1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ Shouldn't the integral be from $\frac{-a}{2}$ to $\frac{a}{2}$? $\endgroup$ – model_checker Sep 13 '16 at 4:12
  • $\begingroup$ 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. $\endgroup$ – Bobbie D Sep 13 '16 at 4:14
  • $\begingroup$ 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. $\endgroup$ – model_checker Sep 13 '16 at 4:18
  • $\begingroup$ 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. $\endgroup$ – Bobbie D Sep 13 '16 at 4:20
  • $\begingroup$ According to my notes: the Electric Field strength for a rod of length $L$ should be $E=\frac{2K\lambda L}{h(4h^2+L^2)^{1/2}}$. $\endgroup$ – model_checker Sep 13 '16 at 4:24
2
$\begingroup$

Your answer is flawless: there is nothing the matter with it. Both formulae are equivalent, and can be shown to be so with this identity:

$tan^{-1}(\frac{2u}{u^2-1})=2tan^{-1}(\frac{1}{u}+/-n\pi)$

Substitute $u^2=1+\frac{2a^2}{z^2}$ and you'll find they are equal.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.