4.1解：??????1??令S?E?H2???1*?m?H?B4???*1???e?E?D4?????*??*???H?J?j?D??????E??j?B?????*???*????E?H?H???E???????E?????*??H????*??????j?H?B?E???*????J?E?j????*??*J?j?D???????*??????*H?B?E?D?????*E?H???1???S???2?*??1???J?E?22?*??1???J?E?22?*?1j??H?4????*?1???B?E?D?4?
????j???m??e????????*??1????即??S??J?E?2j???m??e?2??为频域Poynting定理4.2证明：????????H?J?j???????E??j?B????D??E?????B??H????J??E
????????E1?H2???????????H2???E1?E1???H2????????????H2??j?B1?E1?J2?j?D2???????????????H2??j??H1?E1??E2?E1?j??E2????????????????j??H1?H2??E1?E2?j??E1?E2????????????j??H1?H2??j?????E1?E2???????E2?H1???????????H1???E2?E2???H1????????????H1??j?B2?E2?J1?j?D1???????????????H1??j??H2?E2??E1?E2?j??E1????????????????j??H1?H2??E1?E2?j??E1?E2????????????j??H1?H2??j?????E1?E2?????????????E1?H2???E2?H1??????????即：??E1?H2?E2?H1?0??????????????????????????????
命题得证。4.3解：由电流连续性方程有：????????J?ds???ts?v?vdv???Q?t??t
由高斯散度定理有：?????????E?ds????Edvvs由高斯定理有：???????sD?ds??v?vdv?Q?t?????又D??E?
??s?????1E?ds????ss?????Q?t?D?ds????即?Q?t??t?t????????J?ds??Q?t???s?????Q?t?E?ds????Q?t?＝－??t?0初时条件为：Q?t??可以解方程得：Q?t?＝Q??t?0??????B?????E?t?????又B??H?－?Q?t?????H?t?－t?????11Q????????E????r?4??r2???????－????Q?t4?????r???2??0?r???????H?cont??0????r????又?t?0时,H?0????H?0即，在任意时刻，磁场强度为0。1????1??2?E?D??E22???1???H?B?02?－t??1?Q???22?4??r???e?????2?Q?2－??2t32??r24?m