# Boundary between dielectrics

1. Mar 13, 2008

### plmokn2

1. The problem statement, all variables and given/known data
Say I have a boundary between two dielectrics then it's easy to show using a gaussian pillarbox that:
D(1)-D(2)=free surface charge density=s
where D(1) is the component of the first medium normal to the surface.
But suppose that there's nothing else apart from two infinite dielectrics with a constant free charge density between then, how would I work out what the actual values of D(1) and D(2) are rather than just the difference?

3. The attempt at a solution
It seems reasonable that the D field should be the same on both sides so that D=s/2 but I'm not sure how I'd prove this?

Any help appreciated.

2. Mar 13, 2008

### Mindscrape

I'm having a hard time interpreting what you are asking. Could you draw it?

3. Mar 13, 2008

### plmokn2

I think my question was badly worded/ ive probably misused some terminology.

I suppose my actual qustion is:
Its obvious from the symmetry that for an uncharged nonconducting non-dielectric that if you put a surface charge density s on it the field above is the same magnitude as the field below: D=s/2.

But suppose that the material is a dielectric then I'm not sure how you prove what the D field is above the surface (in air) and below the surface (in the dielectric) (under the same condition of the dielectric being infinite with surface charge s and no other free charge anywhere)

Hope this is a bit clearer, please say if it isn't.

4. Mar 13, 2008

### Mindscrape

Oh, well D is going to be the same for both air and the dielectric, but E will be different because E is D/(epsilon) which will change between the two (epsilon_0 for air and dielectric epsilon in the dielectric). Is this what you are asking about?

5. Mar 13, 2008

### plmokn2

That's it.

Its probably really obvious but how do you know that D has to be the same on both sides of the boundary?
Thanks

6. Mar 13, 2008

### Mindscrape

It comes from the way D is defined. After its whole redefinition then you just get a Gauss's Law for materials which goes as

$$\iint \mathbf{D} \cdot d\mathbf{a} = \sigma_f$$

and that means that the only thing that you care about is the free charge, which will be the same on both sides.

7. Mar 13, 2008

### plmokn2

Thanks.

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