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Published byYandi Dharmawijaya Modified over 6 years ago
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First Term Stuff If you want to make up labs or quizzes from Term 1, see me before the end of this week to make a plan If you have already seen me, you are good
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Electric Flux how many lines?
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Electric Flux Through a closed surface
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Electric Flux FE Conceptually, the number of electric field lines passing through a surface E FE = EA A = area of the surface
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Electric Flux FE Previous formula only works when field is normal to the surface E FE = 0
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Electric Flux FE E q Area in profile is A cos q
q = angle of E field from surface normal FE = EA cos q = E·A A = “Area vector” normal to surface
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Electric Flux Integrate over entire surface FE = ∫E·dA
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Surface Integral Around an entire surface that encloses a region of space Symbol ∫ Area vector points outward
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Gauss’s Law Flux through an enclosing surface is proportional to charge enclosed FE = q/e0 e0 = “vacuum permittivity” = 8.85×10–12 C2/(Nm2)
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Gauss’s law Use flux to find field around a point charge
The answer should be Coulomb’s law
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Field of a Point Charge What is area A of spherical shell?
AP Physics L04_flux Field of a Point Charge What is area A of spherical shell? FE = Qin/e0 = EA Find E
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Field around a Point Charge
Shell Area = 4pr2 FE = q/e0 = EA +q q e0A q e04pr2 R E = = q 4pe0 r2 = kq r2 = if k = 1 4pe0
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Gauss’s Law Example Infinite line of uniform charge density (Q/L) = l
Cylindrical surface with charge on axis Area of surface = 2prL + ends Qin = Ll FE = Qin/e0 = Ll/e0 FE = E(2prL) Ll/e0 = E(2prL) 2pe0 1 l r E = Q
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Gauss’s Law Example Infinite plate of uniform charge density (Q/A) = s
Area of surface = 2A + edges Qin = As FE = Qin/e0 = As/e0 FE = E(2A) As/e0 = E(2A) E = ½ s/e0 Q
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Gauss’s law examples Field within a spherical shell of uniform charge density Field within a sphere of uniform charge density Charge distribution in a conductor Field around charge at center of a neutral, hollow conducting sphere
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