329sp09hw10sol - ECE-329 Spring 2009 Homework 10 Solution...

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ECE-329 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 0 ( x/d ) 4 / 3 where Φ 0 = 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. Let us consider E = 2ˆ x + z ˆ y + y ˆ z V / m . 1
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ECE-329 Spring 2009 a) Since ∇ × E = ˆ x ˆ y ˆ z ∂x ∂y ∂z 2 z y = 0 , the field E is electrostatic.
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