Chapter%202 - Chapter 2 Coulomb's Law and Electric Field...

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Dr. Ray Chen© Chapter 2 Coulomb’s Law and Electric Field Intensity
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Dr. Ray Chen© 2.1 The experimental law of Coulomb R Q 1 Q 2 2 2 1 R Q Q k F = Q 1 , Q 2: positive or negative quantities of charge R: separation , k: proportionality constant Unit in MKS: Q: Coulomb ; R: meter ; F: Newton
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Dr. Ray Chen© 2.1 The experimental law of Coulomb m N / F m N / . k = × = 2 2 2 0 9 12 0 0 C label the has C label the has F/m 10 36 1 10 854 8 4 1 ε π ε πε 2 0 2 1 4 is Law s Coulomb’ R Q Q F = πε
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Dr. Ray Chen© 2.1 The experimental law of Coulomb Vector form of the law: 12 2 12 0 2 1 2 4 a ˆ R Q Q F = πε v sign. same the have Q2 and Q1 where case the for shown is and Q2 on force the is F2 Vector form of C.L. is : Fig. 2.1 12 12 R of direction the in vector unit a is a ˆ 1 2 1 2 12 12 12 r r r r R R ˆ a ˆ v v v v = =
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Dr. Ray Chen© 2.1 The experimental law of Coulomb Example 1: 1 2 4 2 4 1 Q by Q on exerted force the Find 5) 0, (2, of coordinate Cartesian the at located C 10 3) 2, (1, of coordinate Cartesian the at located C 10 3 = × = Q Q
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Dr. Ray Chen© 2.1 The experimental law of Coulomb Solution to Example 1: ( ) 3 2 2 1 3 a 2 a 2 a a a 2 a 2 a a 3) - (5 a 2) - (0 a 1) - (2 2 2 2 z y x 12 z y x z y x 1 2 12 = + + + = + = + + = = ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ ˆ r r R v v v ( ) ( ) z y x 2 z y x z y x 9 4 4 12 2 12 0 2 1 2 a 20 a 20 a 10 3 a 2 a 2 a 30 3 a 2 a 2 a 9 10 36 1 4 10 10 3 4 ˆ ˆ ˆ F N ˆ ˆ ˆ ˆ ˆ ˆ / a ˆ R Q Q F + = + = + × × = = v v π π πε 12 2 12 0 2 1 21 2 12 0 2 1 2 1 4 4 a ˆ R Q Q a ˆ R Q Q F F = = = πε πε v v
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Dr. Ray Chen© 2.1 The experimental law of Coulomb x y z P 1 (1,2,3) P 2 (1,2,10) Example 2: 10) 2, (1, P at located C 5 3) 2, (1, P at located C 10 2 2 1 1 µ µ = = Q Q force no s experience Q charge point a which at P Find b) Q by Q on exerted force vector Find a) 3 3 1 2
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Dr. Ray Chen© 2.1 The experimental law of Coulomb ( ) 10 17 9 7 7 0, , 0 7 4 10 10 10 5 4 a) 3 2 0 6 6 12 2 12 0 2 1 2 N a ˆ . a ˆ R Q Q F z × = × × × × = = πε πε v ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) ( ) why?) rded, 7.10(disca or 9 26 0 191 34 10 10 10 5 3 3 10 10 10 10 5 3 3 10 4 0 z 2, 1, P3 b) 2 3 6 3 5 3 6 3 5 0 3 . z z z z z z z a ˆ z z z z Q z = = + × = × = = πε x y z P 1 (1,2,3) P 2 (1,2,10) Solution to Example 2:
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Dr. Ray Chen© 2.2 Electrical Field Intensity Definition: Force per unit charge (a vector force) t t t t a ˆ R Q Q F 1 2 1 0 1 4 = πε v t t t t a ˆ R Q Q F 1 2 1 0 1 4 = πε v Meter Volt meter Coulomb meter Newton Coulomb Newton Charge Unit Force : Unit = = = m N Joule Coulomb Joule Volt = =
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Dr. Ray Chen© 2.2 Electrical Field Intensity t t Q F E v v = t t a R Q E 1 2 1 0 1 ˆ ˆ 4 = ε π v Electric Field Intensity: R a R πε Q E ˆ 4 2 0 1 = v General Case: - For an arbitrary charge particle Q at the origin of a spherical coordinate.
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  • Spring '08
  • Brown
  • Electrostatics, Electric charge, Fundamental physics concepts, Dr. Ray ChenĀ©