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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