The Electric Potential of a Rectangular Sheet of Charge - Resolved

In the January 9 post of this blog, I challenged readers to find the electrical potential V(z) that will give you the electric field E(z) of Eq. 6.10 in the 5th edition of Intermediate Physics for Medicine and Biology

In other words, the goal is to find V(z) such that E = − dV/dz produces Eq. 6.10. In the comments at the bottom of the post, a genius named Adam Taylor made a suggestion for V(z) (I love it when people leave comments in this blog). When I tried his expression for the potential, it almost worked, but not quite (of course, there is always a chance I have made a mistake, so check it yourself). But I was able to fix it up with a slight modification. I now present to you, dear reader, the potential:

How do you interpret this ugly beast? The key is the last term, z times the inverse tangent. When you take the z derivative of V(z), you must use the product rule on this term. One derivative in the product rule eliminates the leading z and gives you exactly the inverse tangent you need in the expression for the electric field. The other gives z times a derivative of the inverse tangent, which is complicated. The two terms containing the logarithms are needed to cancel the mess that arises from differentiating tan⁻¹.

I don’t know what there is to gain from having this expression for the potential, but somehow it comforts me to know that if there is an analytic equation for E there is also an analytic equation for V.