HW4Solution

HW4Solution - 2. 'cos ( ') ' 4 jkr jkr A e J r e dV r !...

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Unformatted text preview: 2. 'cos ( ') ' 4 jkr jkr A e J r e dV r ! " # $ % &&& ! ! ( ') sin ' ( ') ( ') 2 z m L J r a I k z x y ( ) * + % ,- . / 1 2 3 4 ! ' ' ' ' dV dx dy dz % 2 2 2 ' ' ' ' ' ' 'cos xx yy zz xx yy zz r r x y z ! , , , , % % , , ' ' ' 2 2 sin ' ( ') ( ') ' ' ' 4 2 L xx yy zz jkr jk r z m L e L A a I k z x y e dx dy dz r " # 5 5 , , $ $5 $5 $ ( ) * + % ,- . / 1 2 3 4 & & & ! 2 ' 2 sin ' ' 4 2 L z jkr jk z r z m L e L A a I k z e dz r " # $ $ ( ) * + % ,- . / 1 2 3 4 & ! cos z r 6 % ' ' 2 2 2 cos ' 2 ' 4 2 L L L jk z jk z jkr jk z z m L e e e A a I e dz r j 6 " # * + * + , $ ,- .- . $ 1 2 1 2 $ ( ) $ / % / / 3 4 & ! 7 8 7 8 2 1 cos ' 1 cos ' 2 2 2 1 ' 4 2 L L L jkr jk jk jk z jk z z m L e A a I e e e e dz r j 6 6 " # $ $ , $ $ $ ( ) % $ / 3 4 & ! 7 8 7 8 7 8 7 8 2 2 1 cos ' 1 cos ' 2 2 2 2 1 4 2 1 cos 1 cos L L L L jk jk jk z jk z jkr z m L L e e e e e A a I r j jk jk 6 6 " # 6 6 $ , $ $ $ $ $ 9 : ( ) ( ) ; ; ; ; / / % $ < = / / , $ $ ; ; / / 3 4 3 4 ; ; > ? ! 7 8 7 8 7 8 7 8 7 8 7 8 1 cos 1 cos 1 cos 1 cos 2 2 2 2 2 2 2 2 1 4 2 1 cos 1 cos L L L L L L L L jk jk jk jk jk jk jk jk jkr z m e e e e e e e e e A a I r j jk jk 6 6 6 6 " # 6 6 , $ , $ $ $ $ $ $ 9 : ( ) ( ) $ $ ; ; / / % $ < = / / , $ $ ; ; / / 3 4 3 4 > ? ! 7 8 7 8 cos cos cos cos 2 2 2 2 1 4 2 1 cos 1 cos L L L L jk jk jk jk jkr jkL jkL z m e e e e e e e A a I r j jk jk 6 6 6 6 " # 6 6 $ $ $ $ 9 : ( ) ( ) $ $ ; ; / / % $ < = / / , $ $ ; ; / / 3 4 3 4 > ? ! 2 k # @ % 2 L n @ % kL n # % 2 2 L k n # % If n is odd 1 jkL n e e # A A % % $ If n is even 1 jkL n e e # A A % % For n odd 7 8 7 8 2cos cos 2cos cos 2 2 8 1 cos 1 cos jkr z m n n e A a I r # # 6 6 " # 6 6 $ ( ) * + * +- .- . / 1 2 1 2 / % , , $ / / 3 4 ! For n even 7 8 7 8 2sin cos 2sin cos 2 2 8 1 cos 1 cos jkr z m n n e A ja I r # # 6 6 " # 6 6 $ ( ) * + * +- .- . / 1 2 1 2 / % $ , , $ / / 3 4 ! 3. ( x , y ) B (C/m) D , " E = 5 F = 0 d x y F = V ( x , y ) B (C/m) D , " E = 5 F = 0 d x y F = V To solve this problem we can use the principle of superposition and express the original problem as the superposition of the following two problems: D , " E = 5 d x y 1 ( , ) x y G 1 ( , ) x y G % 1 ( , ) x y V G % D , " E = 5 d x y 1 ( , ) x y G 1 ( , ) x y G % 1 ( , ) x y V G % ( x , y ) B (C/m) D , " E = 5 d x y 2 ( , ) x y G % 2 ( , ) x y G % 2 ( , ) x y G ( x , y ) B (C/m) D , " E = 5 d x y 2 ( , ) x y G % 2 ( , ) x y G % 2 ( , ) x y G + Therefore the overall solution is 1 2 ( , ) ( , ) ( , )...
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This note was uploaded on 12/02/2009 for the course ECE 000 taught by Professor O during the Spring '09 term at UCSB.

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HW4Solution - 2. 'cos ( ') ' 4 jkr jkr A e J r e dV r !...

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