Confusion regarding electric displacement in dielectric I have a little confusion regarding the electric displacement concept for a dielectric.
Let us say that there is a dielectric sphere placed in a uniform electric field. I want to calculate the electric field inside the sphere. Since there is symmetry in the problem, I can use
$$
\nabla \cdot \vec D = \frac{Q_f}{\epsilon_0}
$$
However, since $Q_{free}$ is $0$, $\vec D  =\vec 0$.
I also know that the electric field $\vec E$ is related to $\vec D$ by the relation
$$
\vec D = \epsilon \vec E
$$
from which I can say that the electric field magnitude is also $0$. However, intuitively I feel this is wrong because the electric field magnitude is $0$ for a conductor, not a dielectric. I was hoping someone could find where exactly I am going wrong.
 A: You are correct to conclude that $\nabla \cdot \mathbf D=0$, but it does not follow that the electric displacement vanishes.
There are two ways to see this. One is that the electric displacement needs to default to its no-dielectric value at large distances, i.e. the externally-imposed electric field. The other is that there is free charge in the problem - whatever created the external field, most easily modelled by a large pair of parallel plates far from the sphere. Either way, you need to have $\mathbf D= \epsilon_0 \mathbf E_0$ far from the sphere, along with the divergencelessness condition. 
The solution to this (once you swallow the pill, which isn't that easy) is very simple - it's just  $\mathbf D= \epsilon_0 \mathbf E_0$ everywhere. There's no free charge on the dielectric sphere itself, so it's invisible to the electric displacement field, which keeps its uniform value.
On the other hand, the electric field itself does have a discontinuity, because the permittivity changes. That gives you a layer of charge at the surface of the sphere, where the polarized dielectric meets the vacuum.
A: As I mentioned, you made 2 mistakes:


*

*Spherical Symmetry is broken by the external electric field, so Gauss's isn't simple enough to imply D=0

*The dielectric constant is not homogeneous.  There are different polarizabilities, $\epsilon$ inside and outside the sphere. 


This second condition creates a non trivial boundary condition at the surface of a sphere.
Since $\vec{\nabla} \times \vec{E}$ is still 0 (The bound charge won't make the electric field curl around the object), a loop around the surface of the will show that the component of the electric field tangential (parallel) to the surface will be continuous $E_{||,in}=E_{||,in}$
While Gauss's law for the displacement field will show that the $D_{\perp}$ is continuous across, implying that any bound charge will create a discontinuity in the perpendicular component of the electric field.  For these derivations see the first few slides here
Therefore, the external electric field will polarize the sphere, creating bound charge and reducing the electric field in side the sphere(still in the same direction as the external field).  This bound charge at far distances will make the sphere look like a dipole and generate (in addition to the external electric field) a electric field of a dipole:

You can solve this problem exactly by working with spherical harmonics and solving the boundary conditions written in terms of the electric potential.  Which is done starting on slide 11 here
