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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!

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

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  • $\begingroup$ I just corrected my question saying that it's a dielectric sphere with uniformly distributed excess charge. Since it's a dielectric there wouldn't be any free charges in it, right? So rho would be the excess charge, bound. I don't understand how can there be a "polarization charge density", since polarization is symmetrical, i mean, a dipole is a positive and a negative charge which cancel each other. I think of the charged dielectric molecules as a dipole plus an excess charge. The net charge is only the excess charge. $\endgroup$ – dronkit Oct 13 '18 at 0:29
  • $\begingroup$ @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. $\endgroup$ – Bruno De Souza Leão Oct 14 '18 at 15:59

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