# sp08hw9sol1 - ECE329 Spring2008 Homework 9 Problem 1 E =...

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Unformatted text preview: ECE329 Spring2008 Homework 9 Problem 1. E = − V = −ax ˆ 4 /3 d x V0 dx d = −ax V0 ˆ 41 3 d 4 /3 ·D=ρ d 2 E|x=d/2 = −ax ˆ ρ = −V0 4 3 1 0 4 /3 d −1/3 d 1 /3 4 x = −V0 dx 9 ρ|x=d/4 = − 0 41 9 d 4 /3 d 4 4 1 1 /3 x 3 d 4 /3 0 x−2/3 d 4 /3 −2/3 At the anode, an · (D1 − D2 ) = ρs . Call the dielectric material 1, so an = −ax. ˆ ˆ ˆ ρs = −D1x = Start with Laplace's equation, and c4 are constants. 2 Problem 2. c1 , c2 , c3 , The potential must be continuous at x = d. 4 d 1 /3 04 = 3 d 4 /3 3d The forms of V and E must be as follows where the V = 0. c1 x + c2 c3 x + c4 V (x) = E= 0 0<x<d d < x < x0 −ax c1 ˆ −ax c3 ˆ 0<x<d d < x < x0 c1 d + c2 = c3 d + c4 Use initial conditions at x = 0 and x = x0 . Thus, c2 = 0 and c3 x0 + c4 = V0 . Use the boundary condition on the normal component of D at x = d. an · (D1 − D2 ) = ρs ˆ D1x = D2x 1 c1 Use algebra to solve for the coecients. = 2 c3 2 c1 = c3 1 2 c4 = − 1 c3 d 1 2 c3 x0 + c3 d − c3 d = V0 1 c3 = V0 x0 + 2 1 d−d V0 c1 = x0 + dV0 c4 = V (x) = x0 + 2 1 V0 x0 + V0 x0 + 2 1 d−d 2 1 d−d 2 1 2 1 d−d −1 2 1 2 1 d−d x dV0 x+ x0 + 1 0<x<d 2 −1 1 2 d−d 1 d < x < x0 ...
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