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Unformatted text preview: ECE 329 Spring 2010 Homework 4  Solution Due: Feb. 16, 2010 1. For E = 2 x ˆ x + 2 sin ( z ) ˆ y 2 z ˆ z , we have ∇ × E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 x 2 sin ( z ) 2 z = 2 cos ( z ) ˆ x. Since ∇ × E 6 = 0 , eld E is not electrostatic. 2. a) The bottom plate is holding the positive charge density, because the voltage decreases as z increases, which means that the electric eld is pointing upward. b) E 1 =∇ V 1 = ˆ z ∂ ∂z ( 2 z ) = 2ˆ z ( V/m ) , E 2 =∇ V 2 = ˆ z ∂ ∂z ( 1 z ) = ˆ z ( V/m ) . c) Due to the boundary condition ˆ n · ( D 2 D 1 ) = ρ s , the displacement eld between the two plates is a constant: D = 4 ˆ z ( C/m 2 ) . Therefore, E 1 = D 1 = 4 1 ˆ z ( V/m ) , E 2 = D 2 = 4 2 ˆ z ( V/m ) . By comparing with the result of (b), we have 4 / 1 = 2 and 4 / 2 = 1 , or 1 = 2 and 2 = 4 . d) Since the voltage is continuous at z = d , we have 2 d = 1 d. Thus, d = 1 ( m ) . 3. Consider a pair of parallel conducting plates placed at z = 0 and z = W that sustains an electric eld E = 4ˆ z ( V/m ) . The gap between the plates is originally occupied by vacuum....
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This note was uploaded on 04/20/2010 for the course ECE ECE 329 taught by Professor Goddart during the Spring '10 term at University of Illinois at Urbana–Champaign.
 Spring '10
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