(adsbygoogle = window.adsbygoogle || []).push({}); 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, whereD_{1}= 2a_{x}- 4a_{y}+ 6.5a_{z}nC/m^{2}. In the other region, ε_{r2}= 2.5. FindD_{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 asD_{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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# Find Flux Density On One Side of Dielectric Boundary Given Boundary Conditions

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