Thread: Surface charge

Find the electric field a distance $\displaystyle z $ from the center of a spherical surface, which carries a uniform charge density $\displaystyle \sigma $. Treat the case $\displaystyle z < R $ and $\displaystyle z > R $.

Ok, so I have a feeling that the electric field will only be in the $\displaystyle z $ direction. But I am not quite sure WHY this should be true. What does an electric field physically represent (is it like stalks of wheat being blown by wind?).

From here out I know how to do it: $\displaystyle E_{z} = \frac{1}{4 \pi \epsilon_{0}} \int \frac{\sigma R^{2} \sin \theta \ d \theta \ d \phi (z-R \cos \theta)}{(R^2 + z^2 - 2Rz \cos \theta)^{3/2}} $.

But the question is, how do you know when an electric field has more than one component? And what exactly is it (Griffiths does not really define it, and he says that relativity forces us to abandon the notion that it is an "ether"). Like I know inside the sphere, the electric field is $\displaystyle 0 $ (e.g. $\displaystyle z <R $). But what does this physically mean?

Argue by symetry: since system is symetric about z-axis, electric field must go along z-axis.

I think the field is proportional to the force exerted on a charge. (Exactly how proportional depends on units obv.)

Are you sure you can do that integral? Looks like hell to me.

Well $\displaystyle \int d \phi = 2 \pi $. Then let $\displaystyle u = \cos \theta, \ \ du = -\sin \theta $. Then integrate by partial fractions.

Okay, if it turns out you can't, try question in cylindrical polar co-ordinates. Then go here http://www.mathhelpforum.com/math-he...ver-1-r-r.html and tell her how to do the question.

I have heard she is hot.