At the boundary of an infinite linear dielectric with constant $\kappa$ and a vacuum, a charge q is placed. In order to calculate the electric displacement$(\mathbf{D})$ we can use Gauss's law for dielectrics i.e. $\oint\mathbf{D}\cdot d\mathbf{a}=Q_{fenc}$.
Now since the electric displacement vector does not depend on the dielectric, we can consider a sphere centered at q with radius $r$ and since $\mathbf D$ is spherical symmetric, the integral evaluates to: $$\mathbf D\cdot 4\pi r^2\hat r=q \Rightarrow \mathbf D= \frac{\hat r}{4\pi r^2}$$.
From this we can calculate the electric field for the two parts which is:$$\mathbf E= \frac{\hat r}{4\kappa\pi r^2};\Biggl\{^{\kappa=1\ vacuum}_{\kappa=\kappa\ dielectric}$$
But this is apparently incorrect and I don't understand why. Some guidance would be appreciated.
Best Answer
Careful -- ${\bf D}$ is affected by the presence of a dielectric, just not in an immediately obvious way. You need to use the properties of ${\bf E}$ and ${\bf D}$ fields at dielectric boundaries, specifically \begin{align} {\bf E}_{\parallel}=\text{ continuous}, {\bf E}_{\perp}=\text{ discontinuous}\\ {\bf D}_{\perp}=\text{ continuous}, {\bf D}_{\parallel}=\text{ discontinuous} \end{align} Perhaps this is enough to get you back on track?