Solving Boundary Conditions for Electric Field in Dielectric Media

[Electro]

Greetings everyone:
I'm trying to solve a problem which requires finding the electric field. I've been pondering on this problem for a while but still no results. The book doesn't give any hints or clues on how to tackle this kind of problems so I would really appreciate some of your suggestions.

Assume that the z= 0 plane separates two lossless diaelectric regions with Epsilon(r1) = 2 and Epsilon(r2)=3. If we know that E1 in region 1 is 2y i - 3x j + (5+z) k, what is E2?

 

[siddharth]

What are the boundary conditions on the electric field at the interface between the two dielectric regions?

 

[Electro]

Thanks Siddharth, but if we draw a picture of the two media, geometrically we can conclude that E1t = E2t where E1t=E1-E1n and the same with E2t (I guess you get the picture). I don't see how this would help. Moreover we are dealing with 3-D ( I guess I might have understood it if it was 2-D).
Thanks

I'm editing the post: Thanks for your help, I understand it now.

 

[FrogPad]

Thanks Siddharth, but if we draw a picture of the two media, geometrically we can conclude that E1t = E2t where E1t=E1-E1n and the same with E2t (I guess you get the picture). I don't see how this would help. Moreover we are dealing with 3-D ( I guess I might have understood it if it was 2-D).
Thanks

I'm actually dealing with a similar problem right now. The way I understand it, is that the two boundary conditions are:

$$E_{1t} = E_{2t}$$
$$\vec a _{n2} \cdot (\vec D_1 - \vec D_2) = \rho_s$$

For the first one,
$E_{1t} = E_{2t}$ means the tangential components are the same. This tripped me up a bit... and I'm still a little uneasy doing these problems. Anyhow you have a plane that separates the media $z = 0$. So what can you say about the relationship between all the vectors of $\vec E_1(x,y,z=0)$ and $\vec E_2(x,y,z=0)$.
All of those vectors are tangential to the interface. So you can conclude that only the normal component (in this case it would be $E_{z2},$) is going to change when crossing the interface. Since you know that the tangential components are the same (from the relation above), you are left with:

$$\vec E_2 = \vec a_x E_{1x} + \vec a_y E_{1y} + \vec a_z \bar E_{2z}$$

where the bar was used for emphasis on$\bar E_{2z}$ (the z component of the vector).