工程電磁場(chǎng)第三章課件

EngineeringElectromagnetics

W.H.HaytJr.andJ.A.BuckChapter3:ElectricFluxDensity,Gauss’Law,andDivergenceEngineeringElectromagnetics

W1FaradayExperimentHestartedwithapairofmetalspheresofdifferentsizes;thelargeroneconsistedoftwohemispheresthatcouldbeassembledaroundthesmallersphereFaradayExperimentHestartedw2+QFaradayApparatus,BeforeGroundingTheinnercharge,Q,inducesanequalandoppositecharge,-Q,ontheinsidesurfaceoftheoutersphere,byattractingfreeelectronsintheoutermaterialtowardthepositivecharge.Thismeansthatbeforetheoutersphereisgrounded,charge+Qresidesontheoutsidesurfaceoftheouterconductor.+QFaradayApparatus,BeforeGr3FaradayApparatus,AfterGroundingq=0groundattachedAttachingthegroundconnectstheoutersurfacetoanunlimitedsupplyoffreeelectrons,whichthenneutralizethepositivechargelayer.Thenetchargeontheoutersphereisthenthechargeontheinnerlayer,or-Q.FaradayApparatus,AfterGroun4InterpretationoftheFaradayExperimentq=0Faradayconcludedthatthereoccurredacharge“displacement”fromtheinnerspheretotheoutersphere.Displacementinvolvesafloworflux,

existingwithinthedielectric,andwhosemagnitudeisequivalenttotheamountof“displaced”charge.Specifically:InterpretationoftheFaraday5ElectricFluxDensityq=0Thedensityoffluxattheinnerspheresurfaceisequivalenttothedensityofchargethere(inCoul/m2)ElectricFluxDensityq=0The6VectorFieldDescriptionofFluxDensityq=0Avectorfieldisestablishedwhichpointsinthedirectionofthe“flow”ordisplacement.Inthiscase,thedirectionistheoutwardradialdirectioninsphericalcoordinates.Ateachsurface,wewouldhave:VectorFieldDescriptionofFl7Radially-DependentElectricFluxDensityq=0rAtageneralradiusrbetweenspheres,wewouldhave:ExpressedinunitsofCoulombs/m2,anddefinedovertherange(a≤r≤b)D(r)Radially-DependentElectricFl8PointChargeFieldsIfwenowlettheinnersphereradiusreducetoapoint,whilemaintainingthesamecharge,andlettheoutersphereradiusapproachinfinity,wehaveapointcharge.Theelectricfluxdensityisunchanged,butisdefinedoverallspace:C/m2(0<r<∞)Wecomparethistotheelectricfieldintensityinfreespace:V/m(0<r<∞)..andweseethat:PointChargeFieldsIfwenowl9FindingEandDfromChargeDistributionsWelearnedinChapter2that:Itnowfollowsthat:FindingEandDfromChargeDi10Gauss’LawTheelectricfluxpassingthroughanyclosedsurfaceisequaltothetotalchargeenclosedbythatsurfaceGauss’LawTheelectricfluxpa11DevelopmentofGauss’LawWedefinethedifferentialsurfacearea(avector)aswherenistheunitoutwardnormalvectortothesurface,andwheredSistheareaofthedifferentialspotonthesurfaceDevelopmentofGauss’LawWede12MathematicalStatementofGauss’LawLinecharge:Surfacecharge:Volumecharge:inwhichthechargecanexistintheformofpointcharges:Foravolumecharge,wewouldhave:oracontinuouschargedistribution:MathematicalStatementofGaus13UsingGauss’LawtoSolveforD

EvaluatedataSurfaceKnowingQ,weneedtosolveforD,usingGauss’Law:Thesolutioniseasyifwecanchooseasurface,S,overwhichtointegrate(Gaussiansurface)thatsatisfiesthefollowingtwoconditions:Theintegralnowsimplfies:Sothat:whereUsingGauss’LawtoSolvefor14Example:PointChargeFieldBeginwiththeradialfluxdensity:andconsiderasphericalsurfaceofradiusathatsurroundsthecharge,onwhich:Onthesurface,thedifferentialareais:andthis,combinedwiththeoutwardunitnormalvectoris:Example:PointChargeFieldBe15PointChargeApplication(continued)Now,theintegrandbecomes:andtheintegralissetupas:==PointChargeApplication(cont16AnotherExample:LineChargeFieldConsideralinechargeofuniformchargedensityLonthezaxisthatextendsovertherangezWeneedtochooseanappropriateGaussiansurface,beingmindfuloftheseconsiderations:Weknowfromsymmetrythatthefieldwillberadially-directed(normaltothezaxis)incylindricalcoordinates:andthatthefieldwillvarywithradiusonly:Sowechooseacylindricalsurfaceofradius,andoflengthL.AnotherExample:LineChargeF17LineChargeField(continued)Giving:Sothatfinally:LineChargeField(continued)G18AnotherExample:CoaxialTransmissionLineWehavetwoconcentriccylinders,withthezaxisdowntheircenters.SurfacechargeofdensityS

existsontheoutersurfaceoftheinnercylinder.A-directedfieldisexpected,andthisshouldvaryonlywith(likealinecharge).WethereforechooseacylindricalGaussiansurfaceoflengthLandofradius,wherea<<b.ThelefthandsideofGauss’Lawiswritten:…andtherighthandsidebecomes:AnotherExample:CoaxialTran19CoaxialTransmissionLine(continued)Wemaynowsolveforthefluxdensity:andtheelectricfieldintensitybecomes:CoaxialTransmissionLine(con20CoaxialTransmissionLine:ExteriorFieldIfaGaussiancylindricalsurfaceisdrawnoutside(b),atotalchargeofzeroisenclosed,leadingtotheconclusionthator:CoaxialTransmissionLine:Ext21ElectricFluxWithinaDifferentialVolumeElementTakingthefrontsurface,forexample,wehave:ElectricFluxWithinaDiffere22ElectricFluxWithinaDifferentialVolumeElementWenowhave:andinasimilarmanner:Therefore:minussignbecauseDx0isinwardfluxthroughthebacksurface.ElectricFluxWithinaDiffere23ChargeWithinaDifferentialVolumeElementNow,byasimilarprocess,wefindthat:andAllresultsareassembledtoyield:v=Q(byGauss’Law)whereQisthechargeenclosedwithinvolumevChargeWithinaDifferentialV24DivergenceandMaxwell’sFirstEquationMathematically,thisis:Applyingo