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 $E=0$.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Oct 28, 2016 at 7:50
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    $\begingroup$ @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. $\endgroup$
    – ProfRob
    Oct 28, 2016 at 8:40
  • $\begingroup$ I'm not so interested in the edge effect. Not because it's not there, but because I'm more interested in understanding the basic electrostatics of the problem. That's why I specified that geometry with very large cylinder radius. $\endgroup$ Oct 28, 2016 at 9:24
  • $\begingroup$ 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. $\endgroup$ Oct 28, 2016 at 9:34
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    $\begingroup$ @user3653831 Yes. The example you have given is I think as close as you can come to a cavity where this is possible, but in that regard you are reducing the problem to a capacitor with a dielectric, but with a gap in the dielectric. You keep the voltage constant as you insert the dielectric. In which case it is trivially obvious that the electric field in the gap in the dielectric is not zero, and the same is true for the cavity you propose. You can find examples on the web - the usual one is a spherical cavity in a dielectric. $\endgroup$
    – ProfRob
    Oct 28, 2016 at 10:39

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