# Superficial density of free charges on dielectric as $\sigma=\mathbf{D} \cdot \mathbf{n}$?

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?

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: $$(\mathbf{D}_2-\mathbf{D}_1)\cdot\mathbf{\hat{n}}=\sigma$$