# Electric field of a Hemisphere

1. Sep 29, 2004

### photon_mass

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.

2. Sep 29, 2004

### vsage

Consider the hemisphere is centered at the origin but lies only in the positive y axis. By symmetry the x and z components will cancel out of the electric field so you're only left with the y force. Gauss' law never helped me out much because a lot of what it shows just becomes obvious over time in doing these problems. Anyway yeah now you have to integrate rings of charge stacked on top of each other in the positive y direction (the bottom ring being the largest). Just use the formula in your book for the integral of that constant charge ring: I doubt the teacher wants to get you bogged down in the math as opposed to the physics.

3. Sep 29, 2004

### photon_mass

no integral formula is given. we need to learn ow to get the formula ourselves. this is a problem assignment, not an excersise. excersises are plug and play. problems = formula derivation.
and my trouble is the setting up of the integral.
for the electric field due to a ring of uniform charge i have
E =Kq(z*r /(z^2 + r^2)^3/2), r being radius of the disk. z being the distance from, the disk. call this eqs 1.
i know i need to sum over z from C to zero
i know that r increases as z decreases, and that rmax = C at z=0.
r= C-z eqs 2
substituting 2 into 1 gives:
E = Kq(z(C-z)/((z^2 + (C-z)^2)^3/2) eqs 3
so for the integral i get
dE = [K*dQ*(z(C-z)/((z^2 + (C-z)^2)^3/2)]dz eqs 4
I know i need to integrate from z = C to z = 0.
I don't know what to do with the dQ.
I know that:
dQ = q = charge density*2(Pi)r eqs 5
let s be charge density. substituting 2 into 5, and then 5 into 4 gives:
dE = 2Ks(pi) [z(C-z)^2 /((z^2 + (C-z)^2)^3/2)]dz eqs 6
Is this right?
How am i supposed to integrate this to know if it will work?
this is driving me bonkers.

Last edited: Sep 29, 2004