Question

The electric field for $r<R_1$ is given by $E=\frac{XQ}{\pi {\epsilon }_0{r}^2}\left[\frac{r^3-R_1^3}{R_2^3-R_1^3}\right]$. Find X?

Answer 1

Let us imagine a surface of radius r<$R_1$ inside the cavity. By symmetry the electric field (if any) inside the cavity will be radial and only a function of radius r.
therefore, by Gauss's law,
$E.4\pi r^2=\frac{Q_{in}}{ϵ}=0\Rightarrow E=0forr<R_1$