Lect04 - % #& closed surface E dA Qenclosed 0 E dA...

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Unformatted text preview: % #& closed surface E dA Qenclosed 0 E dA ! " " # E dA $ % E dA E dA EA Qenclosed 0 E 19 Qenclosed A 0 '# ' " " E dA E ( ' " dA " $ and ends E d A E r 2 rL sides E dA 0 E d A E r 2 rL Qenclosed 0 L 0 E r 2 r 0 '# E dA Qenc o E Qenc A 0 A 4 r Qenc E 4 r2 21 2 A 2 rL E 0 A 2 r2 E 0 2 r 2 0 ( * In which case is E at point P the biggest? A) A B) B C) the same + " ' - , 27 ( . , ) Superposition ! NET + Case A + Case B - + 34 / ( * The E field from a charged cube is not constant on any of these surfaces. Gauss's law is true, but it does Gauss' true, not help us to calculate E for this particular problem. (D) The field cannot be calculated using Gauss' Law (E) None of the above How to calculate the field? Go back and find E by superposition. Add the contributions to E from each infinitesimal charge dq in the cube: E 23 dq ^ k 2r r 0 ( &' - . Why ? 2 3 ( " ' -. &5 6 4 * ' -. $ 9 7 # $ $ " " ' . 4 2 3 3 $ ( 2 3 3 8 & E dA Qenc o Qenc 0 1 06 ' -. ( " ! ( " Begin with a neutral conductor: Q=0 No charge inside or on the surface Now bring a positive charge +Q near the conductor. This induces (-) charge on one end and (+) charge on the other. The total charge on the conductor is still Q = 0. The induced charge is on the surface of the conductor. 09 + + + - +Q : ( " ( $ ( ! " # ! $" % &" ' ) $ %" " ( & ( $ % $ % Qouter Qinner 10 . ( * ; 26 ( * ; What is direction of field OUTSIDE the red sphere? 9 " " 3 < 4 3 " 3 < 29 ( * 31 ; y neutral conductor +3Q " =; > " x 6 ' $ 7 5 @ " ' 5 ? ' " Magnitude of E is fcn of (r-r1) ( ? Magnitude of E is fcn of (r-r2) None of the above ( ? Direction of E is along Direction of E is along None of the above ^ x ^ y ^ r 36 y neutral conductor ' " x ^ r ? E +3Q 9 ( " ' A A A B E dA E 4 r 2 Qenc 3Q E 40 5 E E 1 3Q 4 0 r2 1 3Q 4 0 r12 5 E E 1 3Q 4 0 r2 1 4 0 1 3Q 4 0 r2 3Q ( r r2 ) 2 ) ( E 0 ( E 0 neutral conductor A B A A E +3Q 1 3Q 4 0 r2 E 0 6 " 7 Gauss' Law: Gauss' E dA Qenc Similarly: 2 Qenc o 1 5 0 0 E +3Q 0 0 3Q 4 r12 ( 0 3Q 4 r22 / 44 y C 6 6 "$ ' $ " 7 3 + " D> -Q conductor 3 x 7 + + + + + +3Q + +2Q +3Q E dA Qenc o + + + + + + -3Q + + A A + + + A B 5 E 1 3Q 4 0 r2 5 E E E 1 3Q 4 0 r2 1 2Q 4 0 r2 1 4 0 E E 1 2Q 4 0 r2 1 4 0 E 0 ( 46 Q r2 ( Q r2 0 ...
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This document was uploaded on 03/08/2012.

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