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LSU PHYS 2102 - Gauss’ Law I

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Physics 2102 Lecture: 05 FRI 23 JANWhat Are We Going to Learn? A Road MapWhat? — The Flux!Electric Flux: Planar SurfaceElectric Flux: General SurfaceElectric Flux: ExampleSlide 7Electric Flux: ExampleSlide 9PowerPoint PresentationPhysics 2102 Physics 2102 Lecture: 05 FRI 23 JANLecture: 05 FRI 23 JANGauss’ Law IGauss’ Law IVersion: 1/22/07Flux Capacitor (Schematic)Physics 2102Jonathan DowlingQuickTime™ and a decompressorare needed to see this picture.Carl Friedrich Gauss1777 – 1855What Are We Going to What Are We Going to Learn?Learn?A Road MapA Road Map•Electric charge - Electric force on other electric charges- Electric field, and electric potential•Moving electric charges : current •Electronic circuit components: batteries, resistors, capacitors•Electric currents - Magnetic field - Magnetic force on moving charges•Time-varying magnetic field - Electric Field•More circuit components: inductors. •Electromagnetic waves - light waves•Geometrical Optics (light rays). •Physical optics (light waves)What? — The Flux!What? — The Flux!STRONGE-FieldWeakE-FieldNumber of E-Lines Through Differential Area “dA” is a Measure of StrengthdAAngle Matters TooElectric Flux: Planar Electric Flux: Planar SurfaceSurface•Given: –planar surface, area A–uniform field E–E makes angle q with NORMAL to plane•Electric Flux: = E•A = E A cos•Units: Nm2/C•Visualize: “Flow of Wind” Through “Window”EAREA = A=AnnormalElectric Flux: General Electric Flux: General SurfaceSurface•For any general surface: break up into infinitesimal planar patches•Electric Flux  = E-dA•Surface integral•dA is a vector normal to each patch and has a magnitude = |dA|=dA•CLOSED surfaces: –define the vector dA as pointing OUTWARDS–Inward E gives negative flux –Outward E gives positive flux EdAdAEArea = dAElectric Flux: ExampleElectric Flux: Example•Closed cylinder of length L, radius R•Uniform E parallel to cylinder axis•What is the total electric flux through surface of cylinder?(a) (2RL)E (b) 2(R2)E (c) Zero Hint!Surface area of sides of cylinder: 2RLSurface area of top and bottom caps (each):  R2LRE(R2)E–(R2)E=0What goes in — MUST come out! dAdAElectric Flux: ExampleElectric Flux: Example•Note that E is NORMAL to both bottom and top cap• E is PARALLEL to curved surface everywhere • So:  = +  +   R2E + 0 – R2E = 0!•Physical interpretation: total “inflow” = total “outflow”!3dA1dA2dAElectric Flux: ExampleElectric Flux: Example •Spherical surface of radius R=1m; E is RADIALLY INWARDS and has EQUAL magnitude of 10 N/C everywhere on surface•What is the flux through the spherical surface?(a) (4/3)R2 E = 13.33 Nm2/C (b) 2R2 E = 20 Nm2/C (c) 4R2 E= 40 Nm2/C What could produce such a field?What is the flux if the sphere is not centered on the charge?qrElectric Flux: ExampleElectric Flux: Example € dr A = + dA( )ˆ r € r E • dr A = EdAcos(180°) = −EdA € r E = −qr2ˆ r Since r is Constant on the Sphere — RemoveE Outside the Integral! € Φ =r E ⋅dr A ∫= −E dA = −kqr2 ⎛ ⎝ ⎜ ⎞ ⎠ ⎟∫4πr2( )= −q4πε04π( )= −q /ε0Gauss’ Law:Special Case!(Outward!)Surface Area Sphere(Inward!)QuickTime™ and a decompressorare needed to see this


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LSU PHYS 2102 - Gauss’ Law I

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