06 - Gauss Law I

06 - Gauss Law I - Phys 212 - Fall 2009 9/02/2009 Gauss’...

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Unformatted text preview: Phys 212 - Fall 2009 9/02/2009 Gauss’ Law I Reminders HW2 Due Wednesday 1 st Test - Wed. Sept 16 Review - Electric Fields F = k | q 1 || q 2 | r 2 Force between point charges E = F q = k | q | r 2 E feld o¡ a point charge Take symmetry argument ¡urther to develop simple way to solve complex problems Gauss’ Law Used symmetry and superposition to determine felds.- Flux Field Strength Through a Surface Consider air: volume ¡ow rate through a square Φ Need Two Vectors direction of ¢eld direction of surface Φ = v ⋅ A At an angle, ¡ow is diminished. Only perpendicular component contributes v cos( θ ) Φ = [ v cos( θ )] A Maximum Flow when v is perpendicular to the surface Φ = vA Gauss’ Law Relates the Electric fux through a closed surFace to the charge enclosed by that surFace. Φ E = q enc ε Prescription 1) put an arbitrary surFace into a ¡eld 2) divide into individual surFaces 3) add up all fux contribution Carl ¢riedrich Gauss 1777-1855 reminder - surface vector is perpendicular to the surface with magnitude equal to the area of the surface Φ E = E ⋅Δ A closed surface ∑ → E ⋅ d A ∫ Point Charge Choose spherical surface Field is perpendicular to surface everywhere Field is parallel to surface vector...
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This note was uploaded on 10/03/2011 for the course PSYC 212 taught by Professor Ilieva during the Fall '09 term at South Carolina.

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06 - Gauss Law I - Phys 212 - Fall 2009 9/02/2009 Gauss’...

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