# How does the dieletric displacement change among two dieletric materials?

Here's a question I am working on. "A sphere of linear dielectric material with permittivity $\epsilon_1$ and radius $a$ is surrounded by an infinite region of linear permittivity $\epsilon_2$. In the spherical region, there is free charge embedded given by $\rho_{free}=\beta r^2$, $0<r<a$, where $\beta$ is a constant and $r$ is the distance from the center of the sphere. Find the electric displacement $\vec{D}$ in all space."

What I have so far

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

Inside the sphere: $\oint D \cdot da=4 \pi r^2$, $Q_{free,enc}=\begin{cases} \frac{4}{5} \beta \pi r^5 \hat{r}, r<a \\ \frac{4}{5} \beta \pi a^5 \hat{r}, r>a \end{cases}$ Here is where I am confused. I know that inside $\vec{D}=\frac{\beta}{5}r^3 \hat{r}$, but I'm not sure about how to find $\vec{D}$ outside. Normally $\vec{D}$ would be the same inside and out if the sphere were in empty space, but I'm not sure how another dieletric material would affect $\vec{D}$. Any help is appreciated.

## 1 Answer

Your whole derivation is correct. Even in the presence of two different dieletric materials, the $\mathbf{D}$ field will not be affected, but for the free charge density that you already dealt with. So the field will be $$\mathbf{D}=\begin{cases}\dfrac{1}{5}\beta r^3\hat{\mathbf{r}},&\text{if r<a},\\ \dfrac{1}{5}\beta a^5\dfrac{\hat{\mathbf{r}}}{r^2}, & \text{if r>a.} \end{cases}$$ What will change is of course the relation between $\mathbf{D}$ and $\mathbf{E}$, that depends on $\epsilon_i$. This is due to the bound charges that apear in the two dieletric interface at $r=a$.