Lect03 - ! Introduce a new constant: 0 k = 9 x 109 N m2 /...

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Unformatted text preview: ! Introduce a new constant: 0 k = 9 x 109 N m2 / C2 = 8.85 x 10-12 C2 / N m2 04 " # $ # ' $ % " ( & " ! " # ) ' ! * % , ! & % " ( " + & ! 06 . . / . / 0 ( " " ( . . 07 12 - . . ( . / . & 08 % 3 , " ( Field lines are are denser near Q1 so % Q1 > Q2 09 4 , The electric field lines connect the charges. A test charge will move towards one charge and away from the other. So charges 1 and 2 have opposite signs. 10 5 , " ( Density of lines is greater at B than at A. Therefore, magnitude of field at B is greater than at A. 12 6 What charges are inside the red circle? :9 89 :9 8 9 : 9 89 :9 89 $ 13 ; . 7 = "" " " " ! (" " % ! % ( > & ( + % ( & A B C 15 D < % ? S S Flux through surface S Integral of E dA on surface S @ E dA 18 1 A 1 $ ( S S Flux through surface S v dA flowrate Integral of v dA on surface S 20 , An infinitely long charged rod has uniform charge density and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared TAKE s TO BE RADIUS ! with the cylinder in Case 1. L/2 D D D 1 B$C B;C B C B.C 23 , . " " An infinitely long charged rod has uniform charge density l and passes through a cylinder (gray). The cylinder in Case 2 has twice the radius and half the length compared with the TAKE s TO BE RADIUS ! cylinder in Case 1. E dA surface L/2 & & B $C $C " " E barrel dA EAbarrel D B$C E1 2 0s 1 B;C D 2 E D $E1 B C( $= B.C F L 0 E2 2 0 (2s) 26 A1 (2 s) L 2 ( L / 2) 0 A2 (2 (2s)) L / 2 2 sL - . E E dA 0 E " dA dA E dA dA E E E $ ! E E S S 29 E dA 0 3 . E E dA 0 E " dA dA E dA dA E E E $ ! E E S S 30 E dA 0 4 y & " D< GD8 D8 ( z 4 E ^ E0 x 3 . " D 1 2 x dA E dA 0 E $ ; H< D< I< $ ; H< D< I< $ ; H< D< I< $ ; H< D< I< 31 5 y & " D< GD8 D8 ( ( G z J ! $ E ^ E0 x E0 3 89 1 2 x ( 89 ( > B: 1 1 C . " = D $ ; % % $ ; % % $ ; % % 36 6 ! E E dA Q dA E E dA dA E E dA E E E S E dA closed surface Qenclosed o 7 41 , What happens to total flux through the sphere as we move Q ? (A) increases (B) decreases (C) stays same The same amount of charge is still enclosed by the sphere, so flux will not change. 43 < , (A) d A increases d B decreases (B) d A decreases d B increases (C) d A stays same d B stays same 44 , " ! 1 2 " " 45 % ( ( & B " C % BK % %& B& (( C" " C L ( S & E dA Qenclosed ! o closed surface If Qenclosed is the same, the flux has to be the same, which means that the integral must yield the same result for any surface. 47 ( E dA closed surface & Qenclosed o ! In cases of high symmetry it may be possible to bring E outside the integral. In these cases we can solve Gauss Law for E E dA EA closed surface Qenclosed o E Qenclosed A 0 So - if we can figure out Qenclosed and the area of the surface A, then we know E ! This is the topic of the next lecture... 48 - ' ' J % ! , , 0 - 50 3 ...
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This document was uploaded on 03/08/2012.

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