What is the electric field due to a charge inside a hollow sphere? Say, there's a hollow non-conducting sphere with no charge. If a point charge q is placed inside the sphere, how will the electric field behave?
Now inside the sphere I can just apply Gauss's law. But will the electric field exist outside the sphere? If so, will it be disrupted? If the electric field will not be present outside the sphere, why? 
 A: Making some assumptions for the sake of an example. The charge is at the center of the non conducting sphere, the hollow sphere has a shell of some thickness, and the material is a Linear Dielectric. 
I believe you want to use Gauss' Law with the Displacement field since within the shell we have non-conducting material. 
$\vec{D}=\epsilon_0\vec{E}+\vec{P}$
Where $\vec{E}$ is the electric field and  $\vec{P}$ is the Polarization. 
$\nabla \cdot \vec{D}=\rho_f$ where $\rho_f$ is the charge distribution of Free Charge, as opposed to induced Bounded Charge, $\rho_b$ with the total charge being as usual $\nabla \cdot \vec{E}=\rho/\epsilon_0=(\rho_f+\rho_b)/\epsilon_0$. 
In a material, constituent parts re-orient themselves in the presence of the magnetic field effectively changing the Permittivity of free space. 
$\vec{P}$ is frequently proportional to the Electric field in the Medium, it is a Linear Dielectric: $\vec{P}=\epsilon_0\chi_e\vec{E}$. Where $\chi_e$ is the Electric Susceptibility of the Dielectric Material. 
So $\vec{D}=\epsilon_0(1+\chi_e)\vec{E}=\epsilon\vec{E}$. Where $\epsilon$ is the Permittivity of the material. 
Assuming the charge is at the center and applying Gauss' Law:
$$D_r\cdot4\pi r^2=Q$$
So $$\vec{D}=\frac{Q\hat{r}}{4\pi r^2}$$
Note there is no $\epsilon_0$. 
To go from the displacement field to the Electric field, you need to divide $\vec{D}$ by $\epsilon$. In a vacuum $\epsilon=\epsilon_0$.
So in the empty space surrounded by the shell and the empty space outside of the shell, the electric field is as if the shell did not exist at all. 
Within the shell You divide the usual value by $(1+\chi_e)$.
So $\vec{E}=\frac{Q\hat{r}}{4\pi\epsilon r^2}$
That the Field outside the shell is like the shell wasn't there, we expect 0 net charge in the shell. 
This should be demonstrable by applying the proper Boundary Conditions.
