51_Ch22_Gauss

# 51_Ch22_Gauss - Chapter 22 Gauss’s Law • Electric...

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Unformatted text preview: Chapter 22 Gauss’s Law • Electric charge and flux (sec. 22.2 & .3) • Gauss’s 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 Gauss’s 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 φ when the rectangle is tilted at an angle φ . We will next replace the fluid velocity flow vector v with the electric field vector E to get to the concept of electric flux Φ E . Volume flow rate through the wire rectangle. (a) The electric flux through the surface = EA. (b) When the area vector makes an angle φ with the vector E , the area projected onto a plane oriented perpendicular to the flow is A perp . = A cos φ . The flux is zero when φ = 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. Φ E = ∫ E . dA = ∫ E dA cos φ = ∫ E dA = E ∫ dA = E (4&#2; R 2 ) = (1/4&#2; ε o ) q /R 2 ) ( 4 π R 2 ) = q / ε o . So we have the electric flux Φ E = q / ε o ....
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51_Ch22_Gauss - Chapter 22 Gauss’s Law • Electric...

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