What is the Relationship between E, D and P in a Material?

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The discussion centers on calculating the electric field (E), electric displacement (D), and polarization (P) in a thick sheet of polystyrene placed in an electric field. The user successfully calculates E and D but is confused about why the equation for P includes ε0 instead of just ε. It is clarified that ε0 represents the permittivity of free space, which is necessary for the relationship involving D and P. The user seeks further clarification on this point to understand the underlying principles better. The conversation highlights the importance of understanding the roles of different permittivities in electrostatics.
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A problem involving E, D & P. I have the answer but not sure why for the P bit.A bit of clarification if someone could, please.

A thick sheet of polystyrene, of relative permittivity ε = 2.5, is placed
normal to a uniform electric field of magnitude 500Vm−1 in air.

Find the magnitudes of the electric field E, the electric displacement D and the polarization P inside the material.

OK div D = ρf
and D = εε0E
Boundary conditions Doutside = Dinside → εoutEout = εε0Ein

so
Ein = Eout / ε
Ein = 500/2.5
Ein = 200 Vm-1

Dout = Din = ε0 Eout
Dout = 8.85x10-12 x 200
Dout = 1.77x10-9


Fine so far . . .

When I get down to calculating P I have to use

Din = ε0Ein + Pin

why, when everything else has the suffix "in" do I use ε0? Shouldn't it be just plain ε?

Thanks in advance.
 
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