第三課電磁場與電磁波英文版課件

3BoundaryValue

Problems3BoundaryValue

Problems1(1)Distributionproblems(2)Boundaryvalueproblems1.Thetypesofproblemsinelectrostaticfield2.Thecalculationmethodsofboundaryvalueproblems(1)Analyticalmethod(2)NumericalmethodBoundaryconditionsEquationsordq(1)Distributionproblems(2)B3.1TypesofBoundaryConditionsandUniquenessTheorem3.1TypesofBoundaryConditio

ConsideraregionVboundedbyasurfaceS.1.Dirichletboundaryconditionsspecifythepotentialfunctionontheboundary.wherefisacontinuousfunction.2.Neumannboundaryconditionsspecifythenormalderivativeofthepotentialfunctionontheboundary.wheregisacontinuousfunction.3.MixedboundaryconditionsThetypesofboundaryconditionsConsideraregionVbou

TheUniquenessTheoremTheuniquenesstheoremstatesthatthereisonlyone(unique)solutiontoPoisson’sorLaplace’sequationsatisfiedgivenboundaryconditions.

Poisson’sequationLaplace’sequationBoundaryconditionsTheUniquenessTheoremPois

Prove(proofbycontradiction)ConsideravolumeVboundedbysomesurfaceS.SupposethatwearegiventhechargedensitythroughoutVandthevalueofthescalarpotentialonsurfaceS.AssumethatthereexisttwosolutionsandofLaplace’sequationsubjecttothesameboundaryconditions.Then,Prove(proofbycontradict

and

LetthenApplyingGreen’sfirstidentity

Wehave

wherevolumeVboundedbytheenclosedsurfaceS.TheintegralisequaltozerosinceonthesurfaceS.Thus,TheintegralcanbezeroisifisaconstantWeknowthatonthesurfaceS,soweget第三課電磁場與電磁波英文版課件

Thatis,

ThroughoutVandonsurfaceS.OurinitialassumptionthatandaretwodifferentsolutionsofLaplace’sequations,satisfyingthesameboundaryconditions,turnsouttobeincorrect.Hence,thereisauniquesolutiontoLaplace’sequationsatisfiedgivenboundaryconditions.

Thatis,3.2DirectIntegrationNote:Thepotentialfieldisafunctionofonlyonevariable.3.2DirectIntegrationNote:TSolution

Sincethetwoconductorsofradiiaandbformequipotentialsurfaces,thepotentialmustbeafunctionofonly.Example3.2.1TheinnerconductorofradiusaofacoaxialcableisheldatapotentialofUwhiletheouterconductorofradiusbisgrounded.Determine(a)thepotentialdistributionbetweentheconductors,(b)thesurfacechargedensityontheinnerconductor,and(c)thecapacitanceperunitlength.SolutionExample3.2.1

Thus,Laplace’sequationreducesto

Integratingtwice,weobtain(1)whereandareconstantsofintegrationtobedeterminedfromtheboundaryconditions.

Substitutingandinto(1),wehave(2)

Substitutingandinto(1),wehave(3)Thus,Laplace’sequationrHence,thepotentialdistributionwithintheconductorsisTheelectricfieldintensityisThus,ThenormalcomponentofDatisequaltothesurfacechargedensityontheinnerconductor.Thus,Hence,thepotentialdistribut

ThechargeperunitlengthontheinnerconductorisFinally,weobtainthecapacitanceperunitlengthas

1Finally,weobtainthecapaproblemTheinnerconductorofradiusaofacoaxialcableisheldatapotentialofUwhiletheouterconductorofradiusbisgrounded.Thespacebetweentheconductorsisfilledwithtwoconcentriclayersofdielectric.Determine(a)thepotentialdistributionbetweentheconductorsand(b)blemExample3.2.2Asphericalcapacitorisformedbytwoconcentricsphericalshellsofradiiaandb,asshowninFigure3.2.2.TheregionbetweentheinnerandoutersphericalconductorsisfilledwithadielectricofpermittivityIfUisthepotentialdifferencebetweenthetwoconductors,determine(a)thepotentialdistributionbetweentheconductors,and(b)thecapacitanceofthesphericalcapacitor.Example3.2.2Solution

Sincethetwoconductorsofradiiaandbformequipotentialsurfaces,thepotentialmustbeafunctionofonly.Thus,Laplace’sequationreducesto(a<r<b)Integratingtwice,weobtainwhereC1andC2areconstantsofintegrationtobedeterminedfromtheboundaryconditions.(1)Solution(a<r<b)IntegratingSubstitutingboundaryconditionsandinto(1),weobtain

ThenormalcomponentofDatyieldsthesurfacechargedensityontheinnerconductor.Thus,Hence,thecapacitanceofthesystemisSubstitutingboundarycondi通解例體電荷均勻分布在一球形區(qū)域內(nèi)，求電位及電場。解：采用球坐標系,分區(qū)域建立方程邊界條件參考電位圖1.4.2體電荷分布的球體0通解例體電荷均勻分布在一球形區(qū)域內(nèi)，求電位及電場。解：電場強度（球坐標梯度公式）：得到圖隨r變化曲線0電場強度（球坐標梯度公式）：得到圖隨r變化曲線03.3SeparationofVariablesSolveLaplace’sequationTransformthethree-dimensionalpartialdifferentialequationintothreeone-dimensionalordinarydifferentialequations.3.3SeparationofVariablesSo

1.SeparationofVariablesforRectangularCoordinateSystemLaplace’sequationis

(1)Assumethatthepotentialcanbewrittenastheproductofone-dimensionalpotentials.(2)

whereisafunctionofxonly,isafunctionofyonlyandisafunctionofzonly.Substituting(2)into(1),andthendividingby

weobtain1.SeparationofVariables

(3)Sinceeachterminvolvesasinglevariable,theequationisrightifandonlyif

eachtermmustbeaconstant.Thus,welet

where,andarecalledtheseparationconstants.Note:Theformofgeneralsolutiontoeachequationdependsontheseparationconstant.where,andarTheformofsolutionofasfollows.1.Ifisarealnumber,thegeneralsolutionis

2.Ifisaimaginarynumberthegeneralsolutionis

3.Ifkxiszero,then

orwhereA1,A2,B1,B2,C1,C2,D1andD2arearbitraryconstants.hyperbolicsinehyperboliccosineTheformofsolutionof

Example3.3.1Aninfinitelylongrectangulartroughisformedbyfourconductingplanes,locateatandaandandbinair,asshowninFigure3.3.1.ThesurfaceatisatpotentialofU,theotherthreeareatzeropotential.Determinethepotentialdistributioninsidetherectangulartrough.

Two-dimensionalfieldExample3.3.1Two-dimension

Solution

Sincethetroughisinfinitelylonginthez-direction,thepotentialcanonlydependonxandy.Laplace’sequationreducesto(1)Let(2)Substituting(2)in(1)anddividingby(2),weget

LetThen,wehaveorSolutionLetThen,wehaveo

Solution

Sincethetroughisinfinitelylonginthez-direction,thepotentialcanonlydependonxandy.Laplace’sequationreducesto

Usingseparationofvariables,welet

Then,wehaveorSincetheboundaryconditionsareatandwecanassumeandTheformofgeneralsolutionofthepotentialiswhereA1,A2,B1andB2arearbitraryconstants.SolutionwhereA1,A2,B1TheconditionforallrequiresthenTheconditionforallrequiresthenTheconditionforallrequires

Thus,weassumethatthegeneralsolutionisTheconditionforallrequiresTheconditionLet

ThisisaFouriersineseries.Thecoefficientsaredeterminedby第三課電磁場與電磁波英文版課件WecangetThus,thegeneralsolutionofthepotentialis第三課電磁場與電磁波英文版課件第三課電磁場與電磁波英文版課件

Example3.3.2Twosemi-infiniteconductingplatesseparateddaregrounded,asshowninFigure.AconductingplatebetweentwoplatesisheldatapotentialU0.Findthepotentialbetweenthetwoconductingplates.SolutionThepotentialcanonlydependonxandy.Laplace’sequationreducesto(1)Example3.3.2Solution(1)

Let(2)Substituting(2)in(1)anddividingby(2),weget

LetTheformofgeneralsolutionofthepotentialisLet

Usingseparationofvariables,wecanwritetheformofthesolutionas

whereandseparationconstantkaredeterminedbytheboundaryconditions.SolutionThepotentialcanonlydependonxandy.Laplace’sequationreducesto(1)Solution(1)TheconditionforallrequiresTheconditionforallrequires

SincethepotentialisequaltozeroatandweobtainTheformofthegeneralsolutionreducesto

TheconditionforallrequiresThecondition

ThisisaFouriersineseries.Thecoefficientsaredeterminedby

Thus,thegeneralsolutionofthepotentialis

第三課電磁場與電磁波英文版課件

2.SeparationofVariablesforCylindricalCoordinateSystem(two-dimensionalfield)AssumethatthepotentialisthefunctionwithrespecttoandTheLaplace’sequationis

(2)(1)Let(4)Substituting(2)into(1),weobtainandthenmultiplyingbyweobtain(3)2.SeparationofVariablesInmanyproblems,thepotentialisafunctionwithrespecttowithaperiod.Thatis(7)Thus,kmustbeaninteger.Taking(6)becomes(8)Eachtermmustbeaconstant.Thus,welet(5a)

(5b)Solving(5b),wehave(6)Inmanyproblems,thepotentiaSubstitutingnfork,theequation(5a)

becomes

(9)

ThisisanEulerequationandthesolutionis(10)Thus,thegeneralsolutioncanbewrittenasn2Substitutingnfork,the

Example3.3.3Averylongdielectriccylinderofradiusaisalongzaxisinauniformelectricfieldwhichisinthedirectionofxaxis.Determinethepotentialandtheelectricfieldintensity.

Example3.3.3SolutionTheformofthesolutionoftheLaplace’sequationisTheboundaryconditionsareasfollows.(1)when(2)whenisfinite.(3)when(4)when(1)(2)Solution(1)(2)Fromboundarycondition(1),wehaveThus,

LetthenBoundarycondition(2)requiresThus,Usingcondition(3),weobtain(3)Fromboundarycondition(1),wThus,LetwehaveThen,thepotentialisFromcondition(4),wehave(4)(6)(5)(4)(6)(5)Solving(5)and(6),weobtainThus,thepotentialsareUsingSolving(5)and(6),weobtainWeobtainE2E0exWeobtainE2E0ex圖均勻外電場中介質(zhì)圓柱內(nèi)外的電場圖均勻外電場中介質(zhì)圓柱內(nèi)外的電場

3.SeparationofVariablesforSphericalCoordinateSystem(two-dimensionalfield)AssumethatthepotentialisthefunctionwithrespectivetorandTheLaplace’sequationreducesto(1)Let(2)Substituting(2)into(1),weobtain(3)r2r2f(r)3.SeparationofVariablesMultiplyingbyweobtain(4)

Let(5a)(5b)

Let(5b)becomes(6)

ThisisthegeneralizedLegendreequation.（勒讓德方程）MultiplyingbyTake(7)

ThesolutionsareLegendrepolynomials

(8)TakeTheformertermsoftheare(9a)(9b)(9c)

(9d)(9e)

(9f)Theformertermsofthe第三課電磁場與電磁波英文版課件

Consider(5a).(10)

Thesolutionis(11)Thus,thepotentialis(12)

Constantsandaredeterminedbytheboundaryconditions.Consider(5a).

Example3.3.4Adielectricsphereofradiusaisinauniformelectricfieldwhichisinthedirectionofzaxis,asinshowninFigure3.3.5.Determinethepotentialandtheelectricfieldintensity.Example3.3.4

SolutionTheformofthesolutionoftheLaplace’sequationis

and

Useboundaryconditionstodetermineconstants.

（1）when

TheexpressionsofthefieldshaveonlythefirsttermThatis,（1）（2）（3）C1=-E0Solution（1）（2）（3）C1=-E(2)whenisfinitevalue.So(3)when(5)(4)whenweobtain

(7)Solving(5)and(7),weobtainn=1（4）（6）andThepotentialis(2)when

Thus,thepotentialsare

UsingweobtainThus,thepotentialsare圖均勻場中放進了介質(zhì)球的電場E0ez圖均勻場中放進了介質(zhì)球的電場E0ezAconductingsphereofradiusaisinauniformelectricfieldwhichisinthedirectionofzaxis,asinshowninFigure.Determinethepotentialandtheelectricfieldintensity.Aconductingsphereofradius3.4MethodofImages3.4MethodofImages

Example3.4.1Supposethatapointchargeqislocatedabovethesurfaceofaninfiniteconductingplaneandgrounded,asshowninFigure.Determine(a)theelectricpotentialandelectricfieldintensityabovetheplane,and(b)thetotalchargeinducedonthesurfaceoftheconductingplane.yzExample3.4.1yzAnelectricdipole：(1)Thepotentialatanypointonthebisectingplaneiszero,(2)Theelectricfieldintensityisnormaltotheplane.yzTheimaginarycharge–qissaidtobe

theimageoftherealcharge+qAnelectricdipole：yzTheimag

Solution(methodofimages)Theboundaryconditionis.Placeanimaginarycharge

–qat(0,0,-d)andtemporarilyignoretheexistenceoftheplane.ThepotentialatanypointPabovetheplaneis(zeropotentialatinfinity)TheelectricfieldintensityisSolution(methodofimages

whereand

ThenormalcomponentoftheDfieldmustbeequaltothesurfacechargedensityonthesurfaceoftheconductor.+d-dThus,thetotalchargeinducedonthesurfaceis+d-dThus,thetotalcharge

Basicprinciple(1)Imagechargesreplacethetotalchargeinduced.(2)Maintainthefieldequationandoriginalboundaryconditions.(3)Usetherealandimagechargestodeterminethefield.

Note：(1)Imagechargesarelocatedoutsidethefieldregion.(remainfieldequation)(2)Determinethenumber,magnitudeandpositionoftheimagecharges.BasicprincipleThenumberofimagechargesisThenumberofimagechargesis

Example3.4.2Apointchargeqislocatedatadistancedfromthecenterofagroundedconductingsphereofradiusa,asshowninFigure3.4.2.Computethesurfacechargedensityonthesurfaceoftheconductingsphere.Example3.4.2

Solution(methodofimages)TheboundaryconditionisPlaceanimaginarychargeat(0,0,d’)withinthesphere.Nowwetemporarilyignoretheexistenceoftheconductingsphere.ThepotentialoutsidethesphereisSolution(methodofimages

Fromtheboundaryconditions,weobtainSquaringbothsidesThisisanidentitywithrespectto.Thus,

Solvingtheseequations,weget

and第三課電磁場與電磁波英文版課件Chooseand.Thus,thepotentialoutsidethesphereis

SincethenormalcomponentoftheDfieldmustbeequaltothesurfacechargedensityonthesurfaceoftheconductor,wehaveChooseandThus,thetotalchargeinducedonthesurfaceisThus,thetotalchargeinduced圖球外的電場分布圖球外的電場分布（1）Apointchargeqisplacedatadistancedfromthecenterofaconductingsphereofradiusa.圖不接地金屬球的鏡像（2）AconductingsphereofradiusahasachargeofQ.Apointchargeqisplacedatadistancedfromthecenterofthesphere.qdaConsider圖點電荷位于不接地導體球附近的場圖（1）Apointchargeqisplaced

Example3.4.5

Theplaneistheinterfacebetweentwodifferentdielectrics,asshowninFigure(a).Apointchargeqislocatedabovetheinterface.Determinethepotentialintworegions.

(a)(b)(c)Example3.4.5(b)(c)

SolutionTheboundaryconditionsareandattheplaneRegion1:Placeanimaginarychargeat(0,0,-d).Wetemporarilyassumethepermittivityofmedium2isalsoThepotentialatanypointPinregion1isSolutionRegion2:Wesuperposeanimaginarychargeonthechargeq.Wetemporarilyassumethepermittivityofmedium1isalsoThepotentialatanypointPinregion2is

Region2:WesuperposeanAnypointontheinterfacesatisfiesFromboundaryconditions,wehaveSolvingtheseequations,weobtain

AnypointontheinterfacesatThus,thepotentialatanypointinregion1isThepotentialatanypointinRegion2is

Thus,thepotentialatanypoi第三課電磁場與電磁波英文版課件

Example3.4.2Alongstraightlinechargeisparallelalongstraightgroundedwireofradiusa,asshowninFigure.Calculatethepotentialinspace.Example3.4.2第三課電磁場與電磁波英文版課件

Solution(methodofimages)Placeanimaginarylinechargeonthey=0planewithinthewire.isparallelto.Temporarilyremovethewire.ThepotentialatanypointPis

Sincethelongstraightwireofradiusaisgrounded,thepotentialsatpointsAandB

areequaltozero.Solution(methodofimages

Thus,wehaveThus,wehaveLetWeobtainandThus,thepotentialatanypointPis

Trialsolution試探解LetTrialsolution試探解

Thesurfacechargedensityonthesurfaceofthewireis

ThechargeinducedperunitlengthonthesurfaceisMethodofelectricaxisThesurfacechargedensity

Example3.4.4Twoparallelcylindricalconductorsofradiusaseparatedadistance2dformtransmissionline,asshowninFigure.Calculatethecapacitanceperunitlengthoftwocylindricalconductors.yxExample3.4.4yx

Solution

Usingmethodofelectricaxis,thepotentialatanypointoutsideoftheconductorsis

ThepotentialatApointis

ThepotentialatBpointis

Thepotentialdifferenceisyx(C=0zeroreferenceat

yaxis)

Solutionyx(C=0zerorefer

Thecapacitanceperunitlengthoftwocylindricalconductorsis

Whenweobtainyxyx第三課電磁場與電磁波英文版課件Twolongstraightlineswithequalandoppositebutuniformchargedistributionsareseparatedbyadistanceof2b.Findthepotentialinfreespace.(C=0,zeroreferenceat

yaxis)

LetTwolongstraightlineswitheAsetofeccentriccirclesThus,weobtainTheequationofequipotentiallinesiscenterRadiusWhentheradiusisacircleAsetofeccentriccirclesThus根據(jù)，得到Ex和Ey分量LinesofforcesofE根據(jù)，得到Ex和E例不同半徑兩平行長直導線（傳輸線）相距為d,確定電軸位置，單位長度的電容。圖不同半徑傳輸線的電軸位置解：例不同半徑兩平行長直導線（傳輸線）相距為d,確定電軸位ABAB

鏡像法（電軸法）的理論基礎是：鏡像法（電軸法）的實質(zhì)是：鏡像法（電軸法）的關鍵是：鏡像電荷（電軸）只能放在待求場域以外的區(qū)域。疊加時，要注意場的適用區(qū)域。用虛設的鏡像電荷（電軸）替代未知電荷的分布，使計算場域為無限大均勻媒質(zhì)；靜電場惟一性定理；確定鏡像電荷（電軸）的個數(shù)、大小及位置；應用鏡像法（電軸法）解題時，注意：鏡像法（電軸法）的理論基礎是：鏡像法（電軸法）的實質(zhì)是3.5TheFinite-DifferenceMethod(FDM)3.5TheFinite-DifferenceMet

Numericalmethods:

finite-differencemethod(FDM),finiteelementmethod(FEM),methodofmoments(MOM).

Thefinite-differencemethod(FDM)

Numericalmethods:基本思想：將場域離散為許多網(wǎng)格，應用差分原理，將求解連續(xù)函數(shù)的微分方程問題轉(zhuǎn)換為求解網(wǎng)格節(jié)點上的代數(shù)方程組的問題。Dividethesolutiondomainintofinitediscretepointsandreplacethepartialdifferentialequationwithasetofdifferenceequations.基本思想：將場域離散為許多網(wǎng)格，應用差分原理，將求Ifthepotentialvariationisindependentofz,Poisson’sequationsimplifiesto

(1)First,wedividetheregionintoafinitenumberofmeshes.Letusconsiderasquaremeshsurroundingapoint0.Ifthepotentialvariation

Letthepotentialatpoint0beequaltoandatthefoursurroundingpointsandThepotentialatpoint1,usingtheTaylorseriesexpansion,is

wherehisthemeshsize.andthepotentialatpoint3isLetthepotentialatpointThus,wehaveOmittinghighorderterms,weobtain

(2)Similarly,weget(3)Adding(2)and(3),weobtain(4)

Thus,wehaveSubstituting(1)into(4),wehave

Thus,thepotentialatpoint0isIncharge-freeregion,thepotentialatpoint0is

Thepotentialatapointmustbetheaverageofthepotentialatthefoursurroundingpoint.

Substituting(1)into(4),weExample

Arectangulartroughisformedbyfourconductingplanes.Therectangulartroughisinfinitelylonginthez-direction.Thesidesandbottomofthetroughareatzeropotential.Thelidisatapotentialof100V.Wedividetheregioninto16squares.Example第三課電磁場與電磁波英文版課件

Initerativemethod,thepotentialforthe(n+1)thiterationatnode(k

j,)is

Initerativemethod,thep

Successiveover-relaxation(SOR)iterativemethod

whereiscalledaccelerationfactor.Gauss-Seideliterativemethod高斯—賽德爾迭代法逐次超松弛迭代法迭代過程直到節(jié)點電位滿足為止。收斂速度與電位初始值及網(wǎng)格剖分粗細有關；迭代次數(shù)與工程精度有關。Successiveover-relaxation邊界節(jié)點賦已知電位值賦節(jié)點電位初始值累計迭代次數(shù)n=0n=n+1按超松弛法進行一次迭代，求打印NoYes程序框圖下頁上頁返回邊界節(jié)點賦已知電位值賦節(jié)點電位初始值累計迭代次數(shù)n=0上機作業(yè)要求：1.試用超松弛迭代法求解接地金屬槽內(nèi)電位的分布。給定邊值：如圖示；已知：計算：迭代次數(shù)N=?,分布。給定初值：誤差范圍：下頁上頁返回圖接地金屬槽的網(wǎng)格剖分上機作業(yè)要求：1.試用超松弛迭代法求解接地金屬槽內(nèi)電位的有限差分法有限元法邊界元法矩量法積分方程法積分法分離變量法鏡像法、電軸法微分方程法保角變換法計算法實驗法解析法數(shù)值法實測法模擬法邊值問題有限差分法有限元法邊界元法矩量法積分方程法積分法分離變量法鏡3BoundaryValue

Problems3BoundaryValue

Problems109(1)Distribu