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notes7 2317 - ECE 2317 ECE Applied Electricity and...

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Prof. David R. Jackson ECE Dept. Spring 2011 Notes 7 ECE 2317 ECE 2317 Applied Electricity and Magnetism Applied Electricity and Magnetism Notes prepared by the EM Group University of Houston
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Coulomb’s Law Coulomb’s Law [ ] [ ] 1 2 2 2 0 12 0 N F/m ˆ 4 8.854187818 10 q q F r r πε ε - = = × Experimental law: Note: c = speed of light = 2.99792458 × 10 8 [ m/s ] (exactly) 0 0 1 c μ ε = [ ] 7 0 H/m 4 10 (exactly) μ π - = × x y z r q 1 q 2 ˆ r Charles Coulomb
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Coulomb’s Law (cont.) Coulomb’s Law (cont.) Hence: But 1 2 2 2 0 ˆ 4 q q F r r πε = 1 1 2 0 ˆ 4 q E r r πε = ( E 1 : E due to q 1 ) ( 29 2 2 2 1 F q E r = Note: There is no self-force on charge 2 due to its own electric field. x y z r q 1 q 2 ˆ r
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Generalization ( q 1 not at the origin): Coulomb’s Law (cont.) Coulomb’s Law (cont.) r 1 = ( x 1 , y 1 , z 1 ) r 2 = ( x 2 , y 2 , z 2 ) ( 29 ( 29 ( 29 2 1 1 2 2 2 2 2 1 2 1 2 1 R R r r x x y y z z = = - = - + - + - ( 29 1 2 1 2 0 ˆ 4 q E r R R πε = ˆ R R R = 2 1 R r r = - q 2 ( x 2 , y 2 , z 2 ) x y z R ˆ R r 2 r 1 q 1 ( x 1 , y 1 , z 1 )
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Example Example q 1 = 0.7 [ mC ] located at (3,5,7) [ m ] q 2 = 4.9 [ μ C ] located at (1,2,1) [ m ] E 1 ( r 2 ) = electric field due to charge q 1 , evaluated at point r 2 ( 29 2 2 2 1 F q E r = Find: F 1 , F 2 For F 2 : F 1 = force on charge q 1 F 2 = force on charge q 2 R q 2 q 1
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Example (cont.) Example (cont.) ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 ( 29 1 2 1 2 0 2 1 2 2 2 m m ˆ 4 ˆ ˆ ˆ 1 3 2 5 1 7 ˆ ˆ ˆ 2 3 6 [ ] 2 3 6 7 [ ] 2 3 6 ˆ ˆ ˆ ˆ 7 7 7 q E r R R R r r x y z R x y z R R R R x y z R πε = = - = - + - + - = - + - + - = = - + - + - = - - - = = + + R q 2 q 1 q 1 = 0.7 [ mC ] located at (3,5,7) [ m ] q 2 = 4.9 [ μ C ] located at (1,2,1) [ m ] ( 29 2 2 2 1
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