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Lecture07-2011

# Lecture07-2011 - Fall 2010 Prof Yong Chen...

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Unformatted text preview: Fall 2010 Prof. Yong Chen ( [email protected] ) Prof. Michael Manfra ( [email protected] ) Lecture 6 Slide 1 PHYS 272: Matter and Interactions II -- Electric And Magnetic Interactions http://www.physics.purdue.edu/academic_programs/courses/phys272/ PHYSICS 272 Electric & Magnetic Interactions Lecture 7 E-field of distributed charges: rod, ring, disk, sphere [EMI 16.4-16.7] Exam next Monday night 9/26 8pm rooms 112 and 114 Will cover Chapters 14, 15, 16 (through today’s lecture) Fall 2010 Prof. Yong Chen ( [email protected] ) Prof. Michael Manfra ( [email protected] ) Lecture 6 Slide 2 PHYS 272: Matter and Interactions II -- Electric And Magnetic Interactions http://www.physics.purdue.edu/academic_programs/courses/phys272/ Simplified problem: find electric field at the location < x,0,0 > ∑ ∆ = i x i x E E , ∆ E i , x = 1 4 πε Q L x x 2 + y- y i ( 29 2 3 2 ∆ y E x = 1 4 πε Q L x x 2 + y i 2 3 2 ∆ y i ∑ = y E E x = 1 4 πε Q L x 1 x 2 + y i 2 3 2 ∆ y i ∑ Step 3: Add up Contribution of all Pieces ∑ ∆ = + + + = i y i y y y y E E E E E , , 3 , 2 , 1 ... Fall 2010 Prof. Yong Chen ( [email protected] ) Prof. Michael Manfra ( [email protected] ) Lecture 6 Slide 3 PHYS 272: Matter and Interactions II -- Electric And Magnetic Interactions http://www.physics.purdue.edu/academic_programs/courses/phys272/ E x = 1 4 πε Q L x 1 x 2 + y i 2 3 2 ∆ y i ∑ Integration: taking an infinite number of slices E x = lim ∆ y → 1 4 πε Q L x 1 x 2 + y i 2 3 2 ∆ y i ∑ ← definite integral E x = 1 4 πε Q L x 1 x 2 + y 2 3 2 dy- L 2 + L 2 ∫ 1 ( a 2 + x 2 ) 3/2 ∫ dx = x a 2 a 2 + x 2 Step 3: Add up Contribution of all Pieces Fall 2010 Prof. Yong Chen ( [email protected] ) Prof. Michael Manfra ( [email protected] ) Lecture 6 Slide 4 PHYS 272: Matter and Interactions II -- Electric And Magnetic Interactions http://www.physics.purdue.edu/academic_programs/courses/phys272/ E x = 1 4 πε Q L x 1 x 2 + y 2 3 2 dy- L 2 + L 2 ∫ Evaluating integral: E x = 1 4 πε Q L x y x 2 x 2 + y 2 - L 2 + L 2 E x = 1 4 πε Q x x 2 + L / 2 ( 29 2 E y = = 1 4 πε Q r r 2 + L / 2 ( 29 2 Cylindrical symmetry: replace x r Step 3: Add up Contribution of all Pieces Fall 2010 Prof. Yong Chen ( [email protected] ) Prof. Michael Manfra ( [email protected] ) Lecture 6 Slide 5 PHYS 272: Matter and Interactions II -- Electric And Magnetic Interactions http://www.physics.purdue.edu/academic_programs/courses/phys272/ In vector form: r E y = = 1 4 πε Q r r 2...
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Lecture07-2011 - Fall 2010 Prof Yong Chen...

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