# The Electric Field from an Infinite Flat Plate

When you are much closer to touching a plate than you are to its edges, it is approximately an infinite plate.  Suppose the plate has a charge density $\sigma$, what is the electric field that you experience?

For a plate that is truly infinite, with no granularities, every point looks the same.  Since the plate goes on forever in two directions, there’s no means of measuring your distance to the plate.  It would make sense if the electric field was constant throughout the space that we are considering.

I will evaluate this using both the integral approach and through Gauss’s law.  First the integral approach.

Suppose we want to know the electric field a distance $d$ away from a plate of charge density $\sigma$.  We could try an integral over all of the plate, but we could be even smarter about it.

Let’s consider the contributions  to the electric field from each ring of a radius $r$, measured from the point directly beneath the point where we want to find the electric field.  For a ring of width $dr$ and radius $r$ away from the projected point, the electric field will come in at an angle $\theta$ such that $\tan{\theta} = \frac{r}{d}$.  So, the perpendicular contribution $\vec{E}_{\perp}$ will be

$E \left( \cos{\arctan{\frac{r}{d}}} \right) \hat{z} = E\frac{d}{\sqrt{d^2 + r^2}}\hat{z}$

What will the E-field from an infinitesimal disc be then?  The area of the disc $dA = 2\pi r dr$, so $dQ = \sigma dA = 2 \pi r \sigma dr$.  The distance from our point to the ring is $\sqrt{r^2 + d^2}$, so the complete expression for the E-field contribution from an infinitesimal disc is

$\vec{dE} = \frac{1}{4 \pi \epsilon_0} \frac{ 2 \pi r \sigma dr}{\left( r^2 + d^2 \right)^{\frac{3}{2}}}$

If we integrate from $r = 0$ to $r = \infty$ we will cover the entire plane of charge.

\begin{aligned} \vec{E}_{\perp} & = \int_0^{\infty} \frac{ 2 \pi r \sigma dr}{\left( r^2 + d^2 \right)^{\frac{3}{2}}} \\ & = \frac{ 2 \pi \sigma}{4 \pi \epsilon_0} \int_0^{\infty} \left( \frac{\partial}{\partial_r} \frac{ - d}{\sqrt{r^2 + d^2}} \right) \\ & = \frac{\sigma}{2 \epsilon_0} \left[ \frac{-d}{\sqrt{\infty^2 + d^2}} - \frac{-d}{\sqrt{0^2 + d^2}} \right] \\ & = \frac{\sigma}{2 \epsilon_0} \left( 0 + \frac{d}{d} \right) \\ \vec{E}_{\perp} & = \frac{\sigma}{2 \epsilon_0} \end{aligned}