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Unformatted text preview: ! Introduce a new constant: 0 k = 9 x 109 N m2 / C2 = 8.85 x 1012 C2 / N m2 04 "
# $ #
' $ % "
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' ! * % , ! & % " ( " + & ! 06 . . / . / 0 ( " " ( . .
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 . . ( . / . & 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.
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6 What charges are inside the red circle? :9 89 :9 8 9 : 9 89 :9 89 $
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7 = "" " " " ! (" " % ! % ( > & ( + % ( & A B C
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< % ?
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 (
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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...
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 ' ' J % ! , , 0  50 3 ...
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This document was uploaded on 03/08/2012.
 Fall '12

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