# Dielectric Boundary Conditions (Parallel Plate Cap.)

Discussion in 'Homework Help' started by jegues, Oct 12, 2011.

1. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
See figure attached for problem statement, as well the solution.

I'm confused as to how he is writing these equations from the boundary conditions.

What I understand as the boundary condition for D is,

$\hat{n} \cdot \vec{D_{1}} - \hat{n} \cdot \vec{D_{2}} = \rho_{s}$

With the normal vector directed from region 1 to region 2.

With this I only generated 2 equations, as there is only 2 boundarys; one being from d1 to d2 and the other from d2 to d3.

He labels,

$\rho_{s0}, \rho_{s1}, \rho_{s2}, \rho_{s3}$

I'm confused as to what these surface charge densities pertain to? Is the surface charge density with subscript 0 and 3 the charge on the plates of the capacitor? Are 1 and 2 the surface charge densities on the faces of the dielectric material?

He states by conservation of charge that,

$\rho_{s0} = -\rho_{s3}, \quad \rho_{s1} = -\rho_{s2}$

This doesn't seem obvious to me at all. Can someone show me how he is drawing such a conclusion? (Is there some math behind it?)

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2. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
ARGH! Electromagnetic Fields!

I passed No1 by accident and I still struggle for No2. Put it away!

Seriously though, it seems indeed that s1 to s3 refer to the four interfaces, as referenced in the second picture.
And since the slab started uncharged, it is only reasonable that ρ1=-ρ2. The same holds for the battery terminals.

3. ### jegues Thread Starter Well-Known Member

Sep 13, 2010
735
43
What makes you think the slab is uncharged?

We aren't told anything with regards to the slab except that it's conducting and it's permittivity is ε.

Does that imply something about the charge for the slab?

4. ### Georacer Moderator

Nov 25, 2009
5,142
1,266
I can't say you 're wrong, but I don't see any other way to reah the equation ρ2=-ρ3.

Maybe we need a professional opinion here.