Free surface charge density (not bound density)

[Roodles01]

Homework Statement

a parallel plate capacitor has 2 plates separated by a dielectric of rel. permittivity 5.0 are separated by 0.20mm and have area of 5.0 cm2.
Potential difference between the plates is 500V.

I need to be able to calculate the free surface charge density.

Homework Equations

I'm assuming D_out = D_in

From boundary conditions D₁ ┴ = D₂ ┴
and D₂ ┴ - D₁ ┴ = σf

Could someone just confirm what the boundary conditions are so I can say that the free surface charge density is zero or not!

The Attempt at a Solution

I have D_out = D_in = ε₀E_in + P
so from calculations
P = 4.43x10^-9 - (8.85x10^-12 x 100)
P = 3.54x10^-9 Cm^-2

Could I just say;
D₁ ┴ = D₂ ┴ so there is zero free surface charge density!

 

[Roodles01]

Sorry, I should have said.
D1 ┴ is the electric displacement in media 1 perpendicular to a boundary with a different media
D2 ┴ is the electric displacement in media 1 perpendicular to a boundary with a different media

They form the boundary condition

Dout = Din

could have been
D1 = D2.

and P is the polarization of the material.

 

[rude man]

I don't know what you mean by "different media" or "Dout" vs. "Din". There is only one medium, the dielectric of epsilon_relative = 5.

D is continuous from one medium to another along the plates' normal even if there were two or more media, but there aren't.

So V = E x d, d = 0.2mm so you know E.
Then, D = epsilon x E.
And if you know D you know the surface charge density, right?

This ignores fringing effects of course which you have to do since they didn't give you the dimensions of the plates.