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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?

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closed as unclear what you're asking by sammy gerbil, Jon Custer, Emilio Pisanty, Yashas, peterh 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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