lecture 7 - 1 Gausss Law 2 2 4 e o Q Q E k r r = = For r...

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Unformatted text preview: 1 Gausss Law 2 2 4 e o Q Q E k r r = = For r > a Reading: Chapter 28 2 Gausss Law Chapter 28 3 Electric Flux Definition: Electric flux is the product of the magnitude of the electric field and the surface area, A , perpendicular to the field E = EA The field lines may make some angle with the perpendicular to the surface Then E = EA cos E r E EA = E r cos E EA = normal 4 Electric Flux: Surface as a Vector Vector , corresponding to a Flat Surface of Area A, is determined by the following rules: the vector is orthogonal to the surface the magnitude of the vector is equal to the area A 90 o A r normal Area = A The first rule the vector is orthogonal to the surface does not determine the direction of . There are still two possibilities: A r 90 o A r 90 o A r or You can choose any of them 5 Electric Flux: Surface as a Vector If we consider more complicated surface then the directions of vectors should be adjusted, so the direction of vector is a smooth function of the surface point correct or wrong or 1 A r 2 A r 3 A r 6 Electric Flux Definition: Electric flux is the scalar product of electric field and the vector E r cos E EA EA = = r r A r or E r cos E EA EA = = - < r r A r EA = r r A r 7 Electric Flux 1 A r 2 A r 1 2 E = + 1 2 1 area A = 2 area A = E r 2 2 2 2 cos(90 ) sin o EA EA EA = =- = r r 1 1 1 sin EA EA = = r r 1 2 sin sin E EA EA = + 1 A r 2 A r E r if then 1 2 cos cos E EA EA = -- 90 o - 1 A r 2 A r E r 2 = and are orthogonal 1 sin E EA = 2 A r E r flux is positive flux is negative 8 Electric Flux In the more general case, look at a small flat area element In general, this becomes The surface integral means the integral must be evaluated over the surface in question The units of electric flux will be N . m 2 /C 2 cos E i i i i i E A E A = = r r surface lim i E i i A E A E dA = = r r r r r 9 Electric Flux: Closed Surface surface lim i E i i A E A E dA = = r r r r r The vectors point in different directions At each point, they are perpendicular to the surface By convention, they point outward i A r 10 Electric Flux: Closed Surface E r Closed surface 1 2 3 4 1 3 5 1 A r 2 A r 3 A r 4 A r 1 2 3 4 5 6 E = + + + + + 6 2 4 E r E r is orthogonal to , , , and 3 A r 4 A r 5 A r 6 A r Then 3 3 EA = = r r 4 4 EA = = r r 5 5 EA = = r r 6 6 EA = = r r 1 2 E = + 1 1 1 1 cos(90 ) sin o EA EA EA = = + = - r r 90 o + 2 2 2 EA EA = = r r 2 1 ( sin ) E E A A =- but 1 A 2 A 2 1 sin A A = Then E = (no charges inside closed surface) 11 Electric Flux: Closed Surface A positive point charge, q , is located at the center of a sphere of radius r The magnitude of the electric field everywhere on the surface of the sphere is E = k e q / r 2 Electric field is perpendicular to...
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This note was uploaded on 02/15/2012 for the course PHYS 2212 taught by Professor Staff during the Spring '08 term at Georgia State University, Atlanta.

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lecture 7 - 1 Gausss Law 2 2 4 e o Q Q E k r r = = For r...

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