I have spent the past two and a half hours attempting to understand why the electric field on either side of an infinitely sized charged plate is uniform. I get it conceptually, in that as a point moves farther away from the plate, it is able to see a greater amount of charge in a more focused field of view, canceling out the reduction in strength caused by increasing the distance between the point and the plate. My issue is that I want to be able to derive this relationship for myself using calculus and I when doing it by hand, I always arrive at a relationship that reduces the strength of the electric field as the distance increases.

I eventually found a paper online that derives the property of uniformity, but it has a step when evaluating an improper integral that makes no sense to me:

$$ \begin{align} E_P &= \frac{\sigma r}{4\pi\epsilon} \int_{-\infty}^\infty \int_{-\infty}^\infty \frac{1}{(x^2 + y^2 + r^2)^{3/2}} \; dx \; dy \\ &= \frac{\sigma r}{4\pi\epsilon} \int_{-\infty}^\infty \frac{2}{(y^2 + r^2)^{3/2}} \; dy \tag{1} \\ &= \frac{\sigma r}{4\pi\epsilon} \frac{2\pi}{r} \tag{2} \\ &= \frac{\sigma}{2\epsilon} \end{align} $$

What happened between lines (1) and (2)? When I did it by hand, I ended up with, $$ \frac{\sigma r}{4\pi\epsilon} \int_{-\infty}^\infty \frac{2}{(y^2 + r^2)^{3/2}} \; dy = \frac{\sigma r}{4\pi\epsilon} \frac{4}{r^2} = \frac{4\sigma}{r\pi\epsilon} $$

I checked this result with Maxima and Wolfram, and they both confirm my answer. What am I not seeing?


Consider the transition from the line above (1) to line (1). Based on my calculation, line (1) should be $$E_P = \frac{\sigma r}{4 \pi \epsilon} \int_{-\infty}^{\infty} \frac {2}{(r^2 + y^2)} dy $$

This gives the appropriate result in line (2) and the final answer.

  • $\begingroup$ Thank you, I'll see if I can notify the maintainer of the document to fix the typo. $\endgroup$ – Tankobot May 22 '17 at 8:04

You're right, something seems to be mixed up. I'd always integrate this in a different way: $$\begin{align} E_P&=&\frac{r \sigma}{4 \pi \epsilon_0}\int_{-\infty}^\infty\mathrm{d}x\int_{-\infty}^\infty\mathrm{d}y \frac{1}{\sqrt{x^2+y^2+r^2}^3}\\ &=&\frac{r \sigma}{4 \pi \epsilon_0}\int_{0}^\infty2\pi \rho\mathrm{d}\rho \frac{1}{\sqrt{\rho^2+r^2}^3}\\ &=&\frac{r \sigma}{2 \epsilon_0}\int_{0}^\infty \frac{\rho\mathrm{d}\rho}{\sqrt{\rho^2+r^2}^3}\\ &=&\frac{r \sigma}{2 \epsilon_0} \frac{1}{r}\\ &=&\frac{\sigma}{2 \epsilon_0} \end{align} $$ i.e. using symmetry again and changing to cylindrical coordinates.


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