Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question

2 Answers 2

up vote 1 down vote accepted

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 $z>R$.

share|improve this answer
    
"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

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.