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Unformatted text preview: ECE329 Homework 8 Spring2008 Problem 1. The problem was a bit unclear. The solution assumes that the x < range is called material 1, has = 3 , and is on the left side. The boundary condition ˆ a n · ( D 1 D 2 ) = ρ s can be used to nd E 2 x at x = 0 . D 1 x = D 2 x 3 E = 5 E 2 x E 2 x E 1 x = 3 5 From the boundary condition ˆ a n × ( E 1 E 2 ) = 0 and at x = 0 , E 2 z must be zero. Also from this boundary condition, E 1 y = E 2 y . E 2 y E 1 y = 1 At x = 0 , E 2 = E ( 3 5 ˆ a x + 3ˆ a y ) and D 2 = E 0 0 (3ˆ a x + 15ˆ a y ) .  E 2   E 1  = q ( 3 5 ) 2 + 3 2 √ 1 2 + 3 2 = 0 . 97  D 2   D 1  = √ 3 2 + 15 2 √ 3 2 + 9 2 = 1 . 61 Problem 2. First nd ˜ H in both regions using Faraday's law with phasors. ∇ × ˜ E = jωμ ˜ H ˜ H 1 = ˆ a y E β 1 ωμ 1 e jβ 1 z ˜ H 2 = ˆ a y E β 2 ωμ 2 e jβ 2 z Use the boundary condition ˆ a n × ( H 1 H 2 ) = J s at z = 0 . For this case with perfect dielectrics, ˜ H 1 y = ˜ H 2 y ....
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 Spring '08
 FRANKE
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