A point charge $q $ is embedded at the center of a sphere of linear dielectric material (susceptibility $\chi _e $, radius $R $).
Find the electric field, the polarization, and the bound charge densities, $\rho _b $ and $\sigma _b $.
What is the total bound charge on the surface?
Where is the compensating negative bound charge located?
Recall Gauss's law for the electric displacement $\mathbf{D} $:
\[\oint \mathbf{D}\cdot d\mathbf{a}=Q_{f_{\text{enc} }} \]We have spherical symmetry, so we can break down the integral.
\begin{align*} D\cdot (\text{area} ) &= Q_{f_{\text{enc} }} \\ D\cdot (4\pi r^2 ) &= q\\ D &= \frac{q}{4\pi r^2 }\\ \mathbf{D} &= \frac{q}{4\pi r^2 }\hat{\mathbf{r}} \end{align*}The electric field will then be:
\begin{align*} \mathbf{E} &= \frac{1}{\epsilon }\mathbf{D}\tag{Recall $\mathbf{D}=\epsilon \mathbf{E} $} \\ &= \frac{1}{\epsilon }\frac{q}{4\pi r^2 }\hat{\mathbf{r}} \tag{As found above}\\ &= \frac{1}{\epsilon _0 (1+\chi _e)}\frac{q}{4\pi r^2 }\hat{\mathbf{r}} \\ &= \frac{1}{4\pi \epsilon _0 }\frac{q}{(1+\chi _e)r^2 }\hat{\mathbf{r}} \end{align*}The polarization follows from $\mathbf{P}=\epsilon _0 \chi _e\mathbf{E} $:
\[\mathbf{P}= \frac{\chi _e}{4\pi }\frac{q}{(1+\chi _e)r^2 }\hat{\mathbf{r}} \]Now that we know the polarization, we can compute the bound charges. For volume:
\begin{align*} \rho _b &= -\nabla \cdot \mathbf{P} \\ &= -\frac{\chi _e}{4\pi }\frac{q}{(1+\chi _e) } \nabla \cdot \frac{\hat{\mathbf{r}}}{r^2 }\\ &= -\frac{\chi _e}{4\pi }\frac{q}{(1+\chi _e) }4\pi \delta ^3 (\mathbf{r})\\ &= -\frac{\chi _eq}{(1+\chi _e) }\delta ^3 (\mathbf{r}) \end{align*}And for surface:
\begin{align*} \sigma _b &= \mathbf{P}\cdot \hat{\mathbf{r}} \\ &= \frac{\chi _e}{4\pi }\frac{q}{(1+\chi _e)R^2 } \end{align*}The total charge on the surface is:
\begin{align*} Q_{\text{surface} } &= \int \sigma b da \\ &= \sigma _b (4\pi R^2 )\\ &= \frac{\chi _e}{4\pi }\frac{q}{(1+\chi _e)R^2 } (4\pi R^2 )\\ &= \frac{q\chi _e}{1+\chi _e} \end{align*}Total bound charge should be zero, so the rest is in the center as indicated by the dirac delta:
\begin{align*} Q_{\text{center} } &= \int \rho _bd\tau \\ &= \int -\frac{\chi _eq}{(1+\chi _e) }\delta ^3 (\mathbf{r})d\tau \\ &= -\frac{\chi _eq}{(1+\chi _e) }\int \delta ^3 (\mathbf{r}) d\tau \\ &= -\frac{\chi _eq}{(1+\chi _e) } \end{align*}