电磁场与微波技术第2版黄玉兰 习题答案.doc

:ruurg0??6)?5??4?1?(6)?(A?B9ruur相互垂直和?ABruru=0A?Bruru相互平行A和B?(1)?Ax?Ay?AzgA?divA?????x?y?z2222z?72xyx?2?2xy(2)?2222??????g(2x?2xy?dy72xyz)dz?Ads?Ad?dz?????(1)因为闭合路径在xoy平面内,故有:?222dyx?dy)xdx?x?eyez)(dx?e?(A?dl?exeyzyxx???8?A??dl(2)?xdxdy?2))(e2ds??(eyz?2xedxdyAXOYS因为在面内,?xzx?8)?dsA(????s所以,定理成立。(1)由梯度公式?u?u?u?u?e?e?e|(2,1,3)xyz?x?y?z?4e?10e?ezyx22211710?1?方向导数最大值为4?1(4e?10e方向:?e)zyx117()2与梯度垂直最小值为0,?u???0?g????43?a3?sinewrV???sin3qwr?e?V?J??3?''????d场点坐标为P(0,0,z).线电荷元?dlll可以视为点电荷,其到场点的距离矢量uruuruuruurgggaeaz?Er?R?ezRuuruuruurggaeaz?ez22az?Er?,r?得22az?所以P点的电场强度为uuuuuuuuuuuuuuruuuuuuuuuurrurgga?ezeaz?2'????dE?l3022??04?a)(z2uuruuruururu?2'''????gQ?d?eysin0?eacosea=ex0urur??zl?E?ez322?02?a)((1)r?b时urr?2??grE(r)ds?4Es235brrr222???)?4?((b?r)4Eq?rdr530rruEq??g由高斯定理有=dEs?0s532?r4rb2?)即r?E(r)=4(?53032r1br?(?E(r)=)?530()r>b时2rur?2??g)E(4s?Erdrs?8b5222??bdr?)4rEq?(br?150由高斯定理有5b2Eq?E()=r22???>b则,E=Eb-Ea?q??gds=Eb?02??lb?gggl21??Ebr?022??eaber12r?Eb?,同理:Ea???r2r1220022?eebarr12?E?Eb?Ea?(?)?r1r220(2)对于r1<b且在空腔外,E=Eb-Ea?q??g,Ebds=?02??lr1?ggglEb?2r1??0?e1rr1Eb??202?ear2而Ea??r2202?ea2r?E?(r1e??)1r?r220(3)r1?b且在空腔内E=Eb-Eaq???g,Eds=?0??er2r1er2r1?Eb=,Ea???2200???eerr122rr1?E?Eb?Ea???(r1e?r2e)2rr1???(1)r?a时,E?????02a?cos(r-)??E=????A?r>a时r22AaaA??)sinA??e()cos=e(?A??r22rr圆柱是由导体制成的(2)??????cos2表面电荷A???0s0a?=0D的法向量分量连续根据??E?z)??(5Z1rr210?E???,和设内外导体单位长度带电量分别为+-(1)ee利用高斯定理可以求得导体介质的电场为:?lE?er??r2?bb?gQllnE?u?dl??a2a??u2u?eE,得到??rlbbrlnlnaa6uge6E??J?rbrlna(2)?gdsJ??62Iu?gs??e?g?=0B(1)取圆柱坐标系,若为磁场,根据磁场连续性方程,有?ruuur?gQar0=erBru?g所以不是磁场a????B0,0)取直角坐标(2rug,所以是磁场。B?=0ruuru0???H??Eyge?E?H(2)????zjwujwu?x?1jjkx?Eme?e?2zy?5?104?jjkx?e??e2z?1208?7?msin(3?10tv/?)???z(3)均匀平面波,波传播方向是??s/?1020m,Vp?3(1)k?kVpw9Hz10?3??f????22ruruuruu?zj20?4?),jeyex??10e((2)E该波是左旋圆极化波?1EH=-??(3)?jw?j??z203??j20jz?3???e2e1010?e?e2?