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Unformatted text preview: ECE329 Spring 2009 Homework 10 — Solution April 7, 2009 1. The electrostatic potential in a region of space is Φ( x,y,z ) = 4( y 2 + 4) z V . a) The corresponding electrostatic field is E =∇ Φ = 8 yz ˆ y + 4( y 2 + 4)ˆ z V m . b) If the permittivity in the medium is = 2 o , we can use Poisson’s equation to compute the volumetric charge density, ρ = ∇ 2 Φ = 16 o z C m 3 . c) If the permittivity in the medium is = o (1 + e z 2 ) , we have to apply Gauss’ law to compute the volumetric charge density, ρ = ∇ · ( E ) = 8 o z (1 ( y 2 + 3) e z 2 ) C m 3 . 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 Φ = Φ ( x/d ) 4 / 3 where Φ = 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 freecharge 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 ....
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This note was uploaded on 11/16/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.
 Spring '08
 FRANKE
 Electromagnet

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