I'm having some trouble with the following problem:

Consider a plate of a dielectric material homogeneous and isotropic with a dielectric constant equal to er= 2 , in the outside we have a uniform electric field E= 100v\m , that form an angle alpha with the x axis, determine the Electric field in the dielectric and the surface density of the polarization:

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My question is: in order to solve the first question the book uses the relation $$(\mathbf D_2−\mathbf D_1)⋅\hat{\mathbf n}=\sigma,$$ with $D_2$ and $D_1$ are the normal flux density component, and they consider that the free charge $\sigma =0$, because the surface is between a dielectric and a vacuum so there is no free charge are there, and the electric field inside the dielectric is equal to $$ \mathbf E_d=\frac{E}{\epsilon_r}\cos(\alpha)\mathbf u_x + E\sin(\alpha) \mathbf u_y $$ where $\mathbf u_x$ and $\mathbf u_y$ are unit vectors of the $x$ and $y$ axis.

But my question is: do we also have the following relationship $D=\epsilon_0E_d\epsilon_r$ it means that the free charges are not null! so how can we assume that they are null and not null at the same time?


closed as unclear what you're asking by sammy gerbil, Jon Custer, Emilio Pisanty, Yashas, user259412 Dec 10 '17 at 7:07

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Here is the solution to my confusion: the first free charges are those between the dielectric surface and the vacuum in this case they are null. the second type of free charges: are those who generated the field and caused the polarization of the dielectric and they are completely different from the first type! so there is no contradiction.


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