329sum09hw4sol - ECE-329 Summer 2009 Homework 4 —...

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Unformatted text preview: ECE-329 Summer 2009 Homework 4 — Solution June 25, 2009 1. For E = 2ˆ x + z ˆ y V / m , we have that ∇ × E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 z = 1ˆ x. Then, since ∇ × E 6 = , field E is not electrostatic. 2. A vacuum diode consists of a cathode in the x = 0 plane and an anode in the x = d plane. The potential distribution between the plates is given by V ( x ) = V a ( x/d ) 4 / 3 where V a = 1 V . a) The electric field between the plates is E =-∇ Φ =- 4 3 d x d 1 / 3 ˆ x V m . Evaluating at x = d 2 , we get E ( x = d 2 ) =- 2 5 / 3 3 d ˆ x V m . b) The volumetric free-charge density is ρ = ∇ · D = ∇ · ( o E ) =- 4 o 9 d 2 d x 2 / 3 C m 3 . Evaluating at x = d 4 , we get ρ ( x = d 4 ) =- 4 5 / 3 o 9 d 2 C m 3 . c) The surface charge density on the anode (where ˆ n =- ˆ x ) is ρ S = D · ˆ n | x = d =- D x | x = d = 4 o 3 d C m 2 . 3. Consider a pair of parallel conducting plates placed at z = 0 and z = W that sustains an electric field E = 3ˆ z V / m . The gap between the plates is originally occupied by vacuum ( o , μ o ). a) In free space D = o E , then the displacement field between the plates is D = 3 o ˆ z C m 2 . By definition, D = o E + P , thus the polarization field between the plates is P = C m 2 ....
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This note was uploaded on 09/16/2011 for the course ECE 329 taught by Professor Kim during the Fall '08 term at University of Illinois, Urbana Champaign.

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329sum09hw4sol - ECE-329 Summer 2009 Homework 4 —...

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