D and E at boundary (dielectrics)

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SUMMARY

The discussion focuses on solving problem 4.7.3 from "Electromagnetics" by Kraus, 4th edition, which involves two dielectrics with relative permittivities εr = 2 and εr = 5. The electric field E for negative x is given as (3,0,2) V/m. The correct calculation for the displacement field D2 in the positive x region is determined to be D2 = ε0(6,0,10) after correcting the assignment of relative permittivity. The use of Snell's Law-like equations was deemed unnecessary for this problem.

PREREQUISITES
  • Understanding of electromagnetic fields and boundary conditions
  • Familiarity with dielectric materials and their properties
  • Knowledge of vector mathematics, particularly in three dimensions
  • Proficiency in using the Pythagorean theorem in vector calculations
NEXT STEPS
  • Study the derivation and application of boundary conditions for dielectrics
  • Learn about the relationship between electric field E and displacement field D in dielectrics
  • Explore the concept of relative permittivity and its implications in electromagnetic theory
  • Investigate the use of Snell's Law in different contexts beyond optics
USEFUL FOR

This discussion is beneficial for students and professionals in electrical engineering, particularly those focusing on electromagnetics, dielectric materials, and field theory applications.

mishima
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Homework Statement


Electromagnetics, Kraus, 4th edition problem 4.7.3
The y-z plane is the boundary between 2 dielectrics of relative permittivities εr = 2 and εr = 5. For negative values of x, E = (3,0,2) V/m. Find D (magnitude and direction) for positive values of x.

Homework Equations


The one that is similar to Snell's Law, not sure of name:
i) ε2 tan (incident angle to normal) = ε1 tan (refracted angle to normal)

which is derived from boundary conditions for 2 dielectrics with the absence of surface charges

ii)D1(normal) = D2(normal)
iii)E1(tangent) = E2(tangent)


The Attempt at a Solution


The x-axis is parallel to the normal of the boundary. So, I first found the angle the electric field vector makes with the x-axis to be 0.588 radians.

incident angle to normal (x-axis) = inverse tan (2/3)

This value was substituted into equation i) along with 5 and 2 for ε1 and ε2 respectively. Solving for the refracted angle, I obtained 1.03037 radians. I believe all of this is correct.

What I am having trouble with is now getting the magnitude of D2. If I just use equations ii) and iii) with pythagorean theorem I get an answer different than the book, which says: D2 = ε0(6,0,10).
 
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I figured it out, I had originally put the relative permittivity of 5 on the negative x side, but making it positive gives the correct result. Also, the "snell's law"-like equation isn't necessary at all.
 

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