I'm studying EM for the first time, using Griffiths as the majority of undergraduates. So, in studying electric fields in matter, we derived equations for the volume and surface density charge of the bound charges. The first thing we calculated was the "electric field produced by a uniformly polarized sphere of radius R". No problem here, the answer for the field inside the sphere is
$\vec{E} = -\vec{P}/3\epsilon_0$
A remark, there's no mention about the sphere being made of a dielectric material or if it's a conductor, but I guess it doesn't matter since we're only calculating the field due to the polarization of the sphere. OK, some sections later we learn about the electric displacement $\vec{D} = \epsilon_0 \vec{E} + \vec{P}$, where $\vec{E}$ is the total electrical field, and we learn this useful equation:
$\oint_S \vec{D}\,d\vec{a} = Q_{free}$
And there's a nice problem about a thick shell made of dielectric material with a frozen polarization where we're asked to calculate the field using the bounded charges + Gauss Law and using $\vec{D}$ to check if they give the same answer, which they do of course, we realize that $\vec{D} = 0$ everywhere in this problem since there're no free charges in the dielectric.
Now my question: I immediately thought of applying $\vec{D}$ to calculate again the field of the sphere, with the uniform polarization, but I soon ran into some trouble! First I asked myself: are there free charges inside the sphere? Well, if the sphere is made from a dielectric material the answer is no, so $\vec{D}$ = 0 everywhere, which makes me arrive at this conclusion:
$\vec{D} = 0 = \epsilon_0 \vec{E} + \vec{P}\\
\vec{E} = -\vec{P}/\epsilon_0 \, (inside\, the\, sphere)$
Outside the sphere $\vec{P} = 0$ and thus $\vec{E} = 0$, which is not the right answer at all! I'm so much confused by this result because I used the same method that worked in the problem of the thick shell. I don't know what to make of it. This $\vec{E}$ that I calculated is the total electric field, but since I reasoned that there aren't free charges, there is no contribution for the total field from free charges, so the total field is equal to the field generated by the bound charges. What I'm missing here? There are free charges inside the sphere after all? But if there are free charges, why in the problem of the thick shell there are no free charges?
Best Answer
The error occurs at $\mathbf D = 0$. We do know that $\nabla \cdot \mathbf D = 0$, but this will not guarantee that $\mathbf D = 0$ everywhere. Also, the configuration in the problem is not spherically symmetric.
For inside the sphere, $\mathbf D = \epsilon_0\mathbf E + \mathbf P = 2\mathbf P/3$, so $\nabla \cdot \mathbf D = 0$. For outside the sphere, $$\mathbf E = \frac{P}{3\epsilon_0}\frac{R^3\cos \theta}{r^3},$$ where $P$ is the magnitude of the polarization inside the sphere and $R$ is the radius of the sphere. You can check for yourself that $\nabla \cdot \mathbf D = 0$ holds.