《电动力学(第二版)》(郭硕鸿)第四章习题

第四章 习 题

1. ⑴

E??k?dk?z????d??t?1?E0ei E??k?dk?z????d??t?2?E0ei

E?E1?E?E?2i??k?dk?z????d??t?0e?ei??k?dk?z????d??t???E?ei?dk?z?d??t?0?e?i?dk?z?d??t??ei?kz??t?

?E0?2cos?dk?z?d??t?ei?kz??t? ⑵

令kz??t?常数，得

vdzp?dt??k 令dk?z?d??t?常数，得

vg?dzdt?d?dk 2.

sin?sin?''??2?2sin45?1??2，sin?''??1?122 co?s''?1?sin2?''?32 E'??cos???2?2cos?''E?11?1?1cos???2?2cos?''?cos45??2cos?''cos45??2cos?''1?23

?2212?232?1?31?32R???1?3????2?3?1?3? ?2?3E''2?1?1cos?E??1?1cos???2?2cos?''21?21

2?232?21?32T???2?2?1?3???2?3 3.

⑴

?0?6.28?10?5cm，n?1.33，??60?

设产生全反射的临界入射角为?0，则

sin?sin90?0?n?11.33，?0?48.8??60? 因此入射角为60?时将产生全反射。 ⑵ kx\?kx?ksin?

v?p??kx\?ksin??csin60??32c ??1??1??0n2?sin2??n2212?sin2??n2215?6.28?10?1.332?3.14sin260???11.33?2

?1.75?10?5cm4. ⑴

??E?i?B (4.1) ??H??i?D (4.2) ??D?0 (4.3) ??B?0 (4.4)

E?x,t??Ei?k?x??t?0e，E?x??Eik?x0e

D?x,t??Di?kx??t?0e，D?x??Dik?x0e

H?x,t??Hi?kx??t?0e，H?x??Hk?x0ei

B?x,t??B?kx??t?0ei，B?x??Bik?x0e (公式：????f???????f????f) 由式(4.3)得：

0???D?x?????Dik?x0e??ieik?xk?D0?ik?D?x?

同样方法，由式(4.4)得：0?ik?B?x? 因此：k?B?k?D?0

若D与E不同向，则一般k?E?0 (公式????f???????f????f)

B?x???i???E?x???i????Eik?x0e???iik?x0??iik?x???e??E?eik?E0

?1?k?E?x?D?x??i???H?x??i????Hik?x0e??i??eik?x??iik?x?H0??eik?H0 ??11?k?H?x?????k?B?x?(公式：a??b?c??b??c?a??c??a?b?) 因此：B?D?B?E?0 ⑵

(公式：?a?b??c??c?a?b??c?b?a)

D?x???1??k?B?x?

??1??k???1???k?E?x??? ?12?2??kE??k?E?k?⑶ S_?1*1?2Re?E?H??2Re??E*???1k?????????E??????1

??*2??Re?E?k?E?E2k?因一般k?E?0，所以S_一般与E不同向5. ⑴

E1?exE0eikz

Ei?kz??2?2?eyE0e

Ex?E0cos?kz??t?

Ey?E0cos?kz??t??2??E0sin?kz??t?

因此：E222x?Ey?E0，为圆偏振光 tan??EyE?tan?kz??t?

x上式两边对时间t求导，得

sec2???d?dt???sec2?kz??t?，因此：d?dt?0为右旋圆偏振光。 ⑵

一个圆偏振光可以分解为偏振方向互相垂

直、振幅相等、相位差?2的二个线偏振光 ???Ex?E0co?skz???t??Ey??E0sin?kz???

t “+”为右旋圆偏振光

“－”为左旋圆偏振光 6.

(公式：????f???????f????f ) (公式：?a?b??c??c?a?b??c?b?a )

E?E?zi?z0e?e，其中E0?ez H?1i????E?1???zi???eei?z??E0

????i?i??e??zei?zez?E0 ⑴

S_?12Re?E*?H??1???z??2Re???e?i?i??e??zE*0??ez?E0??????2??e?2?zE20ez

z?0处，S_?0???22??eE0ez

⑵

p1?*?E??1L?ReJRe??E222??1

?E2?z20e?2P?L??1E2?2?z02?0edz?E2

?04?由?2????i???????i??2得：

????2??，??_???2?，因此：S?PL

7.

?r?1，??1S?m?1，?0?8.85?10?12F?m?1?70?4??10?H?m?1

⑴f?50Hz

12???????1?2??2??1??????2?2?1?????? ????2??12??2????2??50?4??10?7?1

?71m ⑵f?106Hz

??12??2????2??106?4??10?7?1

?0.50m ⑶f?109Hz

??1??2????22??109?4??10?7?1

?0.016m8.

z?3k②???①??0?0x1?2k1k' 在真空中：k1???0?0??c，

k1?k1xex?k1zez，k1x?ksin?1，k1z?kcos?1 在导电介质中：k??????i???????'，

k?β?iα，α?axex??zez，???xex??zez

在分界面处：k1?x?k?x 得：k1sin?1?i?x??x 因此：?x?0，??x?csin?1

k?β?iα?k1sin?1ex???z?i?z?ez

k2?k2221sin?1???z?i?z???2????i???

?k222 ?1sin2?1??z??z??2??? ?2?zz???? 解上方程得

?2z?1?2??????2??22?c2sin?1????

11?2?

22???????2??????2??2sin2???????c21?????2z??1?2??????2??2c2sin2??1????

1

1?22??????2????????22si2???2???c2n?1??????9.

?2E?k2E?0， k????

?2E2x?kEx?0

Ex?x,y,z??X?x?Y?y?Z?z?，代入上式得

YZd2Xd2Yd2Z2dx2?XZdy2?XYdz2?kXYZ?0

d2Xdx2?k2xX?0 d2Ydy2?k2yY?0 d2Zdz?k22zZ?0 k2k2x?y?k2z??2??

Ex???C1coksxx?D1sinkxx??C2coksyy?D2sinkyy??

?C3cokszz?D3sinkzz?依Ex?x,0,z??0，得C2?0 依Ex?x,b,z??0，得sinkyb?0

kn?y?b，n?0,1,2?

依??E?0得，在x?0,a处?Ex?x?0 得D?1?0，kx?ma，m?0,1,2? 依Ex?x,y,0??0，得C3?0 因此

E?x?A1cosmaxsinn?bysinkzz 同样方法

Em?n?y?A2sinaxcosbysinkzz 另

Ez???C4coskxx?D4sinkxx??C5coskyy?D5sinkyy??

?C6coskzz?D6sinkzz?依Ez?0,y,z??0，得C4?0 依Ez?x,0,z??0，得C5?0 依??E?0得，在z?0处

?Ez?z?0，有D6?0Em?z?A3sinaxsinn?bycoskzz k2??2?????m??2?m??2z?a?????b??

依??E?0得：

m?aAn?1?bA2?kzA3?0 10.

??E?i??0H (10.1.1) ??H??i??0E (10.1.2) 由式(10.1)得(见《电》p.342)

?Ez?y??Ey?z?i??0Hx (10.2.1) ?Ex?z??Ez?x?i??0Hy (10.2.2) ?EyEx?x???y?i??0Hz (10.2.3) ?Hz?y??Hy?z??i??0Ex (10.2.4) ?Hx?z??Hz?x??i??0Ey (10.2.5) ?Hy?Hx?x??y??i??0Ez (10.2.6) 依题意分离变量：E?x,y,z??E?x,y?eikzz，因此有

Ex?x,y,z??Ex?x,y?eikzz Ezy?x,y,z??Ey?x,y?eikz Ez?x,y,z??Ez?x,y?eikzz Hx?x,y,z??Hzx?x,y?eikz Hy?x,y,z??Hy?x,y?eikzz Hz?x,y,z??Hz?x,y?eikzz 以上六式代入式(10.2)得

?Ez?y?ikzEy?i??0Hx (10.3.1)

ikzEx??Ez?i??0Hy (10.3.2) ?x?Eyd2Y2?kyY?0 2dy2222??Ex?i??0Hz (10.3.3) ?x?y?Hz?y?ikzHy??i??0Ex (10.3.4) ikzH?Hx?z?x??i??0Ey (10.3.5) ?Hy?Hx?x??y??i??0Ez (10.3.6) 解(10.3.1)、 (10.3.2) 、(10.3.4)和 (10.3.5)得 E1???????Hz?Ez?x?0i????22????y?kz?x??? ?c2?kz?? E1?y?2?????Hz?Ez?0? i?????c2?k2??x?kz?y??z???? H?1???k?Hz?Ezx??z?i????22????x???0?y?? ?c2?kz?? H1?y?2???k?Hz?Ez?z???0?? i????2????y?x??c2?kz??y bazx

TM波，即Hz?0，磁场无纵向分量

?2E?k2E?0， k????

?2Ez?k2Ez?0

Ez?x,y,z??X?x?Y?y?eikzz，代入上式得

d2Xdx2?k2xX?0 kx?ky?kz????

Ez??C3coskxx?D3sink?xx??Ccoskeik

yy?D4sinkyy??zz4依Ez?0,y,z??0，得：C3?0 依Ez?x,0,z??0，得：C4?0

依Em?z?a,y,z??0，得：kx?a，m?0,1,2?依Ex,b,z??0，得：kn?z?y?b，n?0,1,2?

因此：Em?z?E0sinaxsinn?ikzzbye 13.

f?30?109Hz

①a?0.7cm，b?0.4cm

??cf?3?108 30?109?0.01m?1cm 22???m?cmn?????a?????n??b?? 1?22fcmn?2???m??a?????n??b??22

?c?m??n?2??a?????b???ccmn?f?2cmn?m22

???a??n?????b???cc10?f?2cmn?2?1??a?????0?2?b?? ?2a?1.4cm???c01?cf?2cmn??0?22?a??1?????b?? ?2b?0.8cm?? 12.

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