电动力学习题集答案-1(7)

长 沙 理 工 大 学 备 课 纸

?库仑规范 ??A?0

??2????/?0??? ?2?1?2A1????A?2?2???0Jf?c?t2c?t?5．试从Maxwell方程组出发,给出变化的电磁场矢势和标势的定义，说明何谓电磁场的规范变换，

??并证明电磁场的E和B在这种规范变换下保持不变.

??????????B解：由??B?0 得B???A，将其代入??E??得, ???E??A??0，

???t??t?????A??A故可引入标势?，使得E? ????，即：E??????t?t?设?(x,t)是一个具有连续二阶偏导数的任意标量函数，做变换

????A?A'?A??????——规范变换，则有 ????'?????t??????????B'???A'???A??????A????????A?B?????? ???A'????A????A?E'????'??????????????E?t?t?t?t?t???06．一电量为q的粒子沿z轴作简谐振动，其坐标为z?acos?t。设它的速度为v??c(c为真

空中的光速）求它的辐射场和平均能流密度以及辐射功率.

?ex??sin?cos?cos?cos??sin???er???????cos???e?? 提示：直角坐标基矢与球坐标基矢关系为?ey???sin?sin?cos?sin??e??cos???e??sin?0?z??????解：由定义这个带电粒子对原点的电偶极矩为：P(t)?qacos?tez

振动电偶极矩产生的矢势为

?????0dP(t')?qa??qa?r?A(r,t)???0sin?t'ez??0sin?(t?)ez4?rdt'4?r4?rc???0qa?sin(kr??t)ez 其中，k??/c

4?r

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??0qa?????qa?B???A?[?sin(kr??t)]?ez?0kcos(kr??t)]?r?ez4?r4?r

????0qa?2?0qa?2?cos(kr??t)]er?ez??cos(kr??t)]sin?e?4?cr4?cr?????????D?E122????H????B?c??B??i?E?cik?B??t?t?0?0?????????0qa?22??E?cB?er?E?cB?er??cos(kr??t)]sin?e??er

c4?r???0qa?2?E??cos(kr??t)]sin?e?4?r平均能流密度：

?1S?T?T0222???0qa?4sin2?1E?Hdt?16?2r2c?0T?T0?cos2(kr??t)]dter

?0?0q2a2?4sin2??q2a2?4sin2???er?er2223216??0r2c32??0cr辐射功率：

P????S?ds??2?0d???0q2a2?4sin2?2d?rsin? 23232??0cr014???3? [1?cos?]dcos??cos??cos??????33?????02?????sin3?d???sin2?dcos???0??0?P????q2a2?48?q2a2?4S?ds??2332??0c312??0c3

另解：由定义这个带电粒子对原点的电偶极矩为：

z?er?P(t)?qaexp(?i?t)ez ，

振动电偶极矩产生的矢势为

?e?r?R?e?y???A(r,t)?04?r?R??a???0?0dP(t')?J(x',t')dV'?4?R?4?Rdt'??J(x',t')dV'rx?- 31 -

长 沙 理 工 大 学 备 课 纸

??0qa?exp(?i?t')ez4?Ri?0qa?R? ??exp[?i?(t?)]ez4?Rc??i??i?0qa??exp[i(kR??t)]ez 其中，k??/c

4?R???i?qa??????0qa?2??0B???A?ik??exp[i(kR??t)]ez??exp[i(kR??t)]er?ez4?cR?4?R???0qa?2??exp[i(kR??t)]sin?e?4?cR?????????D?E122????H????B?c??B??i?E?cik?B??t?t?0?0?????????0qa?22??E?cB?er?E?cB?er??exp[i(kR??t)]sin?e??er（2分）

c4?R???0qa?2?E??exp[i(kR??t)]sin?e?4?R平均能流密度：

222?1????qa?4sin2??1?0S?ReE?H?e?e??2216?2R2c?0??2 2??Psin????qa?sin??qa?sin???002e?e?errr32??0r2c32?2?0c3r232?2?0c3r222422242辐射功率：

P????S?ds??2?0d???0q2a2?4sin2?2d?rsin? 23232??0cr0??3??01????sin?d???sin2?dcos??[1?cos2?]dcos???cos??cos3???4??0????03???3???P??2??224224??qa?8?qa?1PS?ds???23332??0c312??0c4??03c3

7．1）写出Maxwell方程组；2）从此方程组出发，引入电磁场的矢势和标势，说明何谓电磁场的规

??范变换，并证明电磁场的E和B在这种规范变换下保持不变.

解：1）Maxwell方程组为

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??????D??B??E??,??D??,??H??J,??B?0

?t?t?????2）由??B?0 得B???A。将其代入??E???B/?t得,

????E??A/?t??0，故可引入标势??长 沙 理 工 大 学 备 课 纸

，使得E??A/?t????，即：

????E??????A/?t

?设?(x,t)是一个具有连续二阶偏导数的任意标量函数，做变换

????A?A'?A??????——规范变换，则有 ????'?????t??????????B'???A'???A??????A????????A?B?????? ???A'????A????A?E'????'??????????????E?t?t?t?t?t???0??8．一电偶极子位于坐标系的原点，它的电偶极矩为P?P0cos?tex。试求1）它在r????2?c/?辐射场的电场强度和磁场强度；2）该处辐射场的能流密度. (15分)

?ex??sin?cos?cos?cos??sin???er???????cos???e?? 提示：直角坐标基矢与球坐标基矢关系为?ey???sin?sin?cos?sin??e??cos???e??sin?0z???????

???1）解：注意：本题电偶极矩P?P0cos?tex沿着x轴，但球坐标选取如常（如r与Z轴间的夹角为?等）这样，振动电偶极矩产生的矢势为

??r???????J(x',t')dV'?0?0dP(t')A(r,t)?0??J(x',t')dV'?4?r4?r?4?rdt'???p0i??p0r???i0exp(?i?t')ex??0exp[?i?(t?)]ex4?r4?rc???i?0?p0exp[i(kr??t)]ex 其中，k??/c

4?r- 33 -

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???i?0?p0?????0p0?2??B???A?ik??exp[i(kr??t)]ex??exp[i(kr??t)]er?ex4?cr?4?r??????ex?sin?cos?er?cos?cos?e??sin?e?)??0p0?2???B?exp[i(kr??t)](sin?e??cos?cos?e?)4?cr?p0?2??H?exp[i(kr??t)](sin?e??cos?cos?e?)

4?cr?????????D?E122????H????B?c??B??i?E?cik?B??t?t?0?0

????????E?c2B?er?E?cB?erc??0p0?2???E?exp[i(kr??t)](sin?e??cos?cos?e?)?er4?r 取实部

??0p0?2???E?exp[i(kr??t)](cos?cos?e??sin?e?)4?r??0p0?2??E?cos(kr??t)(cos?cos?e?sin?e故有??)

4?r?p0?2??H?cos(kr??t)(sin?e??cos?cos?e?) 取实部

4?cr能流密度：

???S?E?H?????0p02?42?cos(kr??t)(cos?cos?e?sin?e)?(sin?e?cos?cos?e????)2216c?r???p2?4cos2(kr??t)?222S?0020(sin??cos?cos?)er16??0cr2242?p0?cos(kr??t)22(1?sin?cos?)er16?2?0c3r2?i?kz??t??9．有一原子团，设其极化率为?(?)，处于电磁场E?E0eez之中, 该原子团位于坐标原点，?其体积为V，且原子线度远小于电磁波波长。试求原子团在远处的辐射电磁场和电偶极辐射的平均能流密度以及辐射总功率。

?解：首先复习一下电偶极矩的计算：p总??x??V''?(x',t')dV'???x????'?q?- 34 -

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