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lec03_PyrakNolte

lec03_PyrakNolte - Point Charge in Uniform E field Lectures...

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1/16/12 1 Point Charge in Uniform E field F = m a = mq E a = q m E v x = v 0 x = v 0 t - - - - - - - - - - - - - - - - - - + + ++ + + + + + + ++ + q E v 0 x y = 1 2 a y t 2 = 1 2 q m E x 2 v o 2 1/16/12 2 R 2 R a a ELECTRIC FIELD ELECTRIC FLUX Lectures 3, 4 & 5 Last Week…..Discrete Charges 1/16/12 3 Magnitude of the Electric Field due to a Point Charge: E = 1 4 πε 0 q r 2 Electric Field due to a Dipole at point P: + - P Dipole center z r + r - E + E q + q - d E = qd 2 πε o z 3 = p 2 πε o z 3 1/16/12 4 Today… More on Electric Field: Continuous Charge Distributions Text Reference: Chapter 22.1 May useful examples 22-1-8 etc. 1
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1/16/12 2 1/16/12 5 1. Extension of point charge approach: E = kq r 2 ˆ r d E = k dq r 2 ˆ r and then integrate. 2. Use Gauss' law: E n dA = 1 ε 0 S Q inside Calculate E -field due to charge distributions 1/16/12 6 Principles (Coulomb’s Law + Law of Superposition) remain the same. Electric Fields from Continuous Charge Distributions + + + + + + - - - - - + + + + + + + + + + Only change: 1/16/12 7 Charge Distributions Problems Step 1:Understand the geometry
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