Answer:
[tex]E=\frac{Q+2A\pi(r^2-a^2)}{4\pi r^2\epsilon_0}=k, A=\frac{(k4\pir^2\epsilon _0-Q)}{2\pi(r^2-a^2)}[/tex]
Explanation:
I will use the Gauss's Law to find the field:
[tex]\int\vec{E}.d\vec{S}=\frac{Q_{in}}{\epsilon_0}[/tex]}
the surface S is a sphere of radio r, the normal vector only has radial coordinates.
E(r)=E(r)r /*The field, based on spherical symmetry, only depends of the radius, and only has radial coordinate*/
if r<a
[tex]\int\vec{E}.d\vec{S}=\int EdS=E\int dS=4\pi r^2E=\frac{Q_{in}}{\epsilon_0}[/tex]
[tex]\vec{E}=\frac{Q}{4\pi r^2\epsilon_0}. \^r[/tex]
if a=<r<b
[tex]\int\vec{E}.d\vec{S}=\int EdS=E\int dS=4\pi r^2E=\frac{Q_{in}}{\epsilon_0}[/tex]
[tex]Q_{in}=Q+\int dq', \rho=\frac{dq'}{dVol}, Q_{in}=Q+\int\rho dVol=Q+\int\limits^{2\pi}_0\int\limits^{\pi}_0\int\limits^r_a {\frac{A}{r}} \, r^2sin(\phi)dr d \phi d \theta[/tex]
[tex]E=\frac{Q+2A\pi(r^2-a^2)}{4\pi r^2\epsilon_0}=k, A=\frac{(k4\pir^2\epsilon _0-Q)}{2\pi(r^2-a^2)}[/tex]
with K=constant
