This preview shows pages 1–12. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
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.
 Spring '08
 Staff

Click to edit the document details