# 329sp09hw2sol - ECE-329 Spring 2009 Homework 2 — Solution...

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

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

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

Unformatted text preview: ECE-329 Spring 2009 Homework 2 — Solution January 29, 2009 1. Charge Q 1 = 8 π o C is placed at r 1 = (- 1 , , 0) and generates E 1 ( x,y,z ) = Q 1 4 π o r- r 1 | r- r 1 | 3 = 2 ( x + 1) ˆ x + y ˆ y + z ˆ z (( x + 1) 2 + y 2 + z 2 ) 3 / 2 V m . Charge Q 2 =- 4 π o C is placed at r 2 = (1 , , 0) and generates E 2 ( x,y,z ) = Q 2 4 π o r- r 2 | r- r 2 | 3 =- ( x- 1) ˆ x + y ˆ y + z ˆ z (( x- 1) 2 + y 2 + z 2 ) 3 / 2 V m . Then, the total electric field generated by these two charges is E ( x,y,z ) = E 1 ( x,y,z ) + E 2 ( x,y,z ) = 2 ( x + 1) ˆ x + y ˆ y + z ˆ z (( x + 1) 2 + y 2 + z 2 ) 3 / 2- ( x- 1) ˆ x + y ˆ y + z ˆ z (( x- 1) 2 + y 2 + z 2 ) 3 / 2 V m . Using this result, we obtain that the field at point r 3 = (0 , 1 , 0) is E ( r 3 ) = 3ˆ x + ˆ y 2 √ 2 V m , and the field at point r 4 = (0 , , 1) is E ( r 4 ) = 3ˆ x + ˆ z 2 √ 2 V m . (- 1 , , 0) (1 , , 0) ~ r 3 = (0 , 1 , 0) Q 1 Q 2 x y ~ E 1 ~ E 2 ~ E ( ~ r 3 ) z (- 1 , , 0) (1 , , 0) ~ r 4 = (0 , , 1) Q 1 Q 2 x z ~ E 1 ~ E 2 ~ E ( ~ r 4 ) y 2. The infinitesimal electric field at position ( x o , , 0) generated by an infinite charge strip along the z-axis with an infinitesimal width dx is found to be d E = ˆ x ρ S 2 π o x o dx, where ρ S is the uniform surface charge density. We can use this result to compute the electric field produced at the same location by a charge strip of finite width W that extends from x = 0 to x = W < x o (see the next figure). 1 ECE-329 Spring 2009 dx x z W d ~ E x o x o- x x Since the contribution of an infinitesimal strip that is placed at a distance x from the origin is simply d E = ˆ x ρ S 2 π o ( x o- x ) dx, the total field produced by the strip can be easily obtained by integrating d E from x = 0 to x = W as follows E = ˆ x W ˆ ρ S 2 π o ( x o- x ) dx = ˆ x ρ S 2 π o [- ln( x o- x )] W = ˆ x ρ S 2 π o (- ln( x o- W ) + ln( x o )) = ˆ x ρ S 2 π o ln x o x o- W ....
View Full Document

## This note was uploaded on 11/16/2009 for the course ECE 329 taught by Professor Franke during the Spring '08 term at University of Illinois at Urbana–Champaign.

### Page1 / 8

329sp09hw2sol - ECE-329 Spring 2009 Homework 2 — Solution...

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

View Full Document
Ask a homework question - tutors are online