WONDERFUL! Thank you so much for the help!
..and thanks to all who've assisted me with this irritating problem.
I've now concluded that E = \frac{k(r-a)}{\epsilon_0 r^{2}} for the region enclosing a < r < b.
wait wait wait wait wait! I think I have it!
Was flipping through Griffiths and spotted a nice little trick for symmetry where E points radially outwards with the same magnitude at all points with the same radius:
\int_S E.da = \int_S |E|da = E4 \pi r^{2} = \frac{Q}{\epsilon_0}
Sub my Q into...
argh!~ This is annoying the heck out of me!
Right, so re-evaluated:
\frac{Q}{\epsilon_0} = \int_V \frac{k}{\epsilon_0} sin(\theta) dr d\theta d\phi
This is calculated down to a nice easy equation Q = 4k \pi (r-a) giving charge enclosed between the shells radius r and a.. correct? This...
Ok, not sure if I'm doing it correctly here, but here's what I've done:
using Gauss' law:
\int_V\frac{\rho}{\epsilon_0} d\tau = \frac{Q}{\epsilon_0},
so i then do:
\frac{Q}{\epsilon_0} = \int^{\pi}_{0}\int^{2\pi}_{0}\int^{r}_{a} \frac{k}{\epsilon_0 r^{2}} dr d\phi d\theta
as it's...
You're absolutely right! I'm getting myself confused here.. Instead of integrating the for the volume, could i perhaps use the shortcut method of V(sphere) = 4/3.pi.r^3 and just deduct the volume of the inner sphere from the larger sphere? would certainly simplify my calculations.
Actually.. I think i have it now:
Using the Flux equations: \Phi = E.A = E.4.pi.r^2 = Q/eps0
and then using Q = rho x Area of sphere, sub into flux and solve for E.
correct? incorrect?
EDIT: Actually.. this is how i worked out the Electric field for r > b.. damn.. back where I started
Cheers for that kuruman, I thought my assumption might have been wrong, but I had a few different conflicting ideas on this.
So for example, if i was to use the differential form of Gauss' law: \nabla.E = \frac{\rho}{\epsilon} , how might I infer the Electric Field itself from this...
Homework Statement
A hollow spherical shell carries charge density \rho = \frac{k}{r^{2}}
in the region a \leq r \leq b. Find the electric field in the three regions:
i. r < a
ii. a < r < b
iii. r > b
Homework Equations
Surface Area of a sphere
Coulombs Law
Gauss's Law
The...