Understanding the E Field Above a Square Loop

[mathnerd15]

this is the E field above a square loop with side=a at distance z on the z axis. by symmetry the Ex, Ey field cancel out. it really kind of bothers me that I can't see this substitution, including the sin(tan^-1(u)) portion. I guess it's an easy technique but I wonder how you get the substitution?

$$Ez=\frac{4\lambda z}{4\pi \varepsilon o}\int_{-a/2}^{a/2}\frac{ dx}{(z^2+x^2+a^2/4)^{3/2}}, x=\sqrt{a^2/4+z^2}tanu, dx=\sqrt{a^2/4+z^2}sec^{2}udu, I=\frac{1}{a^2/4+z^2}\int cosudu=\frac{1}{a^2/4+z^2}sinu\therefore Ez=\frac{8\lambda az}{4\pi\epsilon o \sqrt{2a^2+4z^2}z^2+a^2/4}$$thanks!

 

[arildno]

Thumb rule:
"To remove squares within square roots, use a suitable trig substitution. If stuff is more difficult than that, forget about it"

 

[mathnerd15]

I try to do these by hand- I wonder if some people do all of these by hand?

 

[arildno]

I wonder if some people do all of these by hand?

Those who need to learn them, such as students. Those who are professionally competent don't, but at most, picks up a table of integrals. Only those are professionally competent who has learned them.

 

[mathnerd15]

I'm not sure if maybe it's better to do a lot of problems out of mathematical and Schaum books. I had another career before this that was completely different than physics/mathematics
I've been looking at Apostol and Hubbard Calculus (introduces manifolds) and I'm not sure if I should study these since I've already done Stewart (problems are easy I know)

 

[arildno]

I'm not sure what you're getting at.

Ask yourself:
If lots of the basic tools used within research seems utterly magical to you, because you haven't learned the logic behind them, can you ever succeed as a researcher? There's a reason why an education within physics/math is a fast-track over the evolution of the subjects, ordered in a pedagogically optimal way.
Alternatively: Why should anyone want to employ you, unless you have a proven ability to understand and master previous areas of research?

-------------------------------
Thus, the function behind drilling students with such as nasty integrals is at least three-fold:
To hone the mind of the student, and familiarize him with the technical language, along with being a quality control of the student.

 

[mathnerd15]

thanks, it's not a difficult substitution.

so the integral substitution works well with the tan^2+1 term reducing it to cosu and you just substitute back in for u and solve,
$$sin(arctan(\frac{x}{(a^2/4+z^2)^{1/2}}))=\frac{x}{(a^2/4+x^2+z^2)^{1/2}}$$