329fall10hw2sol - ECE 329 Fall 2010 Homework 2 - Solution...

Info iconThis preview shows pages 1–2. 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 Fall 2010 Homework 2 - Solution Due: Sep. 7, 2010 1. a) According to Gauss' Law, ˛ S E · d S = 1 ˆ V ρdV = 1 ˆ V (- 3) dV =- 3 L 3 ( V · m ) . b) ˛ S E · d S = 1 ˆ V ρdV = 1 ˆ 1 / 2- 1 / 2 ˆ 1 / 2- 1 / 2 ˆ 1 / 2- 1 / 2 ( x 2 + y 2 + z 2 ) dxdydz By symmetry, ˛ S E · d S = 3 ˆ 1 / 2- 1 / 2 ˆ 1 / 2- 1 / 2 ˆ 1 / 2- 1 / 2 x 2 dxdydz = 3 ˆ 1 / 2- 1 / 2 x 2 dx = 1 4 ( V · m ) . c) Since the charge distribution has the same x, y and z dependencies, we know that the electric ux through the 6 surfaces should be equal to each other, namely ˆ S i E · d S = 1 6 ˛ S E · d S = 1 24 ( V · m ) , i = 1 , 2 , ··· , 6 . 2. a) According to Example 4 in Lecture 3 online, the electric elds induced by these 2 slabs are E 1 = - ˆ z ρ W 2 for z <- 2 W ˆ z ρ ( z + 3 2 W ) for- 2 W < z <- W ˆ z ρ W 2 for z >- W , E 2 = ˆ z ρ W 2 for z < W- ˆ z ρ ( z- 3 2 W ) 2 for W < z < 2 W- ˆ z ρ W 2 for z > 2 W , Therefore, the total eld is E = E 1 + E 2 = for z <- 2 W ˆ z ρ ( z +2 W ) for- 2 W < z <- W ˆ z ρ W for- W < z < W ˆ z ρ (- z +2 W ) for W < z < 2 W for z >...
View Full Document

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.

Page1 / 5

329fall10hw2sol - ECE 329 Fall 2010 Homework 2 - Solution...

This preview shows document pages 1 - 2. Sign up to view the full document.

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