# Electric field inside a cavity of a dielectric material subjected to external fields

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 $E=0$.

Also note that the E-field outside the dielectric may no longer equal $E_0$ because of polarisation charge on the surface.
• To me, this becomes even less clear. Assuming I have a polarizable dielectric, does that mean the field outside should be $E_0+\sigma/2\epsilon_0$ with $\sigma$ the polarization charge? Commented Oct 28, 2016 at 7:50
• @user3653831 For the geometry you specify above? Sounds possibly correct if you ignore edge effects and so-on, but that requires the base of the cylinder to be infinitely large. The only point I'm making is that you cannot assume that $\vec{E_0}$ is unchanged when you insert a dielectric into it. Particular examples require the solution of Poisson's equation using the continuity conditions for the field as boundary conditions. Commented Oct 28, 2016 at 8:40
• Now I think the problem is not correctly formulated. It's not the external electric field that's relevant, but the voltage bias applied to the system, i.e. if I put the dielectric inside a capacitor that had $E_0$ without it, the field will change depending on the dielectric's polarizability. Commented Oct 28, 2016 at 9:34