Calculating Bound Surface Charge Density at Dielectric Discontinuity

[doktorwho]

Homework Statement

$e_{r1}=2$
$e_{r2}=10$
$α_1=\pi /4$
$D_1=100 nC/m^2$
$ρ_s=0$
On the borderline of two dielectrics there is no free charge. Calculate the surface bond charge density of both dielectrics.
$ρ_{bs1}=?$
$ρ_{bd2}=?$

Homework Equations

3. The Attempt at a Solution [/B]
So with the information i was given i am able to calculate some things.
Since there is no free charge $D_{1n}=D_{2n}$ so $D_1\cos α_1=D_2 \cos α_2$
And since $E_{1t}=E_{2t}$, $\frac{D_1 \sin α_1}{e_{r1}}=\frac{D_2 \sin α_2}{e_{r2}}$
Dividing these equations i get $\tan α_2=\frac{e_{r1}}{e_{r2}} \tan α_1$ and i get some crazy number.
The bond charge density is $D=e_0E+P$, $P_1=D_1-e_0E_1$, $P=e_0(e_r-1)E_1$
So how do i get this to work?

 

[rude man]

If you express Gauss's law in terms of E instead of D you can determine that there exist different bound surface charge densities at the dielectric discontinuity.