Relationship between polarization density and electric field

Question

Imagine I have an electric field $E$ created due to some free charges. Then I introduce a dielectric material somewhere. What confuses me is the polarization field $P$, which is now proportional to the original $E$ within the dielectric. Does this new field contribute to the total electric field outside of the dielectric? How will the force on a test charge near the dielectric change?

I know with $D=E + 4\pi P$, that the maxwell equation $\operatorname{div}(D) = 4\pi\cdot \rho$ with $\rho$ being the free charge density must be satisfied, but what boundary conditions have I added to calculate $D$?

Answer 1

Polarization P is caused by the presence of E and a dielectric material. It does not "add" to E, instead it adds to flux D. It does not relate to free charges. And the relationship is

$$ D = \epsilon_0 E + P. $$

Polarization reflects what happens to the bound charge-pairs in said dielectric (i.e. the amount the charge-pair separate with applied electric field). By virtue of the separation of the charge-pairs, there is a localized E field created, which distort E locally but since net charge of each pair is still 0, it does not add E globally, And it does not affect D outside the material.

One encounters P in the simple and most common case of linear dielectrics, as contributing to the permittivity. That is,

$$ D = \epsilon_0 E + \epsilon_0\chi E = \epsilon_0 (1+\chi) E. $$

where $\chi$ is called electric susceptibility.

Or to be more accurate, we create a mathematical constant, relative permittivity, i.e.

$$ \epsilon_r = \frac{P}{\epsilon_0 E} - 1, $$

so we can more easily relate P, E and D, vis a vis

$$ D = \epsilon_r \epsilon_0 E. $$

With regards to boundary condition, other than changing relative permittivity on either side of the boundary, there is no other consideration specific to P. All boundary condition of E and D still hold, regardless of P (or $\epsilon_r$).