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电动力学习题解答第1页电动力学答案第一章电磁现象的普遍规律1.根据算符∇的微分性与向量性，推导下列公式：BBBBAAAABBBBAAAAAAAABBBBAAAABBBBBBBBAAAA)()()()()(∇⋅+×∇×+∇⋅+×∇×=⋅∇AAAAAAAAAAAAAAAA)()(221∇⋅−∇=×∇×A解：（1）)()()(ccAAAABBBBBBBBAAAABBBBAAAA⋅∇+⋅∇=⋅∇BBBBAAAABBBBAAAAAAAABBBBAAAABBBB)()()()(∇⋅+×∇×+∇⋅+×∇×=ccccBBBBAAAABBBBAAAAAAAABBBBAAAABBBB)()()()(∇⋅+×∇×+∇⋅+×∇×=（2）在（1）中令BBBBAAAA=得：AAAAAAAAAAAAAAAAAAAAAAAA)(2)(2)(∇⋅+×∇×=⋅∇，所以AAAAAAAAAAAAAAAAAAAAAAAA)()()(21∇⋅−⋅∇=×∇×即AAAAAAAAAAAAAAAA)()(221∇⋅−∇=×∇×A2.设u是空间坐标zyx,,的函数，证明：uufuf∇=∇dd)(，uuudd)(AAAAAAAA⋅∇=⋅∇，uuudd)(AAAAAAAA×∇=×∇证明：（1）zyxzufyufxufufeeeeeeeeeeee∂∂+∂∂+∂∂=∇)()()()(zyxzuufyuufxuufeeeeeeeeeeee∂∂+∂∂+∂∂=dddddduufzuyuxuufzyx∇=∂∂+∂∂+∂∂=dd)(ddeeeeeeeeeeee（2）zuAyuAxuAuzyx∂∂+∂∂+∂∂=⋅∇)()()()(AAAAzuuAyuuAxuuAzyx∂∂+∂∂+∂∂=dddddduuzuyuxuuAuAuAzyxzzyyxxdd)()dddddd(AAAAeeeeeeeeeeeeeeeeeeeeeeee⋅∇=∂∂+∂∂+∂∂⋅++=（3）uAuAuAzuyuxuuuzyxzyxd/dd/dd/d///dd∂∂∂∂∂∂=×∇eeeeeeeeeeeeAAAAzxyyzxxyzyuuAxuuAxuuAzuuAzuuAyuuAeeeeeeeeeeee)dddd()dddd()dddd(∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=zxyyzxxyzyuAxuAxuAzuAzuAyuAeeeeeeeeeeee])()([])()([])()([∂∂−∂∂+∂∂−∂∂+∂∂−∂∂=)(uAAAA×∇=3.设222)'()'()'(zzyyxxr−+−+−=为源点'xxxx到场点xxxx的距离，rrrr的方向规定为电动力学习题解答第2页从源点指向场点。（1）证明下列结果，并体会对源变量求微商与对场变量求微商的关系：rrr/'rrrr=−∇=∇；3/)/1(')/1(rrrrrrr−=−∇=∇；0)/(3=×∇rrrrr；0)/(')/(33=⋅−∇=⋅∇rrrrrrrrrr，)0(≠r。（2）求rrrr⋅∇，rrrr×∇，rrrraaaa)(∇⋅，)(rrrraaaa⋅∇，)]sin([0rrrrkkkkEEEE⋅⋅∇及)]sin([0rrrrkkkkEEEE⋅×∇，其中aaaa、kkkk及0EEEE均为常向量。（1111）证明：222)'()'()'(zzyyxxr−+−+−=○1rzzyyx'xrrzyx/])'()'()()[/1(rrrreeeeeeeeeeee=−+−+−=∇rzzyyx'xrrzyx/])'()'()()[/1('rrrreeeeeeeeeeee−=−−−−−−=∇可见rr'−∇=∇○23211dd1rrrrrrrrrrr−=∇−=∇⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛∇32'1'1dd1'rrrrrrrrrrr=∇−=∇⎟⎠⎞⎜⎝⎛=⎟⎠⎞⎜⎝⎛∇可见()()rr/1'/1−∇=∇○3rrrrrrrrrrrrrrrr×∇+×∇=×∇=×∇)/1()/1(])/1[()/(3333rrrr0301dd43=×−=+×∇⎟⎠⎞⎜⎝⎛=rrrrrrrrrrrrrrrrr○4rrrrrrrrrrrrrrrr⋅∇+⋅∇=⋅∇=⋅∇33331)/1(])/1[()/(rrrr03334=+⋅−=rrrrrrrrrrr，)0(≠r（2222）解：○13])'()'()'[()(=−+−+−⋅∂∂+∂∂+∂∂=⋅∇zyxzyxzzyyxxzyxeeeeeeeeeeeeeeeeeeeeeeeerrrr○20'''///=−−−∂∂∂∂∂∂=×∇zzyyxxzyxzyxeeeeeeeeeeeerrrr○3])'()'()')[(()(zyxzyxzzyyxxzayaxaeeeeeeeeeeeerrrraaaa−+−+−∂∂+∂∂+∂∂=∇⋅aaaaeeeeeeeeeeee=++=zzyyxxaaa○4rrrraaaarrrraaaaaaaarrrraaaarrrrrrrraaaa)()()()()(∇⋅+×∇×+∇⋅+×∇×=⋅∇因为，aaaa为常向量，所以，0=×∇aaaa，0)(=∇⋅aaaarrrr，又0000=×∇rrrr∵，aaaarrrraaaarrrraaaa=∇⋅=⋅∇∴)()(○5)]sin([)sin()()]sin([000rrrrkkkkEEEErrrrkkkkEEEErrrrkkkkEEEE⋅∇⋅+⋅⋅∇=⋅⋅∇0EEEE为常向量，00=⋅∇EEEE，而kkkkrrrrkkkkrrrrkkkkrrrrkkkkrrrrkkkk)cos()()cos()sin(⋅=⋅∇⋅=⋅∇，电动力学习题解答第3页所以)cos()]sin([00rrrrkkkkEEEEkkkkrrrrkkkkEEEE⋅⋅=⋅⋅∇○6)]cos()]sin([)]sin([000rrrrkkkkEEEEkkkkEEEErrrrkkkkrrrrkkkkEEEE⋅×=×⋅∇=⋅×∇4.应用高斯定理证明ffffSSSSffff×=×∇∫∫SVVdd，应用斯托克斯（Stokes）定理证明∫∫=∇×LSϕϕllllSSSSdd证明：（I）设cccc为任意非零常矢量，则∫∫×∇⋅=×∇⋅VVVV)]([ddffffccccffffcccc根据矢量分析公式)()()(BBBBAAAABBBBAAAABBBBAAAA×∇⋅−⋅×∇=×⋅∇，令其中ffffAAAA=，ccccBBBB=，便得ccccffffccccffffccccffffccccffff⋅×∇=×∇⋅−⋅×∇=×⋅∇)()()()(所以∫∫∫×⋅∇=×∇⋅=×∇⋅VVVVVV)(d)]([ddccccffffffffccccffffcccc∫⋅×=SSSSccccffffd)(ffffSSSSccccffffSSSScccc∫∫×⋅=×⋅=d)d(因为cccc是任意非零常向量，所以∫∫×=×∇ffffSSSSffffddVV（II）设aaaa为任意非零常向量，令aaaaFFFFϕ=，代入斯托克斯公式，得∫∫⋅=⋅×∇llllFFFFSSSSFFFFSSSSdd（1）（1）式左边为：∫∫×∇+×∇=⋅×∇SSSSSSaaaaaaaaSSSSaaaad][d)(ϕϕϕ∫∫⋅∇×−=⋅×∇=SSSSSSaaaaSSSSaaaaddϕϕ∫∫∇×⋅=×∇⋅−=SSϕϕSSSSaaaaSSSSaaaadd∫∇×⋅=SϕSSSSaaaad（2）（1）式右边为：∫∫⋅=⋅llllaaaallllaaaaddϕϕ（3）所以∫∫⋅=∇×⋅llllaaaaSSSSaaaaddϕϕS（4）因为aaaa为任意非零常向量，所以∫∫=∇×llllSSSSddϕϕS5.已知一个电荷系统的偶极矩定义为'd'),'()(VttVxxxxxxxxpppp∫=ρ，利用电荷守恒定律0=∂∂+⋅∇tρJJJJ证明pppp的变化率为：∫=VVttd),'(ddxxxxJJJJpppp证明：方法（I）∫∫∂∂==VVVttVttt'd]),(['d),(ddddx'x'x'x'x'x'x'x'x'x'x'x'x'x'x'x'ppppρρ∫∫⋅∇−=∂∂=VVVVtt'd)'('d),(x'x'x'x'JJJJx'x'x'x'x'x'x'x'ρ∫∫⋅∇−=⋅⋅∇−=⋅VVV'xVt'd)'('d)'(dd1111JJJJeeee''''xxxxJJJJeeeepppp'd])'()('[11V'x'xVJJJJJJJJ⋅⋅⋅⋅∇∇∇∇++++⋅⋅⋅⋅−∇−∇−∇−∇====∫∫∫∫电动力学习题解答第4页∫∫+⋅−=VxSVJ'x'd'd1SSSSJJJJ1111因为封闭曲面S为电荷系统的边界，所以电流不能流出这边界，故0'd1=⋅∫S'xSSSSJJJJ，∫=⋅VxVJt'ddd11eeeepppp同理∫=⋅VxVJt'ddd22eeeepppp，∫=⋅VxVJt'ddd33eeeepppp所以∫=VVt'dddJJJJpppp方法（II）∫∫∂∂==VVVttVttt'd]),(['d),(ddddx'x'x'x'x'x'x'x'x'x'x'x'x'x'x'x'ppppρρ∫∫⋅∇−=∂∂=VVVVtt'd)'('d),(x'x'x'x'JJJJx'x'x'x'x'x'x'x'ρ根据并矢的散度公式ggggffffggggfffffgfgfgfg)()()(∇⋅+⋅∇=⋅∇得：JJJJxxxxJJJJxxxxJJJJxxxxJJJJJxJxJxJx+⋅∇=∇⋅+⋅∇=⋅∇')(')(')()'(∫∫+⋅∇−=VVVVt'd'd)('ddJJJJJx'Jx'Jx'Jx'pppp∫∫+⋅−=VV'd)'(dJJJJJxJxJxJxSSSS∫=VV'dJJJJ6.若mmmm是常向量，证明除0=R点以外，向量3/R）（RRRRmmmmAAAA×=的旋度等于标量3/RRRRRmmmm⋅=ϕ的梯度的负值，即ϕ−∇=×∇AAAA，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：3/)/1rrrrrr−=∇（∵])1[()]1([)(3mmmmmmmmrrrrmmmmAAAA×∇×∇=∇××−∇=××∇=×∇∴rrrmmmmmmmmmmmmmmmm])1[()]1([1)(1)(∇⋅∇−∇⋅∇−∇∇⋅+∇⋅∇=rrrrmmmmmmmm]1[1)(2rr∇−∇∇⋅=其中0)/1(2=∇r，（0≠r）r1)(∇∇⋅=×∇∴mmmmAAAA，（0≠r）又)]1([)(3rr∇⋅−∇=⋅∇=∇mmmmrrrrmmmmϕmmmmmmmmmmmmmmmm])1[()1)(()()1()]1([∇⋅∇−∇∇⋅−×∇×∇−∇×∇×−=rrrr)1)((r∇∇⋅−=mmmm所以，当0≠r时，ϕ−∇=×∇AAAA7.有一内外半径分别为1r和2r的空心介质球，介质的电容率为ε，使介质球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）设场点到球心距离为r。以球心为中心，以r为半径作一球面作为高斯面。由对称性可知，电场沿径向分布，且相同r处场强大小相同。电动力学习题解答第5页当1rr时，01=D，01=E。当21rrr时，frrDrρππ)(34431322−=231323)(rrrDfρ−=∴，231323)(rrrEfερ−=，向量式为rrrrEEEE331323)(rrrfερ−=当2rr时，frrDrρππ)(344313232−=2313233)(rrrDfρ−=∴20313233)(rrrEfερ−=向量式为rrrrEEEE30313233)(rrrfερ−=（2）当21rrr时，)()(202202DDDDDDDDEEEEDDDDPPPPεεερ−⋅−∇=−⋅−∇=⋅−∇=pfρεεεε)1()1(020−−=⋅∇−−=DDDD当1rr=时，0)1()()(12020212=−−=−⋅−=−⋅−==rrpDDDDDDDDDDDDnnnnPPPPPPPPnnnnεεεεσ当2rr=时，frrprrrρεεεεσ22313202023)1()1(2−−=−=⋅=