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Lecture 06

# Lecture 06 - Chapter 15 Electric Field of Distributed...

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Chapter 15 Electric Field of Distributed Charges

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Distributed Charges ! E ( x , y , z ) = 1 4 !" 0 Q i r i 2 i = 1 N # ˆ r i ! E ( x , y , z ) = 1 4 0 # ( x ', y ', z ') ˆ rdx ' dy ' dz ' r 2 \$ ( x , y , z ) ( x ', y ', z ') ! ( x ', y ', z ')
Length: L Charge: Q What is the pattern of electric field around the rod? Cylindrical symmetry Uniformly Charged Thin Rod

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Apply superposition principle: Divide rod into small sections Δ y with charge Δ Q Assumptions: Rod is so thin that we can ignore its thickness. If Δ y is very small – Δ Q can be considered as a point charge Step 1: Divide Distribution into Pieces
What variables should remain in our answer? origin location, Q , x, y 0 What variables should not remain in our answer? rod segment location y, Δ Q y – integration variable Vector r from the source to the observation location: ! r = obs . loc ! source = x , y 0 , 0 ! 0, y , 0 ! r = x , y 0 ! y ( ) , 0 r r Q E ˆ 4 1 2 0 ! = ! "# ! Step 2: E due to one Piece

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! r = x , y 0 ! y ( ) ,0 Magnitude of r : ! r = x 2 + y 0 ! y ( ) 2 Unit vector r: ˆ r = ! r ! r = x , y 0 ! y ( ) x 2 + y 0 ! y ( ) 2 Magnitude of Δ E : 2 0 4 1 r Q E ! = ! "# ! E = 1 4 0 ! Q x 2 + y 0 \$ y ( ) 2 r r Q E ˆ 4 1 2 0 ! = ! ! Step 2: E due to one Piece
! ! E = ! E ( ) ˆ r ! ! E = 1 4 "# 0 ! Q x 2 + y 0 \$ y ( ) 2 x , y 0 \$ y ( ) ,0 x 2 + y 0 \$ y ( ) 2 ! E = 1 4 0 ! Q x 2 + y 0 \$ y ( ) 2 ˆ r = x , y 0 ! y ( ) x 2 + y 0 ! y ( ) 2 Vector Δ E : ! ! E = 1 4 0 ! Q x , y 0 \$ y ( ) x 2 + y 0 \$ y ( ) 2 % & ( 3 2 Step 2: E due to one Piece

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! ! E = 1 4 "# 0 ! Q x , y 0 \$ y ( ) ,0 x 2 + y 0 \$ y ( ) 2 % & ( 3 2 Δ Q in terms of integration variable y : ! Q = ! y L " # \$ % & Q !
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Lecture 06 - Chapter 15 Electric Field of Distributed...

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