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电动力学习题解答第一章电磁现象的普遍规律-1-1.根据算符∇的微分性与矢量性推导下列公式BABAABABBArrrrrrrrrr)()()()()(∇⋅+×∇×+∇⋅+×∇×=⋅∇AAAAArrrrr)(21)(2∇⋅−∇=×∇×解1BABAABABBAvvvvvvvvvv)()()()()(∇⋅+×∇×+∇⋅+×∇×=⋅∇首先算符∇是一个微分算符其具有对其后所有表达式起微分的作用对于本题∇将作用于BAvv和又∇是一个矢量算符具有矢量的所有性质因此利用公式bacbcabacvvvvvvvvv)()()(⋅−⋅⋅=××可得上式其中右边前两项是∇作用于Av后两项是∇作用于Bv2根据第一个公式令AvBv可得证2.设u是空间坐标xyz的函数证明.)()()(duAduuAduAduuAududfufrrrr×∇=×∇⋅∇=⋅∇∇=∇证明1ududfezududfeyududfedudfezufeyufexufufzyxxuzyx∇=∂∂⋅+∂∂⋅+⋅=∂∂+∂∂+∂∂=∇∂∂rrrrrr)()()()(2duAduzudzuAdyuduuAdxuduuAdzuzAyuAxuAuAzyxzyxrrrrrrrr⋅∇=∂∂⋅+∂∂⋅+∂∂⋅=∂∂+∂∂+∂∂=⋅∇)()()()()()()(3=∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=∂∂∂∂∂∂=×∇zxyyzxxyzzyuxzyxeyAxAexAzAezAyAuAuAAzyxeeeuArrrrrrrrrrrrrrrr)()()()()()()(电动力学习题解答第一章电磁现象的普遍规律-2-duAdueyuduAdxuduAdexuduAdzuduAdezuduAdyuduAdzxyyzxxyzrrrrrrrrrr×∇=∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=)()()(3.设2'2'2')()()(zzyyxxr−+−+−=为源点'x到场点x的距离r的方向规定为从源点指向场点1证明下列结果并体会对源变数求微商)(''''zeyexezyx∂∂+∂∂+∂∂=∇rrr与对场变数求微商)(zeyexezyx∂∂+∂∂+∂∂=∇rrr的关系)0.(0,0,11,3'333''≠=−∇=⋅∇=×∇−=−∇=∇=−∇=∇rrrrrrrrrrrrrrrrrrrr(最后一式在人r0点不成立见第二章第五节)2求均为常矢量及其中及000,)],sin([)]sin([),(,)(,,EkarkErkErararrrrrrrrrrrrrrrrr⋅×∇⋅⋅∇⋅∇∇⋅×∇⋅∇证明3)()()('''=∂−∂+∂−∂+∂−∂=⋅∇zzzyyyxxxrr0'''=−−−∂∂∂∂∂∂=×∇zzyyxxzyxeeerzyxrrrr])'()'()')][(()[()(zyxzyxzzyyxxezzeyyexxezeyexeaeaearavrvvvvvvvrv−+−+−∂∂+∂∂+∂∂⋅++=∇⋅])'()'()')[((zyxzyxezzeyyexxzayaxavrv−+−+−∂∂+∂∂+∂∂=aeaeaeazzyyxxvvvv=++=ararrararavvvrvvvvvv⋅∇⋅+×∇×+∇⋅+×∇×=⋅∇)()()()()(aararravrvvvvv⋅⋅+×∇×+∇⋅=)()()(araravvvvv⋅∇⋅+×∇×+=)()())(sin()](sin([)]sin([000ErkErkrkErrrrrrrrr⋅∇⋅+⋅⋅∇=⋅⋅∇电动力学习题解答第一章电磁现象的普遍规律-3-0])sin()sin()sin([Eerkzerkyerkxzyxrrrrrrrrr⋅∂∂+⋅∂∂+⋅∂∂=))(cos())(cos(0EkrkEekekekrkzzyyxxrrrrrrrrrr⋅⋅=++⋅=000)sin()]sin([)]sin([ErkErkrkErrrrrrrrr×∇⋅+×⋅∇=⋅×∇4.应用高斯定理证明∫∫×=×∇SVfSdfdVrrr应用斯托克斯Stokes定理证明∫∫=∇×LSldSdφφrr证明1)由高斯定理∫∫⋅=⋅∇SVgSdgdVrrr即∫∫++=∂∂+∂∂+∂∂SzzyyxxVzyxdSgdSgdSgdVzgygxg)(而dVkfyfxjfxfzifzfydVfxyzxyzV])()()[(rrrr∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=×∇∫∫∫−∂∂+−∂∂+−∂∂=dVifjfzkfifyjfkfxyxxzzy)]()()([rrrrrr又])()()[(kSdfdSfjdSfdSfidSfdSffSdySxxyxzzxzyyzSrrrrr∫∫−+−+−=×∫−+−+−=zyxyxzxzydSifjfdSkfifdSjfkf)()()(rrrrrr若令ifjfHkfifHjfkfHyxZxzyzyxrrrrrr−=−=−=,,则上式就是∫∫⋅=⋅∇SVHSddVHrrr,高斯定理则证毕2)由斯托克斯公式有∫∫⋅×∇=⋅SlSdfldfrrrr∫∫++=⋅lzzyyxxldlfdlfdlfldf)(rr∫∫∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=⋅×∇SzxyyzxxyzSdSfyfxdSfxfzdSfzfySdf)()()(rr而∫∫++=lzkyjxildldldlld)(φφφφr电动力学习题解答第一章电磁现象的普遍规律-4-∫∫∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=∇×SyxxzzySkdSxdSyjdSzdSxidSydSzSdrrrr)()()(φφφφφφφ∫∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=zyxdSiyjxdSkxizdSjzky)()()(rrrrrrφφφφφφ若令kzjyixfffφφφ===,,则证毕5.已知一个电荷系统的偶极矩定义为,),()('''∫=VdVxtxtPrrrρ利用电荷守恒定律0=∂∂+⋅∇tJρr证明Pr的变化率为∫=VdVtxJdtPd''),(rrr证明∫∫∇−=∂∂=∂∂VVdVxjdVxttP'''''''rrrrrρ∫∫∫⋅∇−=⋅∇−⋅∇−=∇−=∂∂VxVxdVjxjdVjxjxdVxjtP'''''''''''''''')((])()([)(rrrrr∫∫⋅−=SxSdjxdVjrr'若)0(,0)(,==⋅∞→∫SjSdjxSrrr则同理∫∫=∂∂=∂∂'')(,)(dVjtdVjtzzyyρρrr即∫=VdVtxjdtPd''),(rrr6.若mr是常矢量证明除R0点以外矢量3RRmArrr×=的旋度等于标量3RRmrr⋅=ϕ的梯度的负值即ϕ−∇=×∇Ar其中R为坐标原点到场点的距离方向由原点指向场点证明mrmrrmrmRmRRmAvvvvvvvv])1[()]1([1)(1)()]1([)(3∇⋅∇−∇⋅∇−∇∇⋅+∇⋅∇=∇××−∇=××∇=×∇电动力学习题解答第一章电磁现象的普遍规律-5-)0(,1)(≠∇∇⋅=rrmvrmmrrmrmRRm1)()()1()]1([)]1([)(3∇∇⋅−×∇×∇−∇×∇×−=∇⋅−∇=⋅∇=∇vvvvvvϕrmmr1)(])1[(∇∇⋅−=∇⋅∇−vvϕ−∇=×∇∴Av7有一内外半径分别为r1和r2的空心介质球介质的电容率为ε使介质内均匀带静止自由电荷fρ求1空间各点的电场2极化体电荷和极化面电荷分布解1∫∫=⋅dVSdDfSρrr,(r2rr1)frrrDρππ)(3443132−=⋅即)(,3)(123313rrrrrrrEf−=∴rrερ由)(,)(342313200rrrrQSdEffS−==⋅∫ρεπεrr)(,3)(2303132rrrrrrEf−=∴rrρε01时Errr2)EEEPerrrr)(00000εεεεεεχε−=−=)(3]3)([)()(3310331300rrrrrrrrEPffPrrrrr−⋅∇−−=−⋅∇−−=⋅∇−−=⋅−∇=∴ρεεερεεεεερffρεεερεεε)()03(300−−=−−−=nnPPP21−=σ考虑外球壳时rr2n从介质1指向介质2介质指向真空02=nP电动力学习题解答第一章电磁现象的普遍规律-6-frrfnPrrrrrrrPρεερεεεσ32313203313013)1(3)(2−−=−−===r考虑到内球壳时rr203)(133130=−−−==rrfPrrrrrρεεεσ8内外半径分别为r1和r2的无穷长中空导体圆柱沿轴向流有恒定均匀自由电流Jf导体的磁导率为µ求磁感应强度和磁化电流解flSfISdDdtdIldH=⋅+=⋅∫∫rrrr当0,0,1===BHIrrfrr故时当r2rr1时)(2212rrjSdjrHldHfSfl−=⋅==⋅∫∫ππrrrrrjrrrrrrjBffrrv×−=−=22122122)(2)(µµ当rr2时)(22122rrjrHf−=ππrjrrrBfrrr×−=2212202)(µ)2()1())()(2212000rrrrjHHMJfMM−××∇−=−×∇=×∇=×∇=rrrrrµµµµµχ)(,)1()1(2100rrrjHf−=×∇−=rrµµµµ指向介质从介质21(),(12nMMnMrrrr−×=α在内表面上0)2)1(,012212021=−−===rrrrrMMµµ故)(,012rrMnM==×=rrrα在上表面rr2时)1(22)(0212221211222−−−=×−×−=×−=−×===µµαrfrrfrrMjrrrrjrrrrrMnMnrrrrrrrrrfjrrrr2212202)1(−−−=µµ电动力学习题解答第一章电磁现象的普遍规律-7-9证明均匀介质内部的体极化电荷密度Pρ总是等于体自由电荷密度fρ的倍)1(0εε−−证明ffPEEPρεεερεεεεεερ)1()()()(0000−−=−−=⋅∇−−=−⋅−∇=⋅−∇=rrr10证明两个闭合的恒定电流圈之间的相互作用力大小相等方向相反(但两个电流元之间的相互作用力一般并不服从牛顿第三定律)证明1线圈1在线圈2的磁场中的受力∫×=23121222024lrrldIBvvvπµ21112BldIFdvvv×=∫∫∫∫××=××=∴12123121221210312122211012)(4)(4llllrrldldIIrrldIldIFvrvvvvvπµπµ)()(41221312123121212210∫∫⋅−⋅=llldldrrrrldldIIvvvvvvπµ12线圈2在线圈1的磁场中受的力同1可得∫∫⋅−⋅=21)()(41232121321212121021llldldrrrrldldIIFvvvvvvvπµ2分析表达式1和21式中第一项为0)1()(21221212221212231212123121212=−⋅==⋅=⋅∫∫∫∫∫∫∫llllllrldrdrldrrldldrrldld一周vvvvvvvv同理对2式中第一项∫∫=⋅210)(3212121llrrldldvvv∫∫⋅−==∴12)(421312122102112llldldrrIIFFvvrvvπµ11.平行板电容器内有两层介质它们的厚度分别为l1和l2电容率为21εε和今再两板接上电动势为Ε的电池求1电容器两板上的自由电荷密度fω电动力学习题解答第一章电磁现象的普遍规律-8-2介质分界面上的自由电荷密度fω若介质是漏电的电导率分别为21σσ和当电流达到恒定时上述两问题的结果如何解在相同介质中电场是均匀的并且都有相同指向则,)00f2211212211==−=−Ε=+σεε介质表面上EEDDElElnn故122112122121,εεεεεεllEllE+Ε=+Ε=又根据fnnDDσ=−21n从介质1指向介质2在上极板的交面上121fDDσ=−D2是金属板故D20即12212111εεεεεσllDf+==而02=fσ)0(,'1'1'2'2'13=−=−=DDDDDf是下极板金属故σ13122121ffllσεεεεεσ−=+−=∴若是漏电并有稳定电流时222111,σσjEjErrrr==又