Imagine I have a dielectric material with one cavity placed in a uniform electric field $E_0$. For simplicity, assume it's a cylinder with large radius with its bottom surface perpendicular to the field. And also assume the cavity is a large cylinder whose top and bottom surfaces are parallel to the first's.
I want to know what is the electric field inside the cavity. I could draw a cylindrical Gauss surface that has its bottom inside the cavity and top inside the dielectric. Then, neglecting the flux through the side of the Gauss surface, I could write approximately $D S =\epsilon_0 E S$ with S being the surface of the Gauss cylinder. But I can draw another Gauss cylinder with the bottom in the dielectric and the top outside the dielectric and I have the same type of equation, $DS=\epsilon_0 E_0 S$. This means that the field in the cavity is more or less equal to the applied external field.
Is this reasoning correct, or am I missing something? My colleagues keep telling me $E=0$.
Best Answer
Unless you have free surface charge (which you don't in an insulator) then the normal component of the displacement field is continuous. That leads to a D-field in the dielectric and then a D-field in the cavity.
The value of the E-field inside the cavity will depend on the shape of the object and the shape of the cavity.
Also note that the E-field outside the dielectric may no longer equal $E_0$ because of polarisation charge on the surface.