電磁場與電磁波第12講焦耳定律邊界條件電阻計算及第5章復(fù)習(xí)課件

FieldandWaveElectromagnetic電磁場與電磁波2015.10.3121.CurrentDensityandOhm’sLawReview33.EquationofContinuityandKirchhoff’sCurrentLaw2.ElectromotiveForceandKirchhoff’sVoltageLawOutsidethesourceInsidethesourceEConductingmediumPNEImpressedsourceEi5ForahomogeneousconductingmediumWeknowthatacurl-freevectorfieldcanbeexpressedasthegradientofascalarpotentialfield.LetuswriteSubstitutionofthisequationintoyieldsaLaplace’sequationin;thatis6Twofieldsarefoundtobe

verysimilar

insource-freeregion.

SteadyElectricCurrentField

ElectrostaticField

The

electric

currentdensity

J

correspondstothe

electric

fieldintensity

E,andtheelectric

current

linestotheelectricfield

lines.Insomecases,sincethesteadyelectriccurrentfieldiseasytobeconstructedandmeasured,theelectrostaticfieldcanbeinvestigatedbasedonthesteadyelectriccurrentfieldwiththesameboundaryconditions,andthismethodiscalled

electrostaticsimulation.7CapacitanceJEResistanceBasedontheequationsfortwofields,wecanfindtheresistanceandconductancebetweentwoelectrodesas9Maintopic恒定電流3.

電阻的計算1.

功率耗散和焦耳定律2.

電流密度的邊界條件101.

功率耗散和焦耳定律宏觀上，導(dǎo)體中的電子受電場的影響，發(fā)生漂移運(yùn)動；在微觀上，這些電子與晶格上的原子發(fā)生碰撞。因此，能量從電場傳到作熱振動的原子上。將電荷移動了一段距離?

，電場E

作的功為qE·?，則其所對應(yīng)的功率為：其中

u

為漂移速度。傳遞到體積

dv內(nèi)所有電荷載體的總功率為:11ThetotalelectricpowerconvertedintoheatinvolumeV:ThisisknownasJoule’slaw.ThepointfunctionE·Jisapowerdensityundersteady-currentconditions.Inaconductorofaconstantcrosssection,wecanwrittenas13E2E1

2

1atwhacdban2hS

2

1an2D1D2s14J2J1

2

1atwhacdban2hS2

1an2J1J2sthenormalcomponentofcurrentdensityvector

J

beingcontinuous.theratioofthetangentialcomponentsofcurrentdensityvectorJattwosidesofaninterfaceisequaltotheratiooftheconductivities.15

1,1an2E2,D2,J2s

2,2E1,D1,J1Whenasteady-currentflowsacrosstheboundarybetweentwodifferentlossydielectrics:17Example5-4P214:

Anemfisappliedacrossaparallel-platecapacitorofareaS.Thespacebetweentheconductiveplatesisfilledwithtwodifferentlossydielectricsofthicknessd1

andd2,permittivities1

and2

,andconductivities1

and2

respectively.Determine(a)thecurrentdensitybetweentheplates,(b)theelectricfieldintensitiesinbothdielectrics,and(c)thesurfacechargedensitiesontheplatesandattheinterface.x12++++++++++++---------------yo181.x12++++++++++++---------------yo2.MethodoneOrmethodtwo193.21Example5-6.一導(dǎo)電材料形狀是四分之一的扁平圓墊圈,其均勻厚度為h,電導(dǎo)率為

,墊圈內(nèi)半徑為

a,外半徑b,如下圖所示,求端面之間的電阻.思路V0yxhabr0(r,)022Solution:Thecylindricalcoordinatesystemshouldbeselected.Assumetheelectricpotentialdifferencebetweentwoendfacesis

V0,andlet

SincetheelectricpotentialV

isrelatedtotheangle,itshouldsatisfythefollowingequationThegeneralsolutionisTheelectricpotentialatTheelectricpotentialatV0yxhabr0(r,)023

Basedonthegivenboundaryconditions,wefindThecurrentdensity

J

intheconductingmediumis

ThenthecurrentI

flowingintotheconductingmediumacrosstheendfaceatisConsequently,theresistance

R

betweentwoendfacesis252.BoundaryConditionsforCurrentDensityDifferentialformIntegralformGoverningEquationsforSteadyCurrentDensityWhenasteady-currentflowsacrosstheboundarybetweentwodifferentlossydielectrics:26homeworkThankyou!Bye-bye!答疑安排時間：地點(diǎn)：1401，1403P.5-15294.PowerDissipationandJoule’sLaw6.ResistanceCalculationselectrostaticsimulation305.BoundaryConditionsforCurrentDensityDifferentialformIntegralformGoverningEquations