06 - Gauss Law I

# 06 - Gauss Law I - Phys 212 Fall 2009 Gauss Law I Reminders...

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Phys 212 - Fall 2009 9/02/2009 Gauss’ Law I Reminders HW2 Due Wednesday 1 st Test - Wed. Sept 16

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Review - Electric Fields F = k | q 1 || q 2 | r 2 Force between point charges E = F q 0 = k | q | r 2 E field of a point charge Take symmetry argument further to develop simple way to solve complex problems Gauss’ Law Used symmetry and superposition to determine fields.
- Flux Field Strength Through a Surface Consider air: volume flow rate through a square Φ Need Two Vectors direction of field direction of surface Φ = v A At an angle, flow 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

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Gauss’ Law Relates the Electric flux through a closed surface to the charge enclosed by that surface. Φ E = q enc ε 0 Prescription 1) put an arbitrary surface into a field 2) divide into individual surfaces 3) add up all flux contribution Carl Friedrich 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 Field is constant on the surface Φ E =

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