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My basic understanding of a dielectric material is that the coulomb force between charges in a dielectric must be reduced by some factor since some of the electric field energy gets stored in the medium. The following thought experiment is confusing me and I am wondering where my mistake lies.

Lets consider the border area between two different dielectrics. Now consider a rectangular path where two opposing sides are each in a different dielectric. All sides are short enough that any E field would be constant if it were the same medium and the sides travellng through the border are very short compared to the others.

Now the work done by the E field along this path must be zero. This seems to imply that the component of the E field along the path is equal in the different dielectrics. But arent all components of the E field in the dielectric smaller by some factor?

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  • $\begingroup$ "But arent all components of the E field in the dielectric smaller by some factor?" - smaller relative to what? Think carefully about this. Are you imagining, for example, a charged parallel plate capacitor with two different dielectrics side by side? $\endgroup$ – Alfred Centauri Nov 1 '18 at 15:18
  • $\begingroup$ I meant the E field in the dielectric with a higher dielectric constant should be smaller than the E field in the other dielectric. Is that not the case? $\endgroup$ – fibo11235 Nov 1 '18 at 15:31
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    $\begingroup$ In the case of the charged plate capacitor with the different dielectrics sids by side I would expect the charges on the plates to redistribute until the E field is uniform. But what about the case where all the charges on the plates are completely imobile? $\endgroup$ – fibo11235 Nov 1 '18 at 15:35
  • $\begingroup$ fibo11235, why should it be the case? Yes, if you fix $\vec{D}$ then $\vec{E} = \vec{D}/\epsilon$ is smaller for the dielectric with larger $\epsilon$ but what is it, in your thought experiment that is fixing $\vec{D}$? Think about adding some additional detail to your thought experiment. $\endgroup$ – Alfred Centauri Nov 1 '18 at 15:54
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Let's first consider a parallel plate capacitor with two uniform layers of dielectrics with different dielectric constants and the electric field will be normal to the boundary between the dielectrics.

In this case, the work is performed only in the segments of the path normal to the boundary, since in the segments of the path parallel to the boundary, the electric field is normal to the path and, therefore, does not perform any work.

Within each dielectric, the work performed over one of the segments normal to the boundary will be cancelled by the work performed over the other segment normal to the boundary, since the direction of the field will be along the path in one of the segments and against it in the other.

So, the work performed along the path in each dielectric will be zero and, therefore, the total work performed along the whole path, will be zero.

If we have a more complicated setup, for instance, if the boundary between dielectrics is not parallel to the plates, the electric field lines will bend, probably becoming normal to the boundary between the dielectrics or do whatever it takes to ensure that the line integral $\oint \vec E\vec{ds}$ around any closed loop is zero.

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  • $\begingroup$ why is the electric field normal to the boundary? $\endgroup$ – fibo11235 Nov 1 '18 at 14:22
  • $\begingroup$ @fibo11235 Good point. I've updated the answer to address this case. $\endgroup$ – V.F. Nov 1 '18 at 18:44
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It depends on which quantity is being conserved. If we have a conductor, the potential at its surface must be equal, so there will be a greater surface charge density when there is a higher dielectric constant. In this case, $\mathbf{E}$ is directly determined by the potential, since $\mathbf{E} = -\nabla V$. Thus, $\mathbf{E}$ and $\epsilon$ determine $\mathbf{D}$, and because $\nabla \cdot \mathbf{D} = \rho_f$, we see that there is more charge wherever there is a dielectric present.

However, if we have an insulator, $\rho_f$ is now the independent variable. Since $\nabla \cdot \mathbf{D} = \rho_f$, $\mathbf{D}$ and $\epsilon$ now determine $\mathbf{E}$. The usual electrostatic boundary conditions, $\Delta \mathbf{E}_\bot = \frac{\sigma}{\epsilon_0}$ and $\Delta \mathbf{E}_{||} = 0$, still apply in both cases.

Note that, since $\nabla \times \mathbf{D} \neq \mathbf{0}$, we cannot calculate $\mathbf{D}$ from the usual method of integrating Coulomb's Law. We also cannot define a potential for it. The only way it can be found without first finding $\mathbf{E}$ is by Gauss' Law.

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