}$ from the field point, so we can integrate over ${\theta}$ to get the total field.

$\displaystyle E=\frac{\sigma}{4\pi\epsilon_{0}}\int_{0}^{\pi}\frac{(z-R\cos\theta)(2\pi R\sin\theta)(Rd\theta)}{\left(z^{2}+R^{2}-2zR\cos\theta\right)^{3/2}}$

This integral can be done using Maple, but there are two possibilities. First, if ${z>R}$ so the field point is outside the sphere, we get

$\displaystyle E=\frac{1}{4\pi\epsilon_{0}}\frac{4\pi R^{2}\sigma}{z^{2}}$

Since ${4\pi R^{2}\sigma}$ is the total charge on the sphere, we see that the sphere behaves like a point charge for all field points outside it.

Second, if ${z < R}$ so we are inside the sphere, we get

$\displaystyle E=0$

So anywhere inside a spherical shell with a uniform charge distribution, we feel no field at all.

Example 3

The result of the last example can be used to find the field due to a sphere that contains a uniform volume charge density ${\rho}$ . Since each spherical shell within the sphere behaves as a point charge to all points outside the shell, the field at a point outside the sphere ( ${z>R}$ ) is just

$\displaystyle E=\frac{1}{4\pi\epsilon_{0}}\frac{4\pi R^{3}\rho}{3z^{2}}$ r

At a point inside the sphere, all shells outside the field point contribute nothing, so we get, for ${z < R}$ :

$\displaystyle E$	$\displaystyle =$	$\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{4\pi z^{3}\rho}{3z^{2}}$
$\displaystyle$	$\displaystyle =$	$\displaystyle \frac{z\rho}{3\epsilon_{0}}$

The field thus increases linearly within the sphere and then falls off as an inverse square outside.

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