HW2 #9

If you do #9 by building a plate out of thin rods, you should get this:

E_z=4k\sigma \tan^{-1}\left[\frac{a^2}{2z\sqrt{2a^2+4z^2}}\right]

This can be shown to be equivalent to Griffiths' result, along with the scary arctan identity I posted earlier

\tan^{-1}{\left(\frac{2u}{u^2-1}\right)}=2\tan^{-1}{\left(\frac{1}{u}\right)} \pm n\pi

For instance, try

u^2 = 1 + \frac{a^2}{2z^2}

to make the identity more 'obvious'. Keep in mind the freedom to add or subtract pi from arctan, the boundary conditions (e.g., field is zero at r=infinity) will tell you whether to add or subtract or not.

Griffiths result is

E_z = 8k\sigma\left[\tan^{-1}\left(\sqrt{1+\frac{a^2}{2z^2}}\right)-\frac{\pi}{4}\right]