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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 >...
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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.
 Spring '08
 Kim

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