# Electric field of a Hemisphere [closed]

Consider a nonconducting hemisphere of inner radius R, that has a uniform charge distribution of magnitude Q on its interior surface. Find the magnitude of the electric field at C (the centre of curvature of the hemisphere).

we haven't learned GauB's law yet. That is the next chapter.

What i have so far:

Centre of curvature: isn't a circle a special case of ellipse where the distance between the two foci is zero? so C = R?

as far as what to do, i have nothing. first i was going to consider a spherical shell distribution then devide by 2, but E in the centre of a sphere is 0. because this is a PROBLEM and not an EXCERCISE, i'm guessing it should be alot harder than this. the best i could come up with was to consider a bunch of stacked rings a distance d from C, where r increases as d -> 0, and then sum the field strengths at d. now i'm a victim of mathematical ineptitude. i know that this is done as an integral, but i don't know how. and if the integral turns out to be zero, i'm going to be choked.

-
Notice that at the center of the would be sphere, the electric field from around the hemisphere cancels except for in one direction, since there is rotational symmetry about one of the axis. – kηives Jun 26 '12 at 18:50
"Consider a nonconducting hemisphere of inner radius R" Err...is that supposed to be a hemispherical shell? Otherwise where does a inner radius come into the problem? That said, we don't answer basic exercises on Physics.SE. You are allowed to ask about basic concepts, but perhaps you would be well advised to read about Gauss' law first... – dmckee Jun 26 '12 at 18:51

## closed as too localized by dmckee♦Jun 26 '12 at 18:51

This question is unlikely to help any future visitors; it is only relevant to a small geographic area, a specific moment in time, or an extraordinarily narrow situation that is not generally applicable to the worldwide audience of the internet. For help making this question more broadly applicable, see the FAQ.