# Uniform Polarized Sphere - are there free charges? How to use Electric Displacement?

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?

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.

• You wrote that $\vec{D} = \epsilon_0 \vec{E} + \vec{P} = \vec{P}/2\epsilon_0$ for inside the sphere, how did you conclude that? – Tandeitnik May 22 '19 at 22:25
• Oh oops. It should be $\mathbf D = 2\mathbf P / 3$, since $\mathbf E = -\mathbf P/3 \epsilon_0$. – Hanting Zhang May 22 '19 at 22:55
• I think I understood. Indeed there is no free charge inside the sphere, but since the polarization is uniform, we have that the flux of $\vec{D}$ is also 0, so what we have is 0 = 0 from the "Gauss law" for $\vec{D}$ and I can't deduce the field by that. This makes sense to me. – Tandeitnik May 23 '19 at 2:15

Perhaps, for pedagogical purposes it will be good to talk about one of the exercises from David J. Griffiths 3 ed, which seems to be related to what you are asking.:

Problem 4.15: " A thick spherical shell (inner radius a, outer radius b) is made of dielectric material with a "frozen-in" polarization P(r)=k/r in r direction, where k is a constant and r is the distance from the center. there is no free charge in the problem. Find the electric field in all three regions by two different methods: a) Locate all the bound change, and use Gauss's law to calculate the field it produces. b) use ∫ D •n da = Q_fenc, (where ∫ da is above a closed surface, n, D ∈ R³ and Q_fenc is total free charge enclosed in the volume) to find D, and then get E from D = ε0 E + P "

...as you stated, ∫ D •n da = Q_fen=0 → D = 0 everywhere. D = ε0 E + P =0→ E = - 1/ε0 P so E= 0 for r < a and r > b ; r ∈ R. and E = -( k/ (ε0 r) ) r for a < r < b

Does this help?