# The Unapologetic Mathematician

## Charge Distibutions

The superposition principle for the electric field extends to the realm of continuous distributions, with the sum replaced by an appropriate integral.

For example, let’s say we have a curve $c:[0,1]\to\mathbb{R}^3$, and along this curve we have a charge. It makes sense to measure the charge in units per unit of distance, like coulombs per meter. We can even let it vary from point to point, getting a function $q(t)$ describing the charge per unit length near the point with parameter $t$. To be a little more explicit, if $ds=\lvert c'(t)\rvert dt$ is the “line element” that measures a tiny bit of distance near the point $c(t)$, then $q(t)\lvert c'(t)\rvert dt$ measures a little bit of charge near that point.

We can now use the Coulomb law to see what electric field this tiny bit of charge creates at a point with position vector $p$:

$\displaystyle\frac{1}{4\pi\epsilon_0}\frac{q(t)\lvert c'(t)\rvert dt}{\lvert p-c(t)\rvert^3}\left(p-c(t)\right)$

since the displacement vector from $c(t)$ to $p$ is $p-c(t)$. Now we can take this “differential electric field” and integrate it over the curve, adding up all the tiny contributions to the field at $p$ made by all the tiny bits of charge along the curve.

As an example, let’s consider an infinite line of charge along the $z$ axis with a constant charge density of $\lambda$; a piece of the line of length $l$ will have charge $\lambda l$. Admittedly, this is not a finite-length curve like above, but the same principle applies. We set $c(t)=(0,0,t)$, so $c'(t)=(0,0,1)$ and $\lvert c'(t)\rvert=1$.

Geometric considerations tell us that the electric field generated by the line at a point $p$ is contained in the same plane that contains the line and the point. We can also tell that the field points directly perpendicular to the line; if $p=(x,y,z)$ then the vertical component induced by the chunk at $(0,0,z+d)$ is cancelled out by the component induced by the chunk at $(0,0,z-d)$. Indeed, we can check that the first gives us

$\displaystyle\frac{\lambda}{4\pi\epsilon_0}\frac{(x,y,-d)}{\left(x^2+y^2+d^2\right)^\frac{3}{2}}\,dt$

while the second gives us

$\displaystyle\frac{\lambda}{4\pi\epsilon_0}\frac{(x,y,+d)}{\left(x^2+y^2+d^2\right)^\frac{3}{2}}\,dt$

and the vertical components of these two exactly cancel each other out.

All that remains is to calculate the horizontal component. Without loss of generality we can consider the point $p=(x,0,0)$, and we must calculate the $x$-component of the electric field by taking the integral

$\displaystyle E_x(x,0,0)=\frac{\lambda x}{4\pi\epsilon_0}\int\limits_{-\infty}^\infty\frac{1}{\left(x^2+t^2\right)^\frac{3}{2}}\,dt$

We need an antiderivative of the integrand $\left(x^2+t^2\right)^{-\frac{3}{2}}$, and it turns out that $x^{-2}t\left(x^2+t^2\right)^{-\frac{1}{2}}$ fits the bill. Indeed, we check:

\displaystyle\begin{aligned}\frac{\partial}{\partial t}\frac{t}{x^2\left(x^2+t^2\right)^\frac{1}{2}}&=\frac{1}{x^2}\frac{\partial}{\partial t}\frac{t}{\left(x^2+t^2\right)^\frac{1}{2}}\\&=\frac{1}{x^2}\frac{\left(x^2+t^2\right)^\frac{1}{2}\frac{\partial}{\partial t}t-t\frac{\partial}{\partial t}\left(\left(x^2+t^2\right)^\frac{1}{2}\right)}{\left(x^2+t^2\right)}\\&=\frac{1}{x^2}\frac{\left(x^2+t^2\right)^\frac{1}{2}-\frac{t}{2}\left(x^2+t^2\right)^{-\frac{1}{2}}\frac{\partial}{\partial t}\left(x^2+t^2\right)}{\left(x^2+t^2\right)}\\&=\frac{1}{x^2}\frac{\left(x^2+t^2\right)\left(x^2+t^2\right)^{-\frac{1}{2}}-\frac{t}{2}\left(x^2+t^2\right)^{-\frac{1}{2}}2t}{\left(x^2+t^2\right)}\\&=\frac{1}{x^2}\frac{x^2+t^2-t^2}{\left(x^2+t^2\right)^\frac{3}{2}}\\&=\frac{1}{\left(x^2+t^2\right)^\frac{3}{2}}\end{aligned}

as asserted. Thus we continue the integration:

\displaystyle\begin{aligned}E_x(x,0,0)&=\frac{\lambda x}{4\pi\epsilon_0}\int\limits_{-\infty}^\infty\frac{1}{\left(x^2+t^2\right)^\frac{3}{2}}\,dt\\&=\frac{\lambda x}{4\pi\epsilon_0}\lim\limits_{a,b\to\infty}\left[\frac{t}{x^2\sqrt{x^2+t^2}}\right]_{-a}^b\\&=\frac{\lambda}{4\pi\epsilon_0x}\lim\limits_{a,b\to\infty}\left(\frac{b}{\sqrt{x^2+b^2}}-\frac{-a}{\sqrt{x^2+a^2}}\right)\\&=\frac{\lambda}{4\pi\epsilon_0x}(1+1)\\&=\frac{1}{2\pi\epsilon_0}\frac{\lambda}{x}\end{aligned}

which is a nice, tidy value. More generally, we find

\displaystyle\begin{aligned}E(x,y,z)&=\frac{1}{2\pi\epsilon_0}\frac{\lambda}{\sqrt{x^2+y^2}}\frac{(x,y,0)}{\sqrt{x^2+y^2}}\\&=\frac{1}{2\pi\epsilon_0}\frac{\lambda}{\lvert(x,y,0)\rvert}\frac{(x,y,0)}{\lvert(x,y,0)\rvert}\end{aligned}

pointing directly away from the (positively) charged line, neither up nor down, and falling off in magnitude as the first power of the distance from the line.

January 6, 2012