Electric field inside a uniformly charged dielectric sphere

Question

Every basic course and book on electrostatics has this problem, to find out the electric field inside a uniformly charged sphere. The result always is $\frac{\rho r}{3 \epsilon_0}$, obtained using Gauss. The question is, if only a non conductor body can hold an excess charge density inside it, the sphere is dielectric, and so there must be a polarization effect inside it. Shouldn't the solution be $\frac{\rho r}{3 \epsilon}$, with $\epsilon$ being the dielectric's permitivity? I guess the books are ignoring it and pretending it's a "magical" body-less distribution of charges, or does the polarization effect somehow get cancelled in such a sphere? Thanks!

Answer 1

I would slightly disagree with ZeroTheHero, and I think the answer to your question relies on how you interpret the density $\rho$. The thing is: the books always assert that $\rho$ is the total charge density (free + polarization); the factor or $\epsilon$ would come up if you were considering $\rho$ to be only the free charge density. The reasoning would be to use Gauss' law for a dielectric: $$\nabla\cdot\mathbf{D} = \rho_{free}$$ By the symmetry of the sphere, you'd get $$\mathbf{D}(\mathbf{r}) = \dfrac{\rho_{free}\mathbf{r}}{3}$$ and using the constitutive relation $\mathbf{D} = \epsilon\mathbf{E}$, you'd get $$\mathbf{E}(\mathbf{r}) = \dfrac{\rho_{free}\mathbf{r}}{3\epsilon}$$ So the correct interpretation is that the ratio between the total charge density $\rho$ and the free charge density is given by the relative permissivity $\rho_{free} = \epsilon_{r}\rho$, and since $\epsilon_{r}$ is usually greater than $1$, the total charge density is less then the free charge density -- precisely because of the polarization charges.