Electric field due to nonconducting sphere

Question

For calculating electric field outside a nonconducting sphere with a hollow spherical cavity. When I use the rule (Charge density= $dQ/dV$), I don't know exactly what is $dV$, is the volume here refers to the volume of the Gaussian surface ($V= 4/3 \pi r^3$) so that $dV$ will be = $ \pi r^2 dr$, or the $V$ is the volume containing the charges only, so it will be =$ V_0 – V_1 = 4/3 \pi r_0^3 - 4/3 \pi r_1^3 $. Thus, since $ r_0$ and $r_1$ are constants, therefore $dV$ will be = 0?

Note: $r_0$ is the radius for the whole sphere, $r_1$ is the radius for the cavity, and $r$ is for the Gaussian surface.

Answer 1

The charge density is the charge per unit volume of the object that contains the charge. If the charge density is constant on that object, then the charge density is just the total charge of the object divided by the total volume of the object.

This case case, the volume of the object is $V_0 - V_1$, so if $Q$ denotes its total charge, then the charge density $\rho$ is given by $$ \rho = \frac{Q}{V_0-V_1}. $$