华侨大学电磁场与电磁波复习题及答案2017

计算题25道S=12户+5^+9andf=2r+0.5^9aretwovectorsatpoint(2,^,^/2)and(5,#2,#2),respectively.Determine(a)S+T,(b)S-T,(c)5xfz(d)theunitvectorperpendiculartoSxTand(e)theanglebetweenSandT题2.20-sin(p-sin(pCOS(ffsin0cos。

sin8sin(p

cosOcos0cos。cos0cos。-sin0解:0S=0-1-100125JI=-?rx-\2z0 0T=1 00-1-1002

J0,5元0=2,—05*(q)$十7=(一”工=123)十(23一0.5”日=-江£+25一(12+0・5%”(b)S T=(一不£=123b(22~0.5¥2)=6笈x y z(c)5xf=-兀 0 -12=24x-0.5^p-2/rf0 2 -0.5(d)设a,/?(d)设a,/?为常数,则为aSa-fitaS+jBf\IftheelectricfieldintensityinspaceisgivenasE=Eocos0r-EQsin6瓦find▽・EandVx£.题2.42

解:N・E=-4—(r解:N・E=-4—(r24)+—!———(Xpsin^)4-

产drrrsin<9dO°rsinBd(prsin0dff(-Eosin3sin3)=2rE,0cos0+ (-2E()sin3cos「」 r^in6=0Ir rOrsin&q)V7产16 6 6▽xE=-：—；——I — sin夕dr c6 ctp\E^cos〃一sin0 0=- 〃sin(一E。sin，+sin0),-sin。=0GivenF=x^x+jryy+x2^,verifythedivergencetheoremwhentheregionisboundedbyacylinderx2+y2=16andtheplanesatz=Oandz=2.题2.44解:v.户=必+匕十送=3/+/+/=5/dx-aydzW-PdV=j[广^5x'dxdydz=£££5p~cos诙勿d她JO-PAcos24JO-PAcos24=f[16O0+16Ox」0sin202i、l+cos2。320——； d@dwdz=640/r下面求JFds①z=0,"s=-zdxclyPels=-①z=0,"s=-zdxclyPels=-x2zdxdy=0②z=^ds=zdxdyx2zdxdy=lxx2zdxdy=lx2dxdy=]1202cos2epdpd®-128”③/+/=]6Ms=ppdf^dzJFds=j[(xyxx1yy^ppei^dz「2F盾_a , -e=11 cos'。cos4+"、cos-^sin^sin^)pd^dz二20256cod阿。=512工.•.[户/=128兀+512乃=640%因此[户为=[.▽・户(W,得证。Inelectromagneticfields,wewilldefinetheelectricfieldintensity£asanegativegradientofascalarfunction;Le.£=-V^?.Wewillalsodefinethevolumechargedistributionaspv= Etwhereqisthepermittivityoffreespace.Determinedandpvif(a)(p=VQ(p\n(<p/a)incylindricalcoordinates,whereand。areconstants,(b)</?=V^rcos^insphericalcoordinates,and^?=V^rsm^insphericalcoordinates.题2.38解：（a）在柱坐标下E=—V(p、6中 7d(p.&s=-" 0 Z dp60czIn—0--p1In—6(b)在球坐标系下E=—Vw.fd中-d(p2 d①=—r 8 0 crrd6rsin6^^“C五Krsin夕=—rV{ycosG+0- r=—5匕)cos0+@Vqsin8a=0▽•左=£。［5*'叫)+焉［剑明)+焉富=/［，2'%cosSJ 2%sindcos6［\r* Jrsin0=0(c)在球坐标系下E=K中orrdU/sina@=一5匕sine—I//cos。=—尸匕sin，一夕匕cos， -(sin也)+ -rsin818、 rsin08,匕2sin。 Jrsin0匕cos29=—/(2sin，十cos203.6Anelectroniscirclingaroundnucleusofahydrogenatomataradiusof0.05nm.Determinetheangularvelocityandthetimeperiodoftheelectron.题3.6Twostraightuniformlychargedlinesareextendingalongthezaxisfromz=-toz=+s.Onelinecarriesachargedistributionofl//C//wandissituatedaty=-3//7,andtheotherlineisaty=3〃?andhasachargedistributionof-1//C//K•DeterminetheEfieldatappointonthexaxisatx=4〃?.题3.13Threeinfiniteelectricsheetscarryinguniformchargedensitiesof2,-3and0.5/.iC/irrareeachseparatedbya1-mmairgap,asshowninfigure.Findtheelectricfieldintensityeverywhereinspace.题3.14解：由于两板间距很小，所以可以忽略边缘效应，根据高斯定理有在0.5平面上，有E= =0.5X10-6/（2X8.85X1卜2,825x]04在2平面下，有反=①/2%=2x12x8.85X10"）=L13x1O'在-3和0.5的平面之间1=3-2々）//=-2.5x10-6/（2x8.85x10-12）=-1.412xl05在-3和2的平面之间&=（0L4）及=Tx10~y（2x8用5x10*）=-5.65x1043.16Afinitelineextendsfromz=-10/z?toz=10/wandcarriesachargedistributionofl00^iC//K.DeterminetheEfieldatapoint2metersawayfromthelineinthe^=Oplane.题3.163.16一条从z=-10相向z=10/〃延伸的直线上，电荷分布为100z〃C7，〃，求z=0平面内离直线两米远的点上的E。解：场点（2,60）,场点矢量7=27微线元M二小，微元矢径：r=zz4^()\zz-2p\f^]00处25rio

点0L。oZ-+zdz2525rio

点0L。oZ-+zdz25FfLo(z2+4-4)zz~+dz-Xio25九%f°—4?L。产丁飞一产「HP中4油于被积函数是奇函数,且积分上下限关于原点对称，故=0-25 /E=——21n-25 /E=——21n5+V26)-—V26^2.4xl0,2/7V/m=2.4xl03V/m1Asphericalvolumeofradiusbsupportsachargedistributionofk/r2everywhereexceptatr=0.Determinetheelectricfluxdensityintheregion(a)r<b,and(b)r>b题3.25

半径为b的球体内除r=0外电荷处处为〃//，求r<力和的电位移矢量力。解：①r<b根据高斯定理缶=。得到D47rr2r2sinOdrddd(p=］：££a-sinD47rr2=2笈泳［-cos®］；=4^rrk:.D=—②">b=5-4〃/=f「j—sina/n②">b=5-4〃/=f「j—sina/n■/a/*=f s*nOdrclOd(p一■,r 八M一， bkAlongcoaxialcableconsistsofaninnerconductorofradiusaandanouterconductorofinnerradiusbandouterradiusc.Theinnerconductorcarriesauniformsurfacechargedistributionps.FindtheEfieldatanypointinspacewhen(a)theouterconductorisnotgrounded,and(b)theouterconductorisgrounded.题3.413.41一长直同轴导线，内导休半径为a,外导体的内半径为b,外半径为a内导体带有均匀面电荷2，求下面两种情况中空间任意点的左，(1)外导体按地,(2)外导体不接地。解：设内导体单位长度的电荷量为巧(1)外导体接地时,外导体的外袤面无电荷，故外导体的内表面的面电荷密度当工工〃时，JL)-cls=L)2rvrl=0/.D=0,F=0当力时，f力•4=力2m•/=0=/7八•.5=2-=三竺乌=卫2/rr27r/-r当/”用虫［力.加=方2”〃/=0.•・力=0,月=0FindthemagneticfluxdensityatpointP.题5.5100A10cm

解：以P为坐标原点建立坐标系。对于P右边的半圆有Idl=/pd(p(p.R=-pp/d7xR=Ip2dg*二及1/C=如1/^=”.不5="2=6.28><10-4订4〃为尸4〃力p 4%7 4/7对于P左边的长方形的上下边，利用有限长直线电荷的磁感应强度月=2//(cos，I-cos仇)，由于2m5a〃，可以看成半无限长的导线，Arip9X=90(),cos^="。2=180°,cos4=-1,因为磁感应强度的方向相同，故反=2%&=W®2=4xlOF4邛) Tip对于最左边的竖导线在P点的氏=0,故6=1.028X10-3)Ametalsphereofradiusahasuniformsurfacechargedistribution.Thepermittivityofthesurroundingregionvariesas^=£0(l+«/r).Find(a)D,E,andPeverywhereinspace,(b)theboundchargedensities,and(c)theenergydensity,and(d)showthatthepotentialinthedielectricregionisV=-2―，whereQisthetotalchargeonthesphere.4蔗题3.453.45一个半径为b的金属表面电荷均匀分布，周围介质的电容率£=£。(1+a/〃),求(1)空间处处的%及户(2)束缚电荷密度(3)能量密度(4)证明介质区域内的电位P=—ln(l+o")。式中。为金属球的总电荷。4/4解：(a)r<Z)5jDds=0=>E=O.P=0r>byf)ds=Q=>4兀D,=。=>/),=、Er=—= ,P=D-eQE=£■ ——=一空一，4外・4(1+4 48厂加户。+与41厂(,十〃)—aQ

4rrb~(b+a)体密度e=一▽•户=一—aQ

4rrb~(b+a)体密度e=一▽•户=一1d(aQyr2dr\47T(r+a))-aQ4^r2(r+a)2S)介质面上的束缚电荷面密度〃=户./7=一尸(c)能量密度取二:力.互二(二']---2 京]+与r(d)V=[ Q dr=-^~4q£(1+与 4加%L 130--In =---In =---ln(l+—)/a|_尸+4」，4丹a r4庇声rThemagneticfluxdensityinaregionl^>0,isB=1.5x+0.8y+0.62wT.Ifz=0markstheboundarybetweenregions1and2,determinethemagneticfluxdensityinregion2.Considerregion1asfreespace;regions2hasarelativepermeabilityof100.题5.355.35在z>0的区域1内磁通密度为月=1.5£+0.8夕+0.6如若z=0为区域1与区域2的分界面。区域1为自由空间，区域2的相对磁导率为100。试求区域2的磁通密度。解：由于在空气跟电介质的分界面上面电流，因而月的法向分量连续，，的切向分量连续。%=0.6"%4:%〃2=1。。回B2i=1005,,=IOO(1.5i+0.8用=150.r+80/用=150£+80歹+0.6式/")Acurrentsheetofl2ykA/mseparatestworegionsatz=0.Region2,z>0,hasamagneticfieldintensityof4O.v+50y+12£kA/mandarelativepermeabilityof200.Ifregion1,2<0,hasapermeabilityof1000,determinethemagneticfieldintensityinthisregion.题5.36

5.36一个载电流为〃的薄片，，在z=0处分开两个区域。2>0的区域2的磁场强度为40x+50y+12zkA/m,相对磁导率为2000若z<0的区域1的相对磁导率为1000,求次区域的磁场强度。解：由于启的法向分量连续，所以%=%= =>1000%,=200(122).••%=2车根据电磁场的边界条件，有(―凡)=工=>iX(/7,rX+/fh.-zX(40x+50y)^-\2y=>HXxy-H、vx-40y+50H=-12yn/=28M”50：.Hx=28x+50J+2Az(kA/m)Capacitance?Depth-10cm6=3.6 10cm10cm10cm题3.58解：S]=工=S.=0.1xO,l"=0,01/w2,d=0,00L〃G=% %M10=90,=7.965X10-"。2=向2=36%=3.186乂10的尸区=20%=1.77x10-叮d»因为三个电容并联，所以C=和+Q+g=1292xIO^FCapacitance?Area=lOcm*10cm0.5mm0.5mm题3.59解：S=0.lx0.l仇2=0.01加:4=乩=0.0005掰sC：=苞—X20=180%=L593X10一9产4C=&9=724=6,372x10-"ch■因为两个电容串联，所以C=O二=4,55xIL。G+G17.3.60Capacitance?(a=lcm,b=1.5cm)题3.6017.3.60Capacitance?(a=lcm,b=1.5cm)题3.60解：M=乩=0.0005w,/=10孙b/a=1.51 2叫 50% 夕广G=—x = —3.43x10r2皿〃“)lnl.51 2麻J 25gl1 2麻J 25gl=—x -= 2In(b/a)In1.5=1.71x10^F由于可以看成两个电容并联，C=C+a=5.14xl(r9尸.1 43.61Capacitance?(红色为极板)

（xoy方向俯视图，z方向高度为h=10cm）题3.61解:整个圆柱的电容小届24x3.6x8£整个圆柱的电容小届24x3.6x8£5xl()T2乂

ln3=1.822x10-"45°-3()°・・.C= —xl.822x10一"=7.592x10一心F360°3.62Capacitance?(球形)题3.62题3.62解：设导体球带电分别为土Q,则M 儿4的八入厂 &4叫*-I ok6Jb)4町.2金 c)24丹c哈野.皿Determinethepotentialdistributioninacoaxialcablewhentheinnerconductorofradius10cmisatapotentialoflOOvandtheouterconductorradius20cmisgrounded.Thespacebetweentheconductorsisfilledwithtwoconcentricdielectrics.Thedielectricconstantis3fortheinnerdielectricand9fortheouterdielectric.Thedielectricinterfaceisataradiusof15cm.Determine(a)thepotentialdistribution,(b)theDandEfieldineachregion,(c)thesurfacechargedensityontheinnerconductor,and(d)thecapacitanceofthe100-meter-longcable.题3.663.66同轴电缆内导体上的电位为Y=100V,半径为a=10cm,外接体接半径为b=20cm,谈到体内允满电容率分别为?=3（内层）和々=9（外层）的两种同心电介质，其分界面半径为Co求（a）电位.分布，（b）每个区域内的6和屯，（c）内导体的面电荷密度，（d）电缆单位长度的电容。电缆全长100m。解：⑶根据泊松方程有▽V=（）,由于与原点距离相等的所有点组成的圆柱面是等位面，故尸只是0的函数。当10。加(夕41 = —f02]=0n夕a=G叫q“ 1・・.==一=>匕=Gln〃+Q©PP41no.1+g=10041no.1+g=100同理,当”—吗舒孙。n家暇二5款C=—=>K,=c31»/7+c4而p-'根据边界条件匕(p=0.2)=0,c3\n0.2+c4=0CalculatethecapacitanceoftwosphericalshellsofradiiaandbusingLaplace'sequation.TheinnershellisatapotentialofV0,andtheoutershellisgrounded.Whatisthesurfacechargedensityontheinnershell?Obtaintheexpressionforthecapacitanceofthesystem.题3.67InO.I5+c2=qInO.15+q再根据当夕=0,15再根据当夕=0,15时」昌康=£2薪（2='%）即q二也=>q=-66.5,q=3q=-1995c2=lOO-lnO.lr,=-359.3c4=-ln0,2q=-107,03・.匕=-199,5In0-359.3 V2=-66.5Inp-107.03£1=199.5/p,£2=66.5/x?="=5.3x1。制夕(c)夕,=D(户=O.l)=5.3xl()T(d)c=9=^~=2^-x0Jx5.3x1054(d)c=9=^~=100 =3.33xlOF100ShowthattheelectricfieldintensityasgivenbyE=100sin(10sr+x/>/3pV//wisavalidsolutionofthewaveequationinadielectricmedium.Alsodetermine(a)thedielectricconstantofthemediumassumingthat//,.=1,(b)thevelocityofpropagation,(c)themagneticfieldintensity,(d)theintrinsicimpedance,(e)thewavelength,and(f)theaveragepowerdensity.题848.4证明由E=100sin（10s/+x/V3）zV/m给出的电场强度是电介质，I'的波动方程的一个有效的解。并求（a）媒质的电容率，设〃,=1,（b）波传播速度，（c）磁场通度，（d）本征阻抗，（e）波长，（f）平均功率密度。解：(□)▽*g=~~r=一——sin(1()sZ+x/、G)bx3d2d2E

9r=-108sin(10'z+xM)根据波动方程▽喈-小学为得U^sin(10'/+x/VJ)—/elO,*sin(10，/+x/VJ)3=>3xIO16^=l=>3x =1=>3xI0,6x4^x10-7x8.85x10-|26；=1=>6；=3

\10”三(b)v=—=，厂=10sV3(/w/s)\10”三…-M一 100 f(c)vxE= -y=——^cos小 。3 IDF迎丽100fin8i B1001.rin8x1dt比6\gVJiosIM)=0.46sin10,十=0.46sin10,十100I.H= 尸—sin10/十为5。I平均功率密度二io4JJ«7%siir平均功率密度二io4JJ«7%siir-1

ioiVwS)〃=j4=217.56(0)(?)4--j%-2百不(⑼U')S=ExH= ^sin2[10^+^I1。%l⑸-i-2xlO4V3/zo=22.97郎/TheelectricfieldintensityofauniformplanewaveinfreespaceisgivenbyE=120cos(2^xl09r-/7y)^V//z?.Determine(a)thephaseconstant,(b)themagneticfieldintensity,(c)thewavelength,(d)theaveragepowerdensityinthemedium,(e)theaverageenergydensityintheelectricfield,and(f)theaverageenergydensityinthemagneticfield.题8.6

8.6自由空间均匀平面波的电场强度已知后=120cos（2乃x10〃一4,）2V/m。求（a）相位常数，（b）磁场强度，（c）波长，（d）媒质中的平均功率密度，（c）电场平均能量密度，（f）磁场平均能量密度。解：由于在自由空间，故（。）相位常数尸=竺=-20.94c3x10,9）磁场强度酣相量形式是反⑺=两磁场费.度方= -VxE= x= -E式一印）e~jfiyxjco^l j（D/Ady=*J"必戈=0.318cos(2^xl0yr-20.94y)x{A!m)IO9120i=2x377 ' 7(c)波长/=二IO9120i=2x377 ' 7（e）电场平均能量密度==3.186x10。4（/）磁场平均能量密度二电场平均能量密度=3.186xI0=gAuniformplanewaveatafrequencyof100MHzpropagatesinthezdirectioninaconductingmediumcharacterizedby4r=1J=2.25andcr=9.375"?S/〃?・DetermineYfarp9Sdrvpandtan8.Ifthemaximumvalueoftheelectricfieldintensityintheydirectionis125V//nandoccursatz=0andt=0zwriteexpressionsfortheEandHfield.Obtainageneralexpressionforthepowerdensityinthemedium.Whatisthetotalcurrentdensityinthemedium?题8.108.10频率为100MHz的均匀平面波在导电媒质4=1工=2.25和b=9.375/〃S//〃中沿Z方向传播。求九a/"一、和tan^。若当z=0和t=0时，y方向的电场强度最大值为125/〃〃，学出电场和磁场的表达式。求媒质中功率密度的一般表达式。媒质中的总电流密度是多少？解：co=2/r/=2/rx100x106=2x10srr{rad/s）,乂b 9.375x10-3tanb=——r= —=0.7〉a）82^x108x2.25x8.85x10-121一/二|=1.99）＜10一“（1一/0.75）（oc）/=j①吠=j5口舄灰^也-j°：5="J1-/0.75=乃.5We-‘36.87°=/3.5"8j；3.51（COS18.4°-jsinl8.4°）=1.108+73.33（w-1）a=\.\08（Np/W）,0=333（Np/m）Aglassmaterialiscoatedwithsilver（ct=6.1x107S/m）tomakeitagood