Question

An insulating hollow sphere has inner radius a and outer radius

Answer 1

The solution can be given using the Gauss' Theorem. First let's make the following assumptions:
*This problem has a perfect spherical symmetry so the field for a charge density of $\rho(r)$ that depends only on the radial component(we are of course working in spherical coordinates) will yield a constant electric field pointing radially outwards,that is, the field will be uniform for a<r<b independently of the angles $\theta$ and $\phi$

*As a result of the uniformity of the electric field, E will be constant and it can be taken out of the integral involving Gauss' law.
a)
Now let's get our hands to it, recall that Gauss' Law states:
$\oint\mathbf{E}.d\mathbf{a}=\frac{Q_{enclosed}}{\epsilon_0} \\ \impliesE\oint da=E.4\pi r^2=\frac{Q_{enclosed}}{\epsilon_0}$
Here we exploited the justified assumption that The field is uniform, because the field is pointing in the outward radial direction the scalar product between the field and surface element will yield $\mathbf{E}.d\mathbf{a}=Edacos(0)=Eda$
Now we need to determine the enclosed charge: $Q_{enclosed}=\int_V\rho(r)dV$
In spherical coordinates we thus have:
$Q_{enclosed}=\frac{1}{\epsilon_0}\int_V\rho(r)dV=\int^b_a\int^{2\pi}_0\int^{\pi}_0\frac{\alpha}{r}r^2sin\theta drd\theta d\phi$phi[/tex] over all of the gaussian surface.
Where the integral:
$\int^{2\pi}_0\int^{\pi}_0sin\theta d\theta d\phi=4\pi$
Returning to our integral we have:

$\frac{1}{\epsilon_0}\int^b_a\int^{2\pi}_0\int^{\pi}_0\frac{\alpha}{r}r^2sin\theta drd\theta d\phi=\frac{1}{\epsilon_0}\int^a_b\frac{\alpha}{r}r^2dr\int^{2\pi}_0\int^{\pi}_0sin\theta d\theta d\phi=\frac{1}{\epsilon_0}4\pi \int^b_a\alpha rdr \\ \\ =\frac{1}{\epsilon_0}4\pi\alpha\left[\frac{1}{2}r^2\right]^b_a=\frac{1}{\epsilon_0}2\pi\alpha(b^2-a^2)$
Now it's just a matter of solving for E:
$E=\frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2}$
b)
If a point charge is placed at the center of our system the the resulting field will be the sum of both fields, this field needs to be constant, let's pick for now that the field is zero in the region a<r<b:
$E_p+E=0$
Where $E_p$ is the field due to a point charge.
Again using Gauss's theorem the field of a point charge 1 is:
$E_p\oint da=E.4\pi r^2=\frac{Q_{enclosed}}{\epsilon_0}=\frac{q}{\epsilon_0} \\ \\ \implies E_p=\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$
We then get the following expression:
$E_p+E=0 \\ \implies E=\frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2} +\frac{1}{4\pi\epsilon_0}\frac{q}{r^2} =0$
Solving for q we get:
$q=-2\pi\alpha(b^2-a^2)$

If instead we want a non zero field $E=c$ then we only need to solve 
$E_p+E=c$ which yields:
$E_p+E=c \\ \implies \frac{\alpha}{2\epsilon_0}\frac{b^2-a^2}{r^2} +\frac{1}{4\pi\epsilon_0}\frac{q}{r^2}=c \\ \\ \implies q=-2\pi[\alpha(b^2-a^2)-c]$