Integrating a Difficult Integral: Electric Field of a Spherical Surface

[ELB27]

Homework Statement

While trying to find the electric field at a distance z from the center of a spherical surface that carries a uniform surface charge $\sigma$ I got stuck with the following integral (which I'm quite sure is correct):
\int_0^\pi \frac{(z-Rcos\theta)sin\theta d\theta}{(R^{2}+z^{2}-2Rzcos\theta)^{3/2}}

Homework Equations

The Attempt at a Solution

The only idea I had is to let $u=cos\theta ; du=-sin\theta d\theta$,then the integral becomes:
\int_0^\pi \frac{(Ru-z)du}{(R^{2}+z^{2}-2Rzu)^{3/2}}
and now I'm stuck. This looks very similar to a partial fractions problem except there is a square root in the denominator and if I understand correctly I'm not allowed to use partial fractions in this case.

Any help will be appreciated!

 

[Mogarrr]

You want to integrate with respect to which variable?

 

[ELB27]

preferably $\theta$ but if my substitution of $u$ is of any good than $u$ will do also.

 

[Simon Bridge]

That looks sort-of familiar ... if you're sure, I'll start from there.

Remember that z does not depend on u in your last integral, so it has form $$\int \frac{ax-b}{(c-dx)^{3/2}}\;\text dx$$
... your main problem is to get rid of the square-root in the denominator.
How would you normally do that?

 

[ELB27]

substitute x=\frac{c}{d}sinu
EDIT: no, that's not correct, no 2nd power... perhaps multiply and divide by the square root. Am I allowed to use partial fractions with a square root in the numerator?

 

[Simon Bridge]

You do want to get rid of the root before attempting partial fractions.
How about $u^2=c-dx$ or $dx=c\sin^2\phi$

Aside: I probably should not have used "d" as a constant :(

 

[ELB27]

Alright, substituting $v^2=c-gx$ (replacing $d$ with $g$) I am finally able to integrate and simplify to get the desired result:
\int\frac{\frac{ac-av^{2}}{g}-b}{v^{3}}<br /> \frac{-2v}{g}<br /> dv = \int\frac{2a}{g^{2}}dv<br /> - \int\frac{2ac}{g^{2}v^{2}}dv<br /> + \int\frac{2b}{gv^{2}}dv<br /> =\frac{2av}{g^{2}}<br /> + \frac{2ac}{g^{2}v}<br /> - \frac{2b}{gv}<br /> = \frac{1}{z^{2}}\left(\frac{R+z}{|R+z|}-\frac{R-z}{|R-z|}\right)

Thank you very much Simon!