Electric field of a symetrically charged ball surface I've been trying to solve this for some time, to no avail I must say.

*

*I am to calculate function of intensity of electric field on $z$ axis.

*The problem is:

*

*We have a charged ball surface with radius R.

*The charge density is a function $k\cos(\theta)$, where k is a constant
and theta is the angle of deviation from $z$ axis, so the charge
density is $k$ at the closest point of the ball and $-k$ at the
furthest point.



There's a hint that this can be calculated by Dirac's
delta function, but I think using it isn't necessary. Thanks in
advance to anyone who tries to tackle this problem.
 A: I would start from the basic relation between charge and electric field:
$$E = \iint_{\Omega} \frac{dq}{r^2}$$
where $\Omega$ is the solid angle subtended by a point on the $z$ axis, $dq$ is a point charge on the spherical surface, and $r$ is the distance between the point charge and the point on the $z$ axis.  NB: I am considering only $z>R$ here.
Note that $dq = \sigma(\Omega) R^2 d\Omega$, where $\sigma$ is the local charge density and $d\Omega$ is an element of solid angle.  We have $\sigma(\Omega) = k \cos{\theta}$.  Further, by considering the geometry,
$$r^2=R^2+z^2-2 R z \cos{\theta}$$
We may then write the electric field as
$$E(z) = 2 \pi k R^2 \int_0^{\pi} d\theta \frac{\sin{\theta} \cos{\theta}}{R^2+z^2-2 R z \cos{\theta}}$$
You should be able to evaluate this integral using the substitution $y=\cos{\theta}$:
$$E(z) = \frac{\pi k R}{z} \int_{-1}^1 dy \frac{2 R z y}{R^2+z^2-2 R z y}$$
I will leave further details for the reader.  The result I get is
$$E(z) = \frac{\pi k R}{z} \left [1-\frac{R^2+z^2}{R z} \log{\left ( \frac{z+R}{z-R}\right)} \right ]$$
Note this is valid only for $z>R$. 
A: You may want to use Gauss theorem
