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Jan12Note - PHYS142 Lecture 3 Electric Fields Chapter 23...

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Electric Fields Chapter 23 All sections 23.1 Properties of Electric Charges 23.2 Charging Objects by Induction 23.3 Coulomb’s Law 23.4 The Electric Field 23.5 Electric Field of a Continuous Charge Distribution 23.6 Electric Field Lines 23.7 Motion of a Charged Particle in a Uniform Electric Field PHYS142 Lecture 3 2
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Electric Field of a Continuous Charge Distribution (23.5) Divide the charge distribution into small charge elements, Δ q . Determine the E-field at the point of interest, Δ E , due to Δ q : r r q k E e ˆ 2 Sum/integrate over the entire charge distribution to obtain the E-field: r r dq k r r q k E e i i i i q e i ˆ ˆ lim 2 2 0 3
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For a symmetric surface charge, select a ring of charge. If the width of the ring is dr , then: dr r dA dq ) 2 ( 4 For a spherical volume, select a spherical shell of charge. If the width of the shell is dr , then: ) 4 ( 2 dr r dV dq Example of volume and surface integration
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Example 23.6 (Uniformly charged rod) Calculate the E-field of a uniformly charged rod of length and total charge Q at a point P along the axis of the rod and a distance a from one end. dE at point P due to dq is: 2 2 x dx k x dq k dE e e The direction of dE at P , points away from the charge element dq . From
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Jan12Note - PHYS142 Lecture 3 Electric Fields Chapter 23...

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