Lecture_05

# Lecture_05 - 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 πε 0 Q i r i 2 i = 1 N ˆ r i E ( x , y , z ) = 1 4 0 ρ ( 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 Δ = Δ πε 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 Δ = Δ πε Δ E = 1 4 0 Δ Q x 2 + y 0 y ( ) 2 r r Q E ˆ 4 1 2 0 Δ = Δ Step 2: E due to one Piece
Δ E = Δ E ( ) ˆ r Δ E = 1 4 πε 0 Δ Q x 2 + y 0 y ( ) 2 x , y 0 y ( ) ,0 x 2 + y 0 y ( ) 2 Δ E = 1 4 0 Δ Q x 2 + y 0 y ( ) 2 ˆ r = x , y 0 y ( ) x 2 + y 0 y ( ) 2 Vector Δ E : Δ E = 1 4 0 Δ Q x , y 0 y ( ) x 2 + y 0 y ( ) 2 3 2 Step 2: E due to one Piece

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Δ E = 1 4 πε 0 Δ Q x , y 0 y ( ) ,0 x 2 + y 0 y ( ) 2 3 2 Δ Q in terms of integration variable y : Δ Q = Δ y L Q Δ E = 1 4 0 Q x , y 0 y ( ) L x 2 + y 0 y ( ) 2 3 2 Δ y Step 2: E due to one Piece
Δ E = 1 4 πε 0 Q x , y 0 y ( ) ,0 L x 2 + y 0 y ( ) 2 3 2 Δ y Components of Δ E : Δ E x = 1 4 0 Q L x x

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## This note was uploaded on 10/26/2009 for the course CHM 111 taught by Professor Staff during the Spring '08 term at Purdue.

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Lecture_05 - Chapter 15 Electric Field of Distributed...

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