# Can't seem to derive the formula for the electric field over a square sheet

I'm trying to derive the formula for the electric field a height $h$ above the center of a uniformly charged square sheet with sides $2a$.

To do so I'm using the formula for the electric field above the center of a line segment with length $2L$ and uniform charge density:

$$E = \frac{2k_e\lambda L}{h\sqrt{h^2+L^2}}$$

Then using the fact that $\lambda(2L) = Q$, I get

$$E = \frac{k_eQ}{h\sqrt{h^2+L^2}}$$

Then I can cut the square sheet into line segments with differential charge $dq = \sigma dA$ and differential area $dA = 2adx$. I just have to take into account that I only need the vertical component of the electric fields and integrate:

\begin{align}E &= k_e(\sigma 2a)\int_{-a}^a \frac{dx}{\sqrt{h^2+x^2}\sqrt{h^2+x^2+a^2}}\cos(\theta) \\ &= 4a\sigma k_e\int_0^a \frac{dx}{\sqrt{h^2+x^2}\sqrt{h^2+x^2+a^2}}\frac{h}{\sqrt{h^2+x^2}} \\ &= 4a\sigma hk_e\int_0^a \frac{dx}{(h^2+x^2)\sqrt{h^2+x^2+a^2}}\end{align}

I can't figure out any way to integrate this, but Mathematica is giving me

$$E = 4\sigma k_e\arctan\left(\frac{a^2}{h\sqrt{h^2+a^2+a^2}}\right)$$

However I know the answer is supposed to be

$$E = 2\sigma k_e\left[4\arctan\left(\sqrt{1+a^2/(2h^2)}\right)-\pi\right]$$

I can see these are not just the same answer in different forms by plugging in 1s for all the constants and evaluating. Where am I going wrong?

• Shouldn't the integral be from $\frac{-a}{2}$ to $\frac{a}{2}$? – model_checker Sep 13 '16 at 4:12
• Whoops. It should be from $-a$ to $a$ (I've changed it now). But that doesn't fix the problem because that just gives an extra factor of $2$ to my answer. – Bobbie D Sep 13 '16 at 4:14
• Are you sure that the formula for electric field above the center of a line segment with length $2L$ is correct? I don't think so. – model_checker Sep 13 '16 at 4:18
• Though I did it a different way than him, my answer to that part agrees with this person's answer so I don't think that part's wrong. – Bobbie D Sep 13 '16 at 4:20
• According to my notes: the Electric Field strength for a rod of length $L$ should be $E=\frac{2K\lambda L}{h(4h^2+L^2)^{1/2}}$. – model_checker Sep 13 '16 at 4:24

$tan^{-1}(\frac{2u}{u^2-1})=2tan^{-1}(\frac{1}{u}+/-n\pi)$
Substitute $u^2=1+\frac{2a^2}{z^2}$ and you'll find they are equal.