An infinite Dielectric and a charge!

We know that the displacement vector due to the charge will be in the inverse square form. Let us choose our coordinate system in a manner that the charge is at the origin, with the dielectric at $z<0$ and vacuum in $z>0$ Thus, Let, $\textbf{D} = \frac{qa_1}{4 \pi r^2}$ $\hat{r}$ for $z<0$

$\textbf{D} = \frac{qa_2}{4 \pi r^2}$ $\hat{r}$ for $z>0$

The boundary conditions at z = 0 at all points away from the origin is that the normal component of D should be continuous and that the tangential component of E must be continuous. Since $\hat{z}.\hat{r} = cos θ$ vanishes at θ = π/2 i.e, when z = 0, the continuity of the normal component of D · $\hat{z}$ is established. For the electric field we get,

$a_2 = \frac{a_1}{1.5}$ _ _ (1)

By Gauss' Law for dielectrics we have, Flux of displacement vector = Free charge Thus the flux through a sphere of radius R is given by,

$\frac{q(a_1+a_2)}{2} = q$ _ (2)

Using 1 and 2 we have,

$a_1 = 1.2$ and $a_2 = 0.8$

Thus,

$\textbf{D} = \frac{1.2q}{4 \pi r^2}$ $\hat{r}$ for $z<0$

$\textbf{D} = \frac{0.8q}{4 \pi r^2}$ $\hat{r}$ for $z>0$

and hence,

$\textbf{E} = \frac{1.2q}{(1.5)4 \pi \epsilon r^2}$ = $\frac{0.8q}{4 \pi \epsilon r^2}$ $\hat{r}$ or $z<0$

and,

$\textbf{E} = \frac{0.8q}{4 \pi \epsilon r^2}$ $\hat{r}$ for $z>0$

And hence the field is spherically symmetric in the region!

Those pointing out that the field should be less in the dielectric: that would have happened if the charge was completely surrounded by the dielectric medium.