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Unformatted text preview: ECE 329 Fall 2010 Homework 4  Solution Due: Sep. 21, 2010 1. For E = 2 x ˆ x + 2 sin ( z ) ˆ y 2 z ˆ z , we have ∇ × E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 x 2 sin ( z ) 2 z = 2 cos ( z ) ˆ x. Since ∇ × E 6 = 0 , eld E is not electrostatic. 2. We have 4 regions r < a a < r < b b < r < c r > c , within each one of which we can use Possion's equation to nd the electrostatic potential (notice that within each region, the media, electrostatic source, electrostatic potential and electrostatic eld are continuous). Let us start from the general form of Possion's equation ∇ 2 V = ρ o . Due to the cylindrical symmetry of charge distribution, we know the electrostatic potential V is a function of r and constant with respect to φ and z coordinates. As a result, the general Possion's equation can be reduced to the following simpli ed form when expanded under cylindrical coordinate system 1 r ∂ ∂r r ∂V ∂r = ρ o . Then, we can nd the general solution to the above secondorder di erential equation V = ρ 4 o r 2 + A (ln r ) + B, where A and B are unknown constants to determine. For the convenience, we use subscript i = 1 , 2 , 3 , 4 to denote each region. Therefore, we have V i = ρ i 4 o r 2 + A i (ln r ) + B i , E i =∇ V i = ˆ r ∂V i ∂r = ˆ r ρ i 2 o r A i r . Further, we need to determine the volumn charge density (the numerator of the righthand side in Possion's equation) for each region as follows ρ 1 = 0 , r < a ρ...
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This note was uploaded on 11/06/2011 for the course ECE 329 taught by Professor Kim during the Spring '08 term at University of Illinois, Urbana Champaign.
 Spring '08
 Kim

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