Free surface charge density (not bound density)

AI Thread Summary
To calculate the free surface charge density of a parallel plate capacitor with a dielectric, the boundary condition states that the electric displacement fields (D) are equal across the boundary, leading to the equation D1 ┴ = D2 ┴. Given the dielectric's relative permittivity of 5.0, the polarization (P) can be calculated, which is found to be 3.54 x 10^-9 Cm^-2. The user questions whether this implies zero free surface charge density, as D1 ┴ equals D2 ┴. The discussion clarifies that the electric displacement is continuous, and while fringing effects are ignored, the relationship between D, E, and surface charge density remains valid. The conclusion is that the free surface charge density can be determined from the calculated D value.
Roodles01
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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 Dout = Din

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 Dout = Din = ε0Ein + 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!
 
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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
Roodles01 said:
Dout = Din
could have been
D1 = D2.

and P is the polarization of the material.
 
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.
 
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