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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.

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2 Answers 2

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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 $z>R$.

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"For z<R the field is zero because the charge is on the (outer) surface." No. That only simplification only applies to uniform charge distributions. What you have here is a dipole and it has non-zero field in the interior. –  dmckee Jun 15 '13 at 16:21
@dmckee: i remove the controversial line until I doublecheck. The rest should be right. –  Ron Gordon Jun 15 '13 at 16:46
Ron, What I wrote above is imprecise. Spherically symmetric distributions have net zero field inside. This is basically a consequence of Guass' law plus symmetry. When I wrote the above I was thinking in particular of distributions on a spherical shell which are only spherically symmetric when they are uniform. –  dmckee Jun 15 '13 at 17:12
@dmckee: as I said, we'll get to the bottom of it, but in the meantime I just took the sentence that may not be correct out until I verify myself. –  Ron Gordon Jun 15 '13 at 17:26

You may want to use Gauss theorem

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