# 329sp10hw2sol - ECE 329 Spring 2010 Homework 2 Solution Due...

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Unformatted text preview: ECE 329 Spring 2010 Homework 2 - Solution Due: Feb. 2, 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. According to Example 5 in Lecture 3 online, this problem can be solved by using symmetry. First, we consider Q 1 . By Gauss' Law, ˛ S D 1 · d S = Q 1 holds for any S enclosing Q 1 . Due to symmetry, half of the ux should go to the- ˆ x direction across the ˆ y- ˆ z plane (while the other half should go to the +ˆ x direction across the x = 2 plane),...
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## This note was uploaded on 04/20/2010 for the course ECE ECE 329 taught by Professor Goddart during the Spring '10 term at University of Illinois at Urbana–Champaign.

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329sp10hw2sol - ECE 329 Spring 2010 Homework 2 Solution Due...

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