Find Flux Density On One Side of Dielectric Boundary Given Boundary Conditions

  1. 1. The problem statement, all variables and given/known data
    A dielectric interface is defined as 4x + 3y = 10 m. The region including the origin is free space, where D1 = 2ax - 4ay + 6.5az nC/m2. In the other region, εr2 = 2.5. Find D2 given the previous conditions.


    2. Relevant equations
    an12 = ± grad(f)/|grad(f)|

    D2n = D1n = an(D1 · an)

    D1t = D1 - D1n

    ε = ε0εr

    D2t = (D1t)(ε2)/ε1

    D2 = D2n + D2t


    3. The attempt at a solution
    f = 4x + 3y - 10 = 0

    an12 = ± grad(4x + 3y - 10)/|grad(4x + 3y - 10)| = ± (4ax + 3ay)/5 = ± (.8ax + .6ay) Since the vector points in the positive x and y directions, I choose the plus sign to get: an12 = .8ax + .6ay

    D2n = D1n = (.8ax + .6ay)((2ax - 4ay + 6.5az) · (.8ax + .6ay)) nC/m2 = (.8ax + .6ay)(-.8) nC/m2 = -.64ax - .48ay nC/m2

    D1t = (2ax - 4ay + 6.5az nC/m2) - (-.64ax - .48ay nC/m2) = 2.64ax - 3.52ay + 6.5az nC/m2

    ε1 = ε0εr1 = ε0 = 8.854 pF/m (since the region is free space)

    ε2 = ε0εr2 = (8.854 pF/m)(2.5) = 22.135 pF/m

    D2t = (2.64ax - 3.52ay + 6.5az nC/m2)(22.135 pF/m)/(8.854 pF/m) = (2.64ax - 3.52ay + 6.5az nC/m2)(2.5) = 6.6ax - 8.8ay + 16.25az nC/m2

    D2 = (-.64ax - .48ay nC/m2) + (6.6ax - 8.8ay + 16.25az nC/m2) = 5.96ax - 9.28ay + 16.25az nC/m2

    The answer in the back of the book, however, is given as D2 = .416ax - 1.888ay + 2.6az nC/m2, which is completely different than what I got. I'm not sure where I went wrong. I followed one of my teacher's examples that he has posted (you can find it here), which is very similar to this question, but I still come up with the wrong answer. Can someone please show me where I went wrong? Thanks.
     
    Last edited: Nov 19, 2009
  2. jcsd
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