大学物理期末复习

GeneralRevisionofCollegePhysicsHuangQiongChapter1.ElectrostaticField1.CoulombLawintheVacuum123021124rrqqf(1)Coulombforce(2)TheprincipleofsuperpositionThetotalelectricalforceongivenchargeisthevectorsumoftheelectricalforcescausedbytheothercharges,calculatedasifeachactedalone.ForceDuetotheSystemofPointCharges.(3)FindElectrostaticForceForceDuetoContinuousChargeDistributionsQdVdsdldq3004rrdqqFd3004rrdqqFQdqr+q0P(4).CoulombLawintheDielectric123r021124rrqqf2.ElectricFieldIntensity0qFEMagnitude:theelectricfieldforceonunitpositivechargeDirection:directthedirectionofforceexertedonapositivetestcharge.UnitN/C、V/m(2)Theprincipleofsuperposition(1)DefinitionniinEEEEE121(3).ElectricFieldLinesElectricfieldlinesarenotreal.Fieldlinesarenotmaterialobjects.Theyareusedonlyasapictorialrepresentationtoprovideaqualitativedescriptionofthefield.InelectrostaticfieldE0,electricfieldlinesbeginandendoncharges.InInducedelectricfieldEv,theelectricfieldlinesformclosedloops,withnobeginningandnoend.Theelectricfieldlinesaredenserintheplacewherethefieldintensityisstronger,andtheelectricfieldlinesaresparserintheplacewherethefieldintensityisweaker.istaticqSdE0s10svortexSdEFieldlinesarenevercross.Thefieldatanypointhasauniquedirection.3.GAUSS’SLAWSeSdEiQ01Invacuum,Indielectric,0QSdDSElectricfluxoftheGausssurfaceisrelatedwithchargesinGausssurfaceandisnotrelatedwithchargesoutofGausssurface.ThefieldintensityatapointontheGausssurfaceisrelatedwithchargesintheGausssurfaceandisnotrelatedwithchargesoutoftheGausssurface.IftheelectricfluxofaGausssurfaceequalszero,theremustbenotchargeintheGausssurface.√××IftheelectricfluxofaGausssurfaceequalszero,thenfieldintensityateverypointontheGausssurfaceiszero.×Gausstheoremistenableonlytotheelectrostaticfieldwhosedistributionissymmetricalinspace.×√4.CalculatingtheElectricFieldProblem-solvingstrategy:Analysisthedistributionofcharges.Solution1.Applyingthesuperpositionprincipleofthefieldintensity.Ifthechargesarecountable,theresultantfieldisthevectorsumofthefieldsduetotheindividualcharges.whenconfrontedwithproblemsthatinvolveacontinuousdistributionofcharge,ElementAnalysisMethodSolution2.ApplyingGauss’slawtosymmetricchargedistribution.PlanesymmetrySphericalsymmetryThethreesymmetries:CylindricalsymmetryProblem-solvingstrategy:Selectappropriategaussiansurface.Selectappropriatecoordinates,applyGauss’slaw.Analysisthesymmetryofthefieldintensitydistribution.Example1Example2Example3,,RR,2211ErRErERrTherearetwoconcentricchargedsphericalshellsofradiusR1andR2.Chargequantitiesdistributeuniformly.R1，Q1R2，Q2Example4Example5Example6ThechargelinedensityofaninfiniteuniformchargedcylindricalsurfaceofradiusRis,findtheelectricfieldintensity.rEExample71R2R5.Electricpotentialenergy,Electricpotential00aababa“”＝“０”rPldEQWU0Electricpotential（simplythePotential）unit：V(volt)J/C11VThepotentialatapointequalstheworkrequiredtobringaunitpositivechargefromthispointtothezeropointofelectricpotential.Electricpotentialenergyofq0atapointTherelationshipbetweentheelectricpotentialenergyandElectricpotentialxExUyEyUzEzUIngeneral,theelectricpotentialisafunctionofallthreespatialcoordinates.IfVisgivenintermsofrectangularcoordinates,theelectricfieldcomponentsEx,Ey,andEzcanbefoundfromV(x,y,z)asthepartialderivativesForexample,if,then223Vxyyyz6xExyInacertainregionofspace,theelectricpotentialiszeroeverywherealongthexaxis.Fromthiswecanconcludethatthexcomponentoftheelectricfieldinthisregionis(a)zero(b)inthe+xdirection(c)inthe-xdirection.Inacertainregionofspace,theelectricfieldiszero.Fromthiswecanconcludethattheelectricpotentialinthisregionis(a)zero(b)constant(c)positive(d)negative.2020/1/2035（1）场强相等的区域，电势处处相等？（2）场强为零处，电势一定为零？（3）电势为零处，场强一定为零？（4）场强大处，电势一定高？QQEU势0rrdEURaP0CalculatePotential.TwoMethod:QrdqU040rUEdlApplyingelementanalysismethodApplyingthedefinitionofpotentialExample1Example2ABOq1R2R3RQFindtheelectricfieldintensityandtheelectricpotential.Example3.6.Equipotentialvolumesandsurfaces2、可有EU计算电势的方法（2种）1、微元法QrdQU04iiirQU040rrdEU7.SUMMARY计算场强的方法（3种）1、点电荷场的场强及叠加原理iiirrQE304QrdQrE3042、定义法UE（分立）（连续）（分立）（连续）xExUEU典型电场的电势典型电场的场强均匀带电球面0E304rrqE球面内球面外均匀带电无限长直线rE02均匀带电无限大平面02E均匀带电球面rqU04RqU04均匀带电无限长直线02lnraU均匀带电无限大平面02dEdU方向垂直于直线方向垂直于平面ExampleSolution:MetallicConductorandDielectricinElectrostaticFieldChapter2.1.ElectrostaticEquilibrium(1)Theelectrostaticequilibriumconditionsofconductorinsidetheconductor:0ETheelectricfieldjustoutsidethechargedconductorisperpendiculartotheconductorsurface.0EWhenaconductorisinelectrostaticequilibriumstate,theconductorisabodyofequalpotential,anditssurfaceisasurfaceofequalpotential.(2)ThedistributionofPotential(3)ThedistributionofchargesSolidConductor:Iftheconductorcarriescharge,thechargeresidesentirelyonitsoutsurface.Conductorcavity:+++++++++++++++q++Example1.r0rr0CUQUQC2.Capacita