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1、本文格式为Word版,下载可任意编辑高等电磁场作业 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
2、?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 2