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lecture 7

# lecture 7 - 1 Gauss’s Law 2 2 4 e o Q Q E k πε r r = =...

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Unformatted text preview: 1 Gauss’s Law 2 2 4 e o Q Q E k πε r r = = For r > a Reading: Chapter 28 2 Gauss’s 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 • Φ E = EA • The field lines may make some angle θ with the perpendicular to the surface • Then Φ E = EA cos θ E r E EA Φ = E r cos E EA θ Φ = normal θ θ 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 θ Φ = = r r θ θ A r or E r cos E EA EA θ Φ = = - < r r θ θ A r EA Φ = r r A r 7 Electric Flux 1 A r 2 A r 1 2 E Φ = Φ + Φ 1 2 α 1 area A = 2 area A = E r β 2 2 2 2 cos(90 ) sin o EA EA EA β β Φ = =- = r r 1 1 1 sin EA EA α Φ = = r r 1 2 sin sin E EA EA α β Φ = + 1 A r 2 A r α E r β if then 1 2 cos cos E EA EA α β Φ = -- 90 o β- 1 A r 2 A r α E r 2 Φ = and are orthogonal 1 sin E EA α Φ = 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 Aθ E A ∆Φ = ∆ = ∆ r r surface lim i E i i A E A E dA ∆ Φ = ∆ = r r r r r 9 Electric Flux: Closed Surface surface lim i E i i A E A E dA ∆ Φ = ∆ = 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 α 3 A r 4 A r 1 2 3 4 5 6 E Φ = Φ + Φ + Φ + Φ + Φ + Φ 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 Φ = = r r 4 4 EA Φ = = r r 5 5 EA Φ = = r r 6 6 EA Φ = = r r 1 2 E Φ = Φ + Φ 1 1 1 1 cos(90 ) sin o EA EA EA α α Φ = = + = - r r 90 o α + 2 2 2 EA EA Φ = = r r 2 1 ( sin ) E E A A α Φ =- but 1 A 2 A α 2 1 sin A A α = Then E Φ = (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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lecture 7 - 1 Gauss’s Law 2 2 4 e o Q Q E k πε r r = =...

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