# 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

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

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

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

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 ${R^{2}}$:

 $\displaystyle E$ $\displaystyle \rightarrow$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma\frac{1}{2z^{2}}R^{2}\ \ \ \ \ (4)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{\pi\sigma R^{2}}{z^{2}} \ \ \ \ \ (5)$

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

 $\displaystyle E$ $\displaystyle \rightarrow$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}2\pi\sigma\ \ \ \ \ (6)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{\sigma}{2\epsilon_{0}} \ \ \ \ \ (7)$

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:

$\displaystyle |\mathbf{r}-\mathbf{r}'|=\sqrt{z^{2}+R^{2}-2zR\cos\theta} \ \ \ \ \ (8)$

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

 $\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{z-R\cos\theta}{|\mathbf{r}-\mathbf{r}'|}\ \ \ \ \ (9)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{z-R\cos\theta}{\sqrt{z^{2}+R^{2}-2zR\cos\theta}} \ \ \ \ \ (10)$

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.

$\displaystyle E=\frac{\sigma}{4\pi\epsilon_{0}}\int_{0}^{\pi}\frac{(z-R\cos\theta)(2\pi R\sin\theta)(Rd\theta)}{\left(z^{2}+R^{2}-2zR\cos\theta\right)^{3/2}} \ \ \ \ \ (11)$

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

$\displaystyle E=\frac{1}{4\pi\epsilon_{0}}\frac{4\pi R^{2}\sigma}{z^{2}} \ \ \ \ \ (12)$

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 so we are inside the sphere, we get

$\displaystyle E=0 \ \ \ \ \ (13)$

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

$\displaystyle E=\frac{1}{4\pi\epsilon_{0}}\frac{4\pi R^{3}\rho}{3z^{2}} \ \ \ \ \ (14)$

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

 $\displaystyle E$ $\displaystyle =$ $\displaystyle \frac{1}{4\pi\epsilon_{0}}\frac{4\pi z^{3}\rho}{3z^{2}}\ \ \ \ \ (15)$ $\displaystyle$ $\displaystyle =$ $\displaystyle \frac{z\rho}{3\epsilon_{0}} \ \ \ \ \ (16)$

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