Find Flux Density On One Side of Dielectric Boundary Given Boundary Conditionsby Bizkit Tags: boundary, conditions, density, dielectric, flux 

#1
Nov1909, 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 D_{1} = 2a_{x}  4a_{y} + 6.5a_{z} nC/m^{2}. In the other region, ε_{r2} = 2.5. Find D_{2} given the previous conditions. 2. Relevant equations a_{n12} = ± grad(f)/grad(f) D_{2n} = D_{1n} = a_{n}(D_{1} · a_{n}) D_{1t} = D_{1}  D_{1n} ε = ε_{0}ε_{r} D_{2t} = (D_{1t})(ε_{2})/ε_{1} D_{2} = D_{2n} + D_{2t} 3. The attempt at a solution f = 4x + 3y  10 = 0 a_{n12} = ± grad(4x + 3y  10)/grad(4x + 3y  10) = ± (4a_{x} + 3a_{y})/5 = ± (.8a_{x} + .6a_{y}) Since the vector points in the positive x and y directions, I choose the plus sign to get: a_{n12} = .8a_{x} + .6a_{y} D_{2n} = D_{1n} = (.8a_{x} + .6a_{y})((2a_{x}  4a_{y} + 6.5a_{z}) · (.8a_{x} + .6a_{y})) nC/m^{2} = (.8a_{x} + .6a_{y})(.8) nC/m^{2} = .64a_{x}  .48a_{y} nC/m^{2} D_{1t} = (2a_{x}  4a_{y} + 6.5a_{z} nC/m^{2})  (.64a_{x}  .48a_{y} nC/m^{2}) = 2.64a_{x}  3.52a_{y} + 6.5a_{z} nC/m^{2} ε_{1} = ε_{0}ε_{r1} = ε_{0} = 8.854 pF/m (since the region is free space) ε_{2} = ε_{0}ε_{r2} = (8.854 pF/m)(2.5) = 22.135 pF/m D_{2t} = (2.64a_{x}  3.52a_{y} + 6.5a_{z} nC/m^{2})(22.135 pF/m)/(8.854 pF/m) = (2.64a_{x}  3.52a_{y} + 6.5a_{z} nC/m^{2})(2.5) = 6.6a_{x}  8.8a_{y} + 16.25a_{z} nC/m^{2} D_{2} = (.64a_{x}  .48a_{y} nC/m^{2}) + (6.6a_{x}  8.8a_{y} + 16.25a_{z} nC/m^{2}) = 5.96a_{x}  9.28a_{y} + 16.25a_{z} nC/m^{2} The answer in the back of the book, however, is given as D_{2} = .416a_{x}  1.888a_{y} + 2.6a_{z} nC/m^{2}, 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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