电磁场与电磁波第9讲静电问题的解法wb

FieldandWaveElectromagnetics电磁场与电磁波2012.3.201Maintopic4.SolutionofElectrostaticProblems3.MethodofImages1.Poisson’sandLaplace’sEquations2.UniquenessofElectrostaticSolutions4.Boundary-ValueProblemsinCartesianCoordinates2

The

relationship

betweentheelectricpotentialVandtheelectricfieldintensity

E

is

Takingthedivergenceoperationforbothsidesoftheaboveequationgives

Ina

linear,andisotropic

medium,thedivergenceoftheelectricfieldintensity

E

is1.Poisson’sandLaplace’sEquations3Thedifferentialequationforthe

electricpotential

iswhichiscalled

Poisson’sequation.

Ina

nofreecharge(source-free)

region,andtheaboveequationbecomeswhichiscalled

Laplace’s

equation.

4

Poisson’sequation

statesthattheLaplacian(thedivergenceofthegradient)ofVequals–/forasimplemedium,where

isthepermittivityofthemedium(whichisaconstant)and

isthevolumechargedensityoffreecharges.Operator,

2,theLaplacianoperator,whichstandsfor“thedivergenceofthegradientof,”or“”.Sincebothdivergenceandgradientoperationsinvolve

first-orderspatialderivatives.

Poisson’sequation

isa

second-orderpartialdifferentialequation

thatholdsateverypointinspacewherethesecond-order

derivatives

exist.

Remarks5InCartesiancoordinates:Insphericalcoordinates:Incylindricalcoordinates:6边值问题研究方法计算法解析法积分变换法分离变量法镜像法（电轴法）微分方程法保角变换法实验法作图法实测法模拟法定性定量数学模拟法物理模拟法数值法有限差分法有限元法边界元法矩量法半解析法/半数值法格林函数法7Example1.一维泊松方程的解ThetwometalplateshavinganareaA

andaseparationdformaparallel-platecapacitor.TheupperplateisheldatpotentialofV0

,andthelowerplateisgrounded.Determine(a)thepotentialdistribution(b)theelectricfieldintensity(c)thechargedistributiononeachplate(d)thecapacitanceoftheparallel-platecapacitor8Solution:Choose

anappropriate

coordinatesystem

forthegivengeometry2.Governingequation

forproblemsand

boundarycondition.匀强电场，电位V只是随高度z的变化而变化94.特解（带入边界条件求解未知系数）3.方程的通解1011Example2.

The

inner

conductorofradius

a

ofa

coaxialcable

isheldatapotentialof

V0

whiletheouterconductorofradius

b

is

groundedDetermine(a)the

potentialdistributionbetweentheconductors

(b)the

electricfieldintensity(c)the

chargedensity

ontheinnerconductor

(d)the

capacitanceofthe

perunitlength12Choose

anappropriate

coordinatesystem

forthegivengeometry2.Governingequation

forproblemsand

boundarycondition.Solution:134.特解（带入边界条件求解未知系数）

3.方程的通解141516Example3

Theupperandlowerconductingplatesofalargeparallel-platecapacitorareseparatedbyadistance

d

andmaintainedatpotentials

V0

and

0respectively.

Adielectricslabofdielectricrconstantanduniformthickness0.8d

isplacedoverthelowerplate.EandD

?yxD2D1E2E117(1)

求解区域：平行板电容器之间的区域(2)

分区：由于填充两种介质，因此场量在分界面上会发生突变，因此，分成两个子区域(3)

建立坐标系：竖直向上为y轴方向，建立坐标系(4)场分布分析：在两种介质中都是匀强电场，电位V只是随高度y的变化而变化V(y)，而与x，z无关，(5)

写出场方程与边界条件：待求量是两个区域的电位V1、V2，场方程：泊松方程（有源）or拉普拉斯方程（无源）yxD2D1E2E118区域1：区域2：yxD2D1E2E119

写出通解：一维边值问题BVP电位的边界条件，两个介质的衔接条件：2021yxD2D1E2E122uniquenesstheorem:meansthatasolutionofPoisson’sequation(ofwhichLaplace’sequationisaspecialcase)thatsatisfiesthegivenboundaryconditionsisauniquesolution.

Itdoesnotmeanthatonlyonemethodcanbeusedtoobtainthesolutionoftheelectrostaticproblem.Theimplicationoftheuniquenesstheoremisthatasolutionofanelectrostaticproblemwithitsboundaryconditionsistheonlypossiblesolution

irrespectiveofthemethodbywhichthesolutionisobtained.Asolutionobtainedevenbyintelligentguessingistheonlycorrectsolution2.UniquenessofElectrostaticSolutions23点电荷和带电的球壳、球体在R>a的区域中产生的场是是相等的，称为这三种源是相互等效的.注：在R<a的区域是不等效的，所以等效只是对某一区域等效，对另一区域是不等效的xyzxyzaxyza3.MethodofImages24yQdHalf－spaceproblemExample.

Considerthecaseofa

positivepointcharge

Q,locatedatadistancedabovealarge

grounded(zero-potential)conductingplane.

Theproblemistofindthepotentialateverypointabovetheconductingplane(y>0).(1)chap2：感应电荷很难求(2)直接解方程:25yQdHalf－spaceproblem点电荷＆感应电荷产生的场，静态平衡后，导体表面是等势面，电力线与其正交。而这种电力线的分布与以xoz平面为对称面，在（0，d，0）处点电荷Q，（0，－d，0）处有－Q的一对点电荷在x>0空间的电力线分布相似。(3)另辟蹊径：(等效原理)感应（极化）电荷产生的场，由假想的简单电荷（像点电荷线电荷等）分布产生的场来等效(4)问题：引入像电荷后求得的场，是不是原问题的场？判断的依据

（uniqueness

theorem）是不是满足原问题的场方程＆边界条件？26ImageChargeImagemethod

V(x,0,z)=0yQ–Q根据场叠加原理，写出点电荷和像电荷在上半空间任意一点P处产生的场的表达式BVPB－C（判断的条件）等效问题的场就是原问题的场27MethodofImage

Essence:

Theeffectoftheboundary

isreplacedbyoneorseveral

equivalentcharges,andtheoriginalinhomogeneousregionwithaboundarybecomes

aninfinitehomogeneous

space.

Basis：Theprincipleofuniqueness.Therefore,thesechargesshouldnotchangetheoriginalboundaryconditions.Theseequivalentchargesareattheimagepositionsoftheoriginalcharges,andarecalled

imagecharges,andthismethodiscalledthemethodofimages.

Key：Todetermine

thevalues

and

thepositions

oftheimagecharges.

Restriction：Theseimagechargesmaybedeterminedonlyforsome

specialboundaries

(infiniteplane,infinitelylongwedge,infinitelylongcylindrical,andsphericalboundaries)andchargeswith

certaindistributions.28q

Forthesemi-infinite

wedge

conductingboundary,themethodofimagesisalsoapplicable.However,theimagescanbefoundonlyforconductingwedgeswithanglegivenbywhere

n

isaninteger.Inordertokeepthewedgeboundaryatzero-potential,

several

imagechargesarerequired./3

Whenan

infiniteline

chargeisnearbyaninfiniteconductingplane,themethodofimagescanbeappliedaswell,basedonthe

principleof

superposition./3q29In

rectangularcoordinatesystem,Laplace’sEquationforelectricpotentialisLet

Substitutingitintotheaboveequation,anddividingbothsidesby

X(x)Y(y)Z(z),wehave

Whereeachterminvolves

onlyonevariable.Theonlywaytheequationcanbesatisfiedistohave

eachterm

equalto

aconstant.Lettheseconstantsbe,andwehave4.Boundary-ValueProblemsinCartesianCoordinates30

Thethreeseparationconstantsarenotindependentofeachother,andtheysatisfythefollowingequation

Thethree-dimensional

partial

differentialequationisseparatedtothree

ordinary

differentialequations,andthesolutionsoftheordinarydifferentialequationsareeasiertoobtain.

orwhereA,B,C,D

aretheconstantstobedetermined.where

kx

，ky

，kz

arecalledtheseparationconstants,andtheycouldbe

real

or

imaginary

numbers.

Ifkxis

anrealnumber,

Thesolutionoftheequationforthevariable

x

canbewrittenas31or

Thesolutionsoftheequationsforthevariables

y

and

z

havethe

sameforms.Theproductofthesesolutionsgivesthesolutionoftheoriginalpartialdifferentialequation.

Theseparationconstantscouldbeimaginarynumbers.Ifis

animaginarynumber,writtenas,thentheequationbecomes

Theconstantsinthesolutionsare

also

relatedtotheboundaryconditions.

Itisveryimportanttoselect

theforms

ofthesolutions,whichdependonthegiven

boundaryconditions.32Example.

Twosemi-infinite,groundedconductingplanesareparalleltoeachotherwithaseparationof

d.Thefiniteendisclosedbyaconductingplaneheldatelectricpotential

V0

,andisisolatedfromthesemi-infinitegroundedconductingplanewithasmallgap.Findthe

electricpotential

intheslot

constructedbythethreeconductingplanes.Solution:

Select

rectangular

coordinatesystem.Sincetheconductingplaneisinfinite

inthe

z-direction,thepotentialintheslotmustbe

independent

of

z,andthisisa

two-dimensional

problem.TheLaplace’sEquationfortheelectricpotentialbecomesdxyV=0V=0V=V0O33Usingthemethodof

separationofvariables,andlet

Theboundaryconditionsfortheelectricpotential

intheslot

canbeexpressedas

Inordertosatisfytheboundaryconditionsand,thesolutionof

Y(y)

shouldbeselectedas

Fromtheboundarycondition,wehave

V=0

aty=0,andtheconstant

B=0.Inordertosatisfy,theconstant

ky

shouldbe34WefindSince，weobtainTheconstant

kx

isanimaginarynumber,andthesolutionof

X(x)

shouldbeSince

V

=

0

at

x,theconstant

C=0,andThenWheretheconstant

C=AD

.35Since

V=V0

at

x=0

,andwehave

Therightsideoftheaboveequationisvariable,since

C

and

n

arenotfixed.Tosatisfytherequirementat

x=0,oneneedstotakethe

linearcombination

oftheequationasthesolution,leadingto

I