E導(dǎo)體與電介質(zhì)課件

§1ConductorsintheElectrostaticField§2DielectricsandPolarization

intheElectricField§3CapacitanceandCapacitor§4TheEnergyofElectricfield

Chapter6ConductorsandDielectricsintheElectrostaticField§5Application1ElectrostaticEquilibriumInelectricconditionsallthefreeelectronareatrest2ConditionsofElectrostaticEquilibriumTheelectricfieldiszeroinsideaconductorProving1Notangentialcomponentcanexistab－－－++Proving2§1ConductorsintheElectrostaticFieldCorollary1Theconducingsurfaceisanequipotentialsurface,andallpointsintheinteriorareatthesamepotentialasthesurfaceCorollary2Thefieldatthesurfaceoftheconductor(andjustoutside)isperpendiculartothesurface3DistributionofChargea)AllofthechargesisonthesurfaceoftheconductorProve：b)TheelectricfieldonthesurfaceofaconductorsNote:Thefieldatthesurfaceoftheconductor(andjustoutside)c)DistributionofChargeofSphericalConductorRestrictingfactors：theelectricfieldiszeroinsideaconductor.5ElectrostaticScreening靜電屏蔽ElectrostaticScreening:InsulatingtheinfluenceofelectricfieldusingconductorshellNomatterthechangeofchargeAtheforceonBchargealwaysiszero.A電荷變化,B電荷受力總是為零。Thechargeinsideshellcanbeshielded殼可以屏蔽內(nèi)部IfthechargeBischanged,theforceonthechargeAischangedtoo.B電荷變化，A電荷受力會(huì)變化。ThechargeAoutsideshellcannotbeshielded殼不能屏蔽外邊電荷A。A。BIftheshellistheground,theinfluencebetweenAandBisnotexist.接地導(dǎo)體殼，可以使殼內(nèi)外的電荷彼此不影響。例：球形殼C，電荷A帶＋q,B帶＋Q，

r遠(yuǎn)大于R。求:A，B，C彼此之作用力。解：(1)A、B之間作用力：(2)A、C對(duì)B來說是帶電量q的點(diǎn)電荷，B受力：(3)靜電屏蔽，A受力 F=0+QBA+qCrR(4)A、C之間作用力：(5)B、C之間作用力F＝0(6)C接地，A、B受力為零問題：導(dǎo)體殼是否真的能夠屏蔽掉外電場(chǎng)使其不能進(jìn)入腔內(nèi)？答：接地導(dǎo)體殼return電介質(zhì)中不存在自由電荷，在電場(chǎng)中會(huì)發(fā)生極化現(xiàn)象。極性介質(zhì)：HCl、SO2、H2O非極性介質(zhì)：O2H2CO21Dielectricpolarization無外電場(chǎng)時(shí)，電偶極矩取向雜亂或?yàn)榱悖麄€(gè)介質(zhì)呈電中性。當(dāng)有外電場(chǎng)時(shí)，非極性分子產(chǎn)生極化，極性分子趨向外場(chǎng)排列，介質(zhì)表面產(chǎn)生束縛電荷。E---2Polarization

極化的強(qiáng)弱程度用電極化強(qiáng)度矢量描述。由于每個(gè)分子為偶極子，偶極矩P越大，單位體積內(nèi)偶極子越多，極化越強(qiáng)。線性各向同性介質(zhì)§2DielectricsandDielectricPolarization解：QR例1一點(diǎn)電荷Q被半徑為R的介質(zhì)1包圍，球外是2的介質(zhì)，求空間電場(chǎng)強(qiáng)度。例2平行極板間極化強(qiáng)度矢量，求介質(zhì)表面束縛電荷密度解：選取圓柱形高斯面return110-310-610-910-12FmFF微nF納pF皮1ThecapacitanceofanIsolatedConductorUnit：Farads法拉TheradiusoftheearthR＝6370kmC=7.08×10-4F§3CapacitanceCapacitorCapacitanceisameasureofamountofchargeaconductorholdsatagivenpotentialdifference2CapacitanceofaCapacitorCapacitor：Itisconsistsoftwoconductorplatesbeingveryclosed3CombinationsofCapacitors電容器的聯(lián)接(1)ConnectedinSeries串聯(lián)(2)ConnectedinParallel并聯(lián)Forcapacitorsconnectedinseriesthechargeoneachcapacitoristhesame.C1C2

C1

C2CapacitorsconnectedinParallelhavethesamepotentialdifferenceacrosstheirplateComparing彈簧的串并聯(lián)，電阻的串并聯(lián)，物理中由幾個(gè)相同量綱量組成另一個(gè)同量綱量時(shí)，只有上面的組合。(3)ACylindricalCapacitor-取高為h圓柱形高斯面RARBr

hNote：1Thecapacitanceofacapacitordependsonthegeometryofthecapacitorandthepermittivityε,independentof

thepotentialdifferenceandthecharge2ThesphericalcapacitorisastandardcapacitorEx1Aparallel-platecapacitorconsistsoftwosquareplates,eachofsides4.0cmseparatedby2.0mm.(a)Calculatethecapacitance.(b)Howmanyelectronscanbestoredonaplateifthecapacitorischargedto50V?Solution(a)(b)Q=CU=7.1×10-12×50=3.5×10-10CThemagnitudeofthechargeofoneelectronis1.6×10-19C,sothenumberNofelectronsthatcanbestoredis=7.1×10-12F=7.1pF.Ex2Asphericalcapacitorconsistsoftwothinconcentricsphericalshellsofradiiaandb,witha<b.AchargeQisplacedontheinnershell,and–Qontheoutershell,whichisgrounded.Calculatethecapacitance.InordertouseC=,weneedfirsttocalculatethepotentialdifferencebetweentheshells,whichinturndependsonthefield.Aconcentricsphereofradiusr,wherea<r<b,containschargeqso,byGauss’slaw,theelectricfieldonthisspherehasintensityE=UsingandsubstitutingforE`return2ThedensityofelectricfieldenergyTheenergystoredinacapacitorisintheformofelectricpotentialenergy.Whereistheenergy?Considerthespecificexampleofaparallel-platecapacitorItmeansthattheenergyisstoredintheelectricfield.Althoughthisdiscussionreferstoaparallel-platecapacitor,theconceptofenergydensityisdefinedbythesameequationforcapacitorsofanyshapeEnergydensity:theenergyperunitvolumeForacapacitorofanyshape,theenergystored解1用電場(chǎng)能量公式:在r處取半徑為r,r+dr的球殼,在球殼內(nèi)電場(chǎng)能量:解2導(dǎo)體球電容公式:例求半徑為R,帶電為+Q的導(dǎo)體球所具有的靜電能..B.A問題帶電平行板電容器充滿均勻介質(zhì),中間挖去一扁平的圓柱,A點(diǎn)和B點(diǎn)的D和E誰大?答案:EA>EBDA=DBreturn§5Application例1無限長均勻帶電圓柱面內(nèi)外的場(chǎng)強(qiáng)分布：>0，R面內(nèi)面外方向垂直柱面向外解：電荷分布具有柱對(duì)稱性，電場(chǎng)分布具有柱對(duì)稱性；利用高斯定理求場(chǎng)強(qiáng)分布，選取柱形高斯面rR

例3無限長均勻帶電圓柱面，半徑為R，單位長帶電量為+，試求其電勢(shì)分布。解：電場(chǎng)強(qiáng)度分布：r>R：r<R：

P1P2??電場(chǎng)強(qiáng)度方向垂直于帶電圓柱面沿徑向。若仍選取無窮遠(yuǎn)處為電勢(shì)零點(diǎn)，則導(dǎo)致電勢(shì)為無窮大選取距軸線nR處為零電勢(shì)點(diǎn)：

例4如圖，在一電荷體密度為的均勻帶電介質(zhì)球體中，挖出一個(gè)以O(shè)’為球心的球狀小空腔。試證：球形空腔內(nèi)的電場(chǎng)是均勻電場(chǎng)，其表達(dá)式為：證明：用填補(bǔ)法證明。介質(zhì)球在空腔內(nèi)任一點(diǎn)的場(chǎng)為：P點(diǎn)的合場(chǎng)強(qiáng)：??OO’空間任意一點(diǎn)的電場(chǎng)強(qiáng)度可看做體電荷密度為的介質(zhì)球體和一個(gè)空腔處