This is an example problem from my E&M textbook that I don't quite understand:
A metal sphere of radius a carries a charge $Q$. It is surrounded, out to radius b, by linear dielectric material of permittivity $\epsilon$. Find the potential at the centre (relative to infinity).
Using $$\iint \mathbf{D}\cdot\ d\mathbf{A}=Q_{free}$$ I get $\mathbf{D}=\dfrac{Q}{4\pi r^2}\hat{\mathbf{r}}$. What I don't understand is that the author claims that this is only valid for $r>a$. Why is this the case? Couldn't I simply draw a gaussian surface for the inner surface to compute $\mathbf{D}$ in the same way? Why does this only work for the outer surface but not on the inside? I guess I'm having trouble visualising what $\mathbf{D}$ actually is since for the electric field, I can picture the 'flux' of the electric field lines when using Gauss's law but I'm lost on what I'm supposed to picture for electric displacement.
Best Answer
Inside the metal sphere, where $r<a$, there is no free charge so your Gaussian sphere will enclose $Q_{free}=0$. As a result, both $\vec E$ and $\vec D$ are $0$.
In addition but separately from this, your expression is problematic as it would produce $D\to\infty$ as $r\to 0$.