# Electric field inside a uniformly charged dielectric sphere

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!

• Sorry, i had a confused concept of "free charge", I meant "excess charge" since it's a dielectric – dronkit Oct 13 '18 at 0:25

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.
• @dronkit Maxwell's equations for a dielectric impose that the total charge in a dielectric must be proportional to the free charge. Just take the equation $\nabla\cdot\mathbf{D} = \rho_{free}$, as well as $\nabla\cdot\mathbf{E} = \dfrac{\rho}{\epsilon_0}$, and the constitutive relation $\mathbf{D} = \epsilon\mathbf{E}$. As for the "polarization charge density", read "bound charge density" -- I just called it differently while writing the answer. – Bruno De Souza Leão Oct 14 '18 at 15:59