Physics 2102Physics 2102Lecture 3Lecture 3GaussGauss’’ Law ILaw IMichael Faraday 1791-1867Version: 1/22/07Flux Capacitor (Schematic)Physics 2102Jonathan DowlingWhat are we going to learn?What are we going to 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? What? —— The Flux! The Flux!STRONGE-FieldWeakE-FieldNumber of E-LinesThrough DifferentialArea “dA” is aMeasure of StrengthdAθAngleMatters TooElectric Flux: Planar SurfaceElectric Flux: Planar Surface• Given:– planar surface, area A– uniform field E– E makes angle θ with NORMAL toplane• Electric Flux:Φ = E•A = E A cosθ• Units: Nm2/C• Visualize: “Flow of Wind”Through “Window”θEAREA = A=AnnormalElectric Flux: General SurfaceElectric Flux: General Surface• For any general surface: break up intoinfinitesimal planar patches• Electric Flux Φ = ∫E•dA• Surface integral• dA is a vector normal to each patch andhas a magnitude = |dA|=dA• CLOSED surfaces:– define the vector dA as pointingOUTWARDS– Inward E gives negative flux Φ – Outward E gives positive flux ΦEdAdAEArea = dAElectric Flux: ExampleElectric Flux: Example• Closed cylinder of length L, radius R• Uniform E parallel to cylinder axis• What is the total electric flux throughsurface of cylinder?(a) (2πRL)E(b) 2(πR2)E(c) ZeroHint!Surface area of sides of cylinder: 2π RLSurface area of top and bottom caps (each): πR2LRE(πR2)E–(πR2)E=0What goes in — MUST come out! dAdAElectric Flux: ExampleElectric Flux: Example• Note that E is NORMALto both bottom and top cap• E is PARALLEL tocurved surface everywhere• So: Φ = Φ1+ Φ2 + Φ3= πR2E + 0 - πR2E= 0!• Physical interpretation:total “inflow” = total“outflow”!3dA1dA2dAElectric Flux: ExampleElectric Flux: Example• Spherical surface of radius R=1m; E is RADIALLYINWARDS and has EQUAL magnitude of 10 N/Ceverywhere on surface• What is the flux through the spherical surface?(a) (4/3)πR2 E = −13.33π Nm2/C(b) 2πR2 E = −20π Nm2/C(c) 4π R2 E= −40π Nm2/CWhat could produce such a field?What is the flux if the sphere is not centeredon the charge?qrElectric Flux: ExampleElectric Flux: Example dr A = + dA( )ˆ r r E • dr A = EdA cos(180°) = !EdA r E = !qr2ˆ r Since r is Constant on the Sphere — RemoveE Outside the Integral! ! =r E " dr A #= $E dA = $kqr2% & ' ( ) * #4+r2( )= $q4+,04+( )= $q /,0Gauss’ Law:Special Case!(Outward!)Surface Area Sphere(Inward!)GaussGauss’’ Law: General Case Law: General Case• Consider any ARBITRARYCLOSED surface S -- NOTE:this does NOT have to be a“real” physical object!• The TOTAL ELECTRIC FLUXthrough S is proportional to theTOTAL CHARGEENCLOSED!• The results of a complicatedintegral is a very simpleformula: it avoids longcalculations! ! "r E # dr A =±q$0Surface%S(One of Maxwell’s 4 equations!)ExamplesExamples!="#$Surface0%qAdErrGaussGauss’’ Law: Example Law: ExampleSpherical symmetrySpherical symmetry• Consider a POINT charge q & pretend that youdon’t know Coulomb’s Law• Use Gauss’ Law to compute the electric field at adistance r from the charge• Use symmetry:– draw a spherical surface of radius R centeredaround the charge q– E has same magnitude anywhere on surface– E normal to surface0!q="rqE24|||| rEAE!=="2204||rkqrqE ==!"GaussGauss’’ Law: Example Law: ExampleCylindrical symmetryCylindrical symmetry• Charge of 10 C is uniformly spreadover a line of length L = 1 m.• Use Gauss’ Law to computemagnitude of E at a perpendiculardistance of 1 mm from the center ofthe line.• Approximate as infinitely longline -- E radiates outwards.• Choose cylindrical surface ofradius R, length L co-axial withline of charge.R = 1 mmE = ?1 mGaussGauss’’ Law: cylindrical Law: cylindricalsymmetry (cont)symmetry (cont)RLEAE!2|||| =="00!"!Lq==#RkRRLLE!"#!"#!222||00===• Approximate as infinitely longline -- E radiates outwards.• Choose cylindrical surface ofradius R, length L co-axial withline of charge.R = 1 mmE = ?1 mCompare with Example!Compare with Example!!"+=2/2/2/322)(LLyxadxakE#if the line is infinitely long (L >> a)…2242LaaLk+=!2/2/222LLaxaxak!"#$%&'+=(akLaLkEy!!222==SummarySummary• Electric flux: a surface integral (vector calculus!);useful visualization: electric flux lines caught by thenet on the surface.• Gauss’ law provides a very direct way to computethe electric
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