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329sum11hw3sol

329sum11hw3sol - ECE 329 Homework 3 Solution Due 5PM 1...

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ECE 329 Homework 3 — Solution Due: June 24, 2011, 5PM 1. Given a vector field A = ( x - y x +( x + y y , let us verify if A satisfies the following vector identity ∇ × ∇ × A = · A - ∇ 2 A. Solving the left-hand side gives us ∇ × ∇ × A = ∇ × ˆ x ˆ y ˆ z x y z x - y x + y 0 = ∇ × (2ˆ z ) = 0 , and solving the right hand side give us · A - ∇ 2 A = ( · (( x - y x + ( x + y y )) - ∇ 2 (( x - y x + ( x + y y ) = (2) - 0 = 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 0 = 1ˆ x. Then, since ∇ × E = 0 , 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 = 0 . z y x P = ( x, y, z ) (0 , 0 , 0) 1

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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 0 E · dl = - x ˆ 0 E x ( x, 0 , 0) dx - y ˆ 0 E y ( x, y, 0) dy - z ˆ 0 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 . 4. Let us compute the circulation ¸ C E · dl over the triangular path C shown below. x y ( - 1 , - 1 , 0) z (1 , - 1 , 0) (1 , 1 , 0) l 1 l 3 l 2 Given E = y ˆ x + x ˆ y V / m , we find that ˛ C E · dl = ˆ l 1 E (1 , y, 0) · dl 1 + ˆ l 2 E ( x, x, 0) · dl 2 + ˆ l 3 E ( x, - 1 , 0) · dl 3 = 1 ˆ - 1 ( y ˆ x
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