# Electric Field from Dielectric Shell

This is a question taken from a past E&M exam

A thick spherical shell (inner radius $R_1$ and outer radius $R_2$) is made of a dielectric material with a "frozen in" polarization $$P(r)=\frac{k}{r}\hat{r}$$ where $k$ is a constant and $r$ is the distance from the center. Find the electric field everywhere in space.

attempt: I tried using Gauss' law

$$\int D\cdot n \,da=Q_{enc\,,free}$$

However, there is no free charge, thus $D=0$ and therefore

$$\vec{D}=\epsilon_o \vec{E}+\vec{P}\,,\vec{D}=0\rightarrow \vec{E}=-\frac{\vec{P}}{\epsilon_0}=-\frac{k}{\epsilon_0r}\hat{r}$$

Is the electric field for all space. This doesn't seem right, I would think that the electric field differs inside the material of the shell vs. outside the shell or inside the cavity. What am I doing wrong?

• Actually your calculation is correct. The electric field satisfies $E = - P/\epsilon_0$ everywhere. Outside the shell, the polarization $P$ vanishes, so also $E$ vanishes. – Ján Lalinský Oct 31 '13 at 14:08

You are correct: there is no free charge so $\vec{D}=0$ which means $$\vec{E}=-\frac{1}{\epsilon_0}\vec{P}=-\frac{k}{\epsilon_0r}\hat{r}$$ But this is for $R_1\leq r\leq R_2$.
Inside the shell, $r<R_1$, there are no enclosed charges, so $\vec{E}=0$ there. Outside the shell, there is also no charge. Recall that the total charge for dielectrics can be expressed as $$Q_{tot}=\oint_\mathcal{S}\sigma_b da-\int_\mathcal{V}\rho_b dV = \oint_\mathcal{S}\vec{P}\cdot d\vec{a}-\int_\mathcal{V}\vec{\nabla}\cdot\vec{P} dV$$ where $\rho_b$ and $\sigma_b$ are the volume and surface charge densities of the bound charges. From the divergence theorem, the two terms involving $\vec{P}$ are identical, so $Q_{tot}=0$.
Thus, your answer should be $$\vec{E}\left(r\right)=\begin{cases}-\frac{k}{\epsilon_0r}\hat{r} & R_1\leq r\leq R_2 \\ 0 & {\rm otherwise}\end{cases}$$