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Consider a dielectric in a electric field $\mathbf{E}$. How can I find the superficial density of free charghes $\sigma$?

It is related with the vector $\mathbf{D}$ but I'm quite confused about a minus sign. I will indicate with $\mathbf{n}$ the versor going out from the dielectric.

Can I say the following?

$$\sigma=\mathbf{D} \cdot \mathbf{n}$$

This looks ok but it also means that free charghes and polarization charghes (with density $\sigma_P$) in the dielectric have the same sign.

$$\sigma_P=\mathbf{P} \cdot \mathbf{n} =\frac{\epsilon_r-1}{\epsilon_r} \mathbf{D} \cdot \mathbf{n}=\frac{\epsilon_r-1}{\epsilon_r} \sigma$$

While on textbook I found that these two are opposites, but I think that in that case $\mathbf{n}$ is taken to be the versor going out of the conductor near the dielectric (for istance in the case in which the dielectric is in a capacitor), where there is a chage density $\sigma$.

So is this the correct way to express the free charges on the surface of the dielectric?

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If you have a free charge at the interface of different dielectrics (including vacuum), according to Gauss law, the relation of the surface charge to the electric displacements is given by: $$(\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{\hat{n}}=\sigma$$

The minus sign in one of the displacements has its origin in Gauss' law which considers normal displacement vectors pointing outward from the considered volume. Free interface (surface) charges occur at the interface of dielectrics with different conductivities. See my answers to the following questions: Free charge in a dielectric and Surface charge density in conducting plate .

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