329sum11hw3sol - ECE 329 Homework 3 — Solution Due: June...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

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

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 left-hand 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 ....
View Full Document

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.

Page1 / 5

329sum11hw3sol - ECE 329 Homework 3 — Solution Due: June...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online