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Unformatted text preview: Chapter 22 Gausss Law Electric charge and flux (sec. 22.2 & .3) Gausss Law (sec. 22.4 & .5) Charges on conductors (sec. 22.6) C 2009 J. Becker Learning Goals  we will learn: Ch 22 How you can determine the amount of charge within a closed surface by examining the electric field on the surface! What is meant by electric flux and how you can calculate it. How to use Gausss Law to calculate the electric field due to a symmetric distribution of charges. A c ha rg e ins ide a b o x c a n b e pro b e d with a te s t c ha rg e q o to measure E fie ld The volume (V) flow rate (dV/dt) of fluid through the wire rectangle (a) is vA when the area of the rectangle is perpendicular to the velocity vector v and (b) is vA cos when the rectangle is tilted at an angle . We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric flux E . Volume flow rate through the wire rectangle. (a) The electric flux through the surface = EA. (b) When the area vector makes an angle with the vector E , the area projected onto a plane oriented perpendicular to the flow is A perp . = A cos . The flux is zero when = 90 o because the rectangle lies in a plane parallel to the flow and no fluid flows through the rectangle A flat surface in a uniform electric field. E = E . dA = E dA cos = E dA = E dA = E (4&#2; R 2 ) = (1/4&#2; o ) q /R 2 ) ( 4 R 2 ) = q / o . So we have the electric flux E = q / o ....
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 '09
 JBecker
 Physics, Charge

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