# Math Help - EM field integral

1. ## EM field integral

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!

2. ## Re: EM field integral

Originally Posted by mathlover10
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!
I assume your integral is (as it's a bit hard to read) \displaystyle \begin{align*} \frac{4\lambda \, z}{4\pi \, \varepsilon \, \omega } \int_{-\frac{a}{2}}^{\frac{a}{2}}{ \frac{dx}{\left( x^2 + z^2 + \frac{a^2}{4} \right) ^{\frac{3}{2}} } } \end{align*}

Because it has \displaystyle \begin{align*} x^2 + \textrm{something}^2 \end{align*} in the denominator, that means a substitution of the form \displaystyle \begin{align*} x = \textrm{something}\cdot \tan{(\theta)} \implies dx = \textrm{something}\cdot \sec^2{(\theta)}\,d\theta \end{align*} is appropriate, as the function will simplify with the identity \displaystyle \begin{align*} \tan^2{(\theta)} + 1 \equiv \sec^2{(\theta)} \end{align*}. Make the substitution and see what you can do...

3. ## Re: EM field integral

it only takes a few seconds...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,
$sin(arctan(\frac{x}{(a^2/4+z^2)^{1/2}}))=\frac{x}{(a^2/4+x^2+z^2)^{1/2}}$[/QUOTE]