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Unformatted text preview: ECE 329 Homework 3 — Solution Due: June 24, 2011, 5PM 1. Given a vector field A = ( x y )ˆ x +( x + y )ˆ y , let us verify if A satisfies the following vector identity ∇× ∇× A = ∇ ∇· A∇ 2 A. Solving the lefthand side gives us ∇× ∇× A = ∇× ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z x y x + y = ∇× (2ˆ z ) = , and solving the right hand side give us ∇ ∇· A∇ 2 A = ∇ ( ∇· (( x y )ˆ x + ( x + y )ˆ y ))∇ 2 (( x y )ˆ x + ( x + y )ˆ y ) = ∇ (2) 0 = , Since both sides are the same, the identity is verified. 2. Given E = 2ˆ x + z ˆ y V / m , we have that ∇× E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 z = 1ˆ x. Then, since ∇× E = , field E is not electrostatic. 3. Given E = 2ˆ x + 2 y ˆ y + 3ˆ z V / m, which is an electrostatic field since ∇× E = ˆ x ˆ y ˆ z ∂ ∂x ∂ ∂y ∂ ∂z 2 2 y 3 = . z y x P = ( x, y, z ) (0 , , 0) 1 Using the path of integration shown in the above figure, we can calculate the electrostatic potential V ( x,y,z ) at any point in space P = ( x,y,z ) as follows V ( P ) V ( 0) = P ˆ E · d l = x ˆ E x ( x, , 0) dx y ˆ E y ( x,y, 0) dy z ˆ E z ( x,y,z ) dz = (2 x + y 2 + 3 z ) V . If V ( 0) = 0 V , the potential at P = (1 , 2 , 3) is V (1 , 2 , 3) = (2 + 2 2 + 3 2 ) = 15 V ....
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This note was uploaded on 11/06/2011 for the course ECE 329 taught by Professor Kim during the Summer '08 term at University of Illinois, Urbana Champaign.
 Summer '08
 Kim

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