电磁场与电磁波第8讲边界条件电容能量

11.ElectricPotentialReview22.ConductorsinStaticElectricField3.DielectricsinStaticElectricFieldInsideaconductor(understaticconditions)BoundaryConditions(ataConductor/FreeSpaceInterface)34.ElectricFluxDensityandDielectricConstant4Maintopic1.

静电场的边界条件2.

电容和电容器3.

静电场的能量51.BoundaryConditionsforElectrostaticFields

电磁问题通常涉及具有不同物理特性的媒质,并且需要关于两种媒质之间分界面的场量关系的知识.例如,我们希望知道场矢量E

和D

在穿过两个界面时是如何变化的.现在我们来考虑两种一般媒质之间的分界面.6(1)电场强度E的切向分量

(tangentialcomponent)Letusconstructasmallpath

abcdawithsidesab

andcdinmedia1and2respectively,bothbeingparalleltotheinterfaceandequalto

w.Theintegralformisassumedtobevalidforregionscontainingdiscontinuousmedia,isappliedtothispath.Ifweletsidesbc=da=

h

approachzero,theircontributionstothelineintegralofEaroundthepathcanbeneglected.ThuswehaveE1E2

2

1at

w

hacdban27

电场强度在分界面处的切向分量是相等的.换句话说,电场强度的切向分量连续.对于线性各向同性的一般介质,有:

an2

的方向是有媒质②

指向媒质

①.

8(2)电通密度D的法向分量(normalcomponent)Inordertofindarelationbetweenthenormalcomponentsofthefieldsataboundary,weconstructasmallpillboxwithitstopfaceinmedium1andbottomfaceinmedium2.Thefaceshaveanarea

S,andtheheightofthepillbox

hisvanishinglysmall.ApplyingGauss’slawtothepillbox,wehave

h

S

2

1an2D1D2

s9边界的法线方向由媒质②指向媒质①.

S

是边界上的面电荷密度.D场的法向分量在通过存在面电荷的分界面时是不连续的,不连续的量等于面电荷密度.10讨论(1)、BoundaryConditionsfor

Dielectric-conductorInterface

讨论(2)、

Intheabsenceofnetsurfacefreecharge,onehasTheboundaryconditionsthatmustbesatisfiedforstaticelectricfieldsareasfollows:11例1:两理想介质的分界面为Z=0的平面，如图所示，在介质2中的场强为求介质1中分解面上的场分量。

2=0

r2

1=0

r1xzy12132.

电容和电容器导体比例常数C

称为孤立导体的电容.电容是指电位每增加一单位所必须施加于物体的电荷量.14

电容器的电容是双导体系统的一种物理性质,其依赖于导体的形状和它们之间媒质的介电常数,与电荷Q

和导体之间的电位差V12都无关.电容器即使没有电压提供或导体上没有自由电荷它也是存在的.15CapacitanceCcanbedeterminedfromaboveequationbyeither(1)assumingaV12

anddeterminingQintermsofV12,or(2)assumingaQanddeterminingV12intermsofQ.Atthisstage,sincewehavenotyetstudiedthemethodsforsolvingboundary-valueproblems(whichwillbetakenupinChapter4),wefindCbythesecondmethod.Theprocedureisasfollows:1.对已知的几何形状,选择合适的坐标系.2.假设导体上携带了电荷+Qand-Q.3.通过高斯定理或其它关系根据假设的电荷量Q来确定E.4.求出导体两端的电位差V12.5.FindCbytakingtheratioQ/V12.16Example3-18P12417Example3-19P12518(1)Series＆ParallelConnectionsofCapacitorsParallelConnectionsofCapacitors电容器的电压相等

Seriesconnectionsofcapacitors(电容器的电量相等)(2)CapacitancesInMulticonductorSystems19静电能来源：外力克服电场力做功转化而来，静电场能仅与带电体的最终带电状态有关而与到达这一状态的中间过程无关。静电能：当电荷放入电场中，就会受到电场力的作用，电场力做功使电荷位移，这说明电场具有能量。静电场内储存着能量，这种能量通常被称为静电能。电场越强，对电荷的力就越大，做功的能力就越强，说明电场具有的能量就越大。3.ElectrostaticEnergy能量的零点：

最初电荷都分散在彼此相距很远(无限远)的位置上。通常规定，处于这种状态下的静电能为零。静电场能量We等于于把各部分电荷从无限分散的状态聚集成现有带电体系时抵抗静电力所作的全部功。20BringachargeQ2frominfinityagainstthefieldofachargeQ1

infreespacetoadistanceR12(1)TwochargeswhereV2isthepotentialatP2establishedbychargeQ1,chosethereferencepointforthepotentialatinfinity;Thisworkisstoredintheassemblyofthetwochargesaspotentialenergy.AnotherformwhereV1isthepotentialatP1establishedbychargeQ2.21Bringanothercharge

Q3frominfinitytoapointthatis

R13

tocharge

Q1

andR23

from

charge

Q2

infreespace,anadditionalworkisrequiredthatequals

whereV3

isthepotentialatP3

establishedbychargesQ1andQ2,W3,whichisstoredintheassemblyofthethreechargesQ1,Q2

,andQ3

,is(2)

Threecharges22WecanrewriteW3

inthefollowingformwhereV1

isthepotentialatQ1

establishedbychargesQ2andQ3,similarly,V2andV3arethepotentialsatQ2andQ3,respectively,inthethree-chargeassembly.23Extendingthisprocedureofbringinginadditionalcharges,wearriveatthefollowinggeneralexpressionfor

thepotentialenergyofagroupofN

discretepointchargesatrest.(ThepurposeofthesubscripteonWeistodenotethattheenergyisofanelectricnature.)WehavewhereVk

,

theelectricpotentialatQk,

iscausedbyalltheotherchargesandhasthefollowingexpression:(3)

AgroupofNdiscretepointchargesatrest24Tworemarksareinorderhere.First,Wecanbenegative.Inthatcase,workisdonebythefield(notagainstthefield).Second,Weinthisequationrepresentsonlytheinteractionenergy(mutalenergy)anddoesnotincludetheworkrequiredtoassembletheindividualpointchargesthemselves(self-energy).(4)acontinuous

chargedistributionWereplaceQkby

dvandthesummationbyanintegrationandobtain:NotethatWeinthisequationincludesthework(self-energy)requiredtoassemblethedistributionofmacroscopiccharges,becauseitistheenergyofinteractionofeveryinfinitesimalchargeelementwithallotherinfinitesimalchargeelements.25Example3-23P136Findtheenergyrequiredtoassembleauniformsphereofchargeofradiusbandvolumechargedensity

.

RbdR26(4)ElectrostaticEnergyinTermsatFieldQuantities

Recallingthevectoridentity,Wecanwriteas27SinceV’canbeanyvolumethatincludesallthecharges,wemaychooseittobeaverylargespherewithradiusR(R→

).Foralinearmedium,wehaveD=

E,28Wecanalwaysdefineanelectrostaticenergydensity

we

mathematically,However,thisdefinitionofenergydensityisar