The Unapologetic Mathematician

Mathematics for the interested outsider

Charged Rings and Planes

Let’s work out a couple more examples that may come in handy in the future, if only to get the practice. We’ll start with a “charged ring” which is a charge distribution on a circle. Specifically, we may as well consider the circle of radius R in the x-y plane: c(t)=(R\cos(t),R\sin(t),0). If the charge density is \lambda, then the total charge is 2\pi R\lambda.

First of all, symmetry tells us that we may as well just consider the points of the form (x,0,z) for x\geq0, and we will first specialize to the points (0,0,z). Along this line, the field generated by a little piece of charge on one side of the circle points across to the other side of the circle and out further along the z axis; the piece on the other side cancels the horizontal contribution, but adds to the outward push. This outward direction along the z axis is all that we need to calculate in this case.

The Coulomb law tells us that the differential element of charge at c(t) is q(t)\lvert c'(t)\rvert dt=\lambda Rdt. The displacement vector is (-R\cos(t),-R\sin(t),z), and its length is \sqrt{R^2+z^2}. And so the integral is

\displaystyle\begin{aligned}E_z(0,0,z)&=\int\limits_0^{2\pi}\frac{1}{4\pi\epsilon_0}\frac{\lambda Rz}{\left(R^2+z^2\right)^\frac{3}{2}}dt\\&=\frac{1}{4\pi\epsilon_0}\frac{(2\pi R\lambda)z}{\left(R^2+z^2\right)^\frac{3}{2}}\end{aligned}

So there’s some extra push near the origin, but when z gets much bigger than R, this is effectively the Coulomb law again for a point source with charge $latex 2\pi R\lambda — the total charge on the ring.

Pushing off of the center line is a bit rougher. The differential element of charge is the same, of course, but now the displacement vector is (x-R\cos(t),-R\sin(t),z). Its magnitude is \sqrt{x^2+z^2+R^2-2xR\cos(t)}, leading to the integral

\displaystyle E(x,0,z)=\int\limits_0^{2\pi}\frac{1}{4\pi\epsilon_0}\frac{\lambda R}{\left(x^2+z^2+R^2-2xR\cos(t)\right)^\frac{3}{2}}(x-R\cos(t),-R\sin(t),z)dt

This is, not to put too fine a point on it, really ugly, and it would take us way too far afield to go into it.

However, we can use what we know to determine the electric field generated by a charged plane with a charge density of \sigma, measured in charge per unit area. At any point (x,y,z) we can drop a perpendicular to the plane. We cut the plane into charged rings of radius R and width dR, which gives us a (linear) charge density of \lambda=\sigma dR. The result above says that the ring of radius R has only a z component, which is

\displaystyle E_z(x,y,z)=\frac{1}{4\pi\epsilon_0}\frac{(2\pi R\sigma dR)z}{\left(R^2+z^2\right)^\frac{3}{2}}=\frac{z\sigma}{2\epsilon_0}\frac{RdR}{\left(R^2+z^2\right)^\frac{3}{2}}

So we integrate this out over all radii:

\displaystyle\begin{aligned}E_z(x,y,z)&=\frac{z\sigma}{2\epsilon_0}\int\limits_0^\infty\frac{R\,dR}{\left(R^2+z^2\right)^\frac{3}{2}}\\&=\frac{z\sigma}{2\epsilon_0}\lim\limits_{a\to\infty}\int\limits_0^a\frac{R\,dR}{\left(R^2+z^2\right)^\frac{3}{2}}\\&=\frac{z\sigma}{2\epsilon_0}\lim\limits_{a\to\infty}\left[-\frac{1}{\sqrt{R^2+z^2}}\right]_0^a\\&=\frac{z\sigma}{2\epsilon_0}\lim\limits_{a\to\infty}\left(\frac{1}{\sqrt{z^2}}-\frac{1}{\sqrt{a^2+z^2}}\right)\\&=\frac{z\sigma}{2\epsilon_0}\frac{1}{\lvert z\rvert}\\&=\frac{1}{2\epsilon_0}\sigma\frac{z}{\lvert z\rvert}\end{aligned}

So the field points out from the plane, and it does not fall off with distance at all!

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January 10, 2012 - Posted by | Electromagnetism, Mathematical Physics

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