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Finding the Electric Field

  1. Sep 30, 2006 #1
    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?
  2. jcsd
  3. Sep 30, 2006 #2


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    What are the boundary conditions on the electric field at the interface between the two dielectric regions?
  4. Oct 1, 2006 #3
    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).

    I'm editing the post: Thanks for your help, I understand it now.
    Last edited: Oct 1, 2006
  5. Oct 1, 2006 #4
    I'm actually dealing with a similar problem right now. The way I understand it, is that the two boundary conditions are:

    [tex] E_{1t} = E_{2t} [/tex]
    [tex] \vec a _{n2} \cdot (\vec D_1 - \vec D_2) = \rho_s [/tex]

    For the first one,
    [itex] E_{1t} = E_{2t} [/itex] 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 [itex] z = 0 [/itex]. So what can you say about the relationship between all the vectors of [itex] \vec E_1(x,y,z=0) [/itex] and [itex] \vec E_2(x,y,z=0) [/itex].
    All of those vectors are tangential to the interface. So you can conclude that only the normal component (in this case it would be [itex] E_{z2}, [/itex]) 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:

    [tex] \vec E_2 = \vec a_x E_{1x} + \vec a_y E_{1y} + \vec a_z \bar E_{2z} [/tex]

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