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

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Bizkit
#1
Nov19-09, 09:50 PM
P: 23
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.
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