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

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SUMMARY

The discussion centers on calculating the electric displacement field D2 across a dielectric boundary defined by the equation 4x + 3y = 10 m. The initial conditions specify D1 as 2ax - 4ay + 6.5az nC/m² and εr2 as 2.5. However, the correct relative permittivity for medium 2 should be εr2 = 0.4 to match the textbook answer D2 = 0.416ax - 1.888ay + 2.6az nC/m². The discrepancy arises from an error in the provided value of εr2, which was clarified through a YouTube video demonstrating a similar problem.

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  • Understanding of electric displacement fields (D-fields)
  • Familiarity with dielectric materials and their properties
  • Knowledge of vector calculus, specifically gradient operations
  • Basic principles of electromagnetism, including boundary conditions
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  • Study the impact of relative permittivity on electric fields
  • Learn about boundary conditions for electric fields at dielectric interfaces
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Homework Statement


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.


Homework Equations


an12 = ± grad(f)/|grad(f)|

D2n = D1n = an(D1 · an)

D1t = D1 - D1n

ε = ε0εr

D2t = (D1t)(ε2)/ε1

D2 = D2n + D2t


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 http://montoya.sdsmt.edu/ee381/examples/tilt_dielectric_boundary.pdf" ), 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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This question is 12+ years old at the time of answering, but maybe the following will be helpful to someone.

It appears that the question has a mistake. The given value of relative permittivity for medium 2 is wrong.

If we use ##\epsilon_{r2} = 0.4## instead of the stated value (##\epsilon_{r2} = 2.5##) then using the OP’s method gives the ‘official’ answer exactly.

I discovered this from a YouTube video where an (almost) identical problem is solved. The presenter explicitly changes the stated value of ##\epsilon_{r2}## early on but doesn't immediately say why. But near the end of the video, the presenter explains that there is a mistake with the textbook’s given value of ##\epsilon_{r2}## which is why she changed it to 0.4,.

 
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