sp08hw9sol1

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 coecients. = 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 ...
View Full Document

Ask a homework question - tutors are online