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1、Field and Wave Electromagnetic 電磁場與電磁波 2015. 10. 10 2 1. Electric Potential EV 2 1 21 P P WV V E d l q 0 ( )PpV P V P V V E d l V 0 ( V)4 qV R 10 1 V 4 R R n k k k qV 0 1 ( V ) 4 VV dvR 0 1 ( V ) 4 s SV dsR 0 1 (V) 4 l LV d lR Review 3 2. Conductors in Static Electric Field 3. Dielectrics in Static
2、Electric Field Inside a conductor (under static conditions) 0 0E Boundary Conditions (at a Conductor/Free Space Interface) 0 0t s n E E 21 0 l i m ( C/ m ) nv k k v p P v p s nPa p P 4 4. Electric Flux Density and Dielectric Constant 20 0 0( 1 ) ( C/m )erD E P E E E = D 0E 0C E d l S D d s Q 5 Main
3、topic 1. 靜電場的邊界條件 2. 電容和電容器 3. 靜電場的能量 6 1. Boundary Conditions for Electrostatic Fields 電磁問題通常涉及具有不同物理特性的媒質(zhì) ,并且需要關(guān) 于兩種媒質(zhì)之間分界面的場量關(guān)系的知識 . 例如 ,我們希望知道 場矢量 E 和 D 在穿過兩個(gè)界面時(shí)是如何變化的 . 現(xiàn)在我們來考 慮兩種一般媒質(zhì)之間的分界面 . D 0E 0C E d l S D d s Q 7 (1) 電場強(qiáng)度 E的切向分量 (tangential component) Let us construct a small path abcda wi
4、th sides ab and cd in media 1 and 2 respectively, both being parallel to the interface and equal to w. The integral form is assumed to be valid for regions containing discontinuous media, is applied to this path. If we let sides bc=da=h approach zero, their contributions to the line integral of E ar
5、ound the path can be neglected. Thus we have E1 E2 2 1 at w h a c d b an2 1 2 1 t 2 t 1 t 2 t d d d d d 0 d d ( ) 0 b c d a l a b c d bd ac E w E w EE 12 12 E l E l E l E l E l E l E l E w E w 8 電場強(qiáng)度在分界面處的切向分量是相等的 . 換句話說 ,電場強(qiáng)度 的切向分量連續(xù) . 對于線性各向同性的一般介質(zhì) ,有 : 2 2t 1 t1 DD 12ttEE 2 1 2 ( ) 0 na E E an2 的
6、方向是有媒質(zhì) 指向媒質(zhì) . 9 (2) 電通密度 D的法向分量 (normal component) In order to find a relation between the normal components of the fields at a boundary, we construct a small pillbox with its top face in medium 1 and bottom face in medium 2. The faces have an area S, and the height of the pillbox h is vanishingly s
7、mall. Applying Gausss law to the pillbox, we have h S 2 1 an2 D 1 D2 s 21 1 2 1 222 2 1 2 1 n 2 n d ( ) ( s S b o tto m s id e to p nn to p b o tto m ns D d S D d S D d S S D d S D d S D a S D a S a D D S D D S S DS 10 2 1 2 1n 2n ( ) o r n s sa D D D D 邊界的法線方向由媒質(zhì) 指向媒質(zhì) . S 是邊界上的面電荷密度 . D場的法向分量在通過存在面
8、電荷的分界面時(shí)是不連續(xù)的 ,不連續(xù)的量 等于面電荷密度 . 11 2 1 2 1 2 2 1 2 1 2 ( ) 0 o r ) o r n t t n s n n s a E E E E a D D D D （ 討論 (1)、 Boundary Conditions for Dielectric-conductor Interface 討論 (2)、 In the absence of net surface free charge, one has The boundary conditions that must be satisfied for static electric fiel
9、ds are as follows: 2 1 1 2 1 1 0 o r o r nt n s n s a E E a D D 0 2 1 2 1 2 2 1 2 1 2 ( ) 0 o r ) 0 o r n t t n n n a E E E E a D D D D （ 12 例 1:兩理想介質(zhì)的分界面為 Z=0的平面 ， 如 圖所示 ， 在介質(zhì) 2中的場強(qiáng)為 2 ( 5 )x y zE a y a x a z 求介質(zhì) 1中分解面上的場分量 。 2= 0 r2 1= 0 r1 x z y 122 ( ) 0na E E 122 ()nsa D D 1 1 2 2 1 1 1 2 2 2 2
10、 2 2 0 1 ( 5 ) ( 5 ) ( ( 5 ) ) ( ) 0 ( | ( 5 ) ) x x y y z z x x y y z z x y z x y z z x y z x x y y z z z x y z x z x E a E a E a E D a E a E a E E a y a x a z D a y a x a z a a y a x a z a E a E a E a a y a x a z a E a 011 | )0 y y z z z E a E 13 1 2 1 2 1 2 1 21 2 1 2 1 ( ) ( ) ( 5 ) 0 ( ) ( ) (
11、5 ) 0 ( ) ( ) 0 ( 5 ) 0 5 ; 5 z x x y y z z z x x y y z z y x x y z x y z x y z a a y E a x E a E a a y E a x E a E a y E a x E E E y E x E E a y a x a 14 2. 電容和電容器 導(dǎo)體 Q V EV 0 1 ( V ) 4 s SV dsR SS Q k QQ k Q k V k V V k V 不 變 SS Q k QV k V E k E k Q k Q V k V 不 變 0 s nE ( c /v or F )QC V 比例常數(shù) C 稱為
12、孤立導(dǎo)體的電容 . 電容是指電位每增加一單位所必須施加 于物體的電荷量 . 15 12 (c / v o r F )QC V 電容器的電容是雙導(dǎo)體系統(tǒng)的一種物理性質(zhì) ,其依賴于導(dǎo)體的 形狀和它們之間媒質(zhì)的介電常數(shù) ,與電荷 Q 和導(dǎo)體之間的電位差 V12 都無關(guān) .電容器即使沒有電壓提供或?qū)w上沒有自由電荷它也是存 在的 . 16 Capacitance C can be determined from above equation by either (1) assuming a V12 and determining Q in terms of V12, or (2) assuming a
13、 Q and determining V12 in terms of Q. At this stage, since we have not yet studied the methods for solving boundary-value problems (which will be taken up in Chapter 4), we find C by the second method. The procedure is as follows: 1. 對已知的幾何形狀 ,選擇合適的坐標(biāo)系 . 2. 假設(shè)導(dǎo)體上攜帶了電荷 +Q and -Q. 3. 通過高斯定理或其它關(guān)系根據(jù)假設(shè)的電
14、荷量 Q來確定 E. 4. 求出導(dǎo)體兩端的電位差 V12. 5. Find C by taking the ratio Q/V12. 1 12 2V E d l s nE 17 Example 3-18 P124 18 Example 3-19 P125 19 (1) Series Parallel Connections of Capacitors Parallel Connections of Capacitors 電容器的電壓相等 nCCCC 21/ Series connections of capacitors(電容器的電量相等 ) nnsr CCCC 1111 1 (2) Capa
15、citances In Multiconductor Systems 20 Q q P 靜電能來源： 外力克服電場力做功轉(zhuǎn)化而 來 ， 靜電場能僅與帶電體的最終帶電狀態(tài)有關(guān)而與 到達(dá)這一狀態(tài)的中間過程無關(guān) 。 靜電能：當(dāng)電荷放入電場中 ， 就會受到 電場力 的作用 ， 電場力做功使電 荷位移 ， 這說明電場具有能量 。 靜電場內(nèi)儲存著能量 ， 這種能量通常被 稱為靜電能 。 電場越強(qiáng) ， 對電荷的力就越大 ， 做功的能力就越強(qiáng) ， 說明 電場具有的能量就越大 。 2 12 1 2 1 = ( ) PPW W Q E dl Q V V 3. Electrostatic Energy 能量的零點(diǎn)
16、： 最初 電荷 都分散在彼此相距很遠(yuǎn) (無限遠(yuǎn) )的位 置上 。 通常規(guī)定 ， 處于這種狀態(tài)下的靜電能為零 。 靜電場能 量 We等于于把各部分電荷從無限分散的狀態(tài)聚集成現(xiàn)有帶電 體系時(shí)抵抗靜電力所作的全部功 。 21 Bring a charge Q2 from infinity against the field of a charge Q1 in free space to a distance R12 120 1 2222 4 R QQVQW 11 120 2 12 4 VQR QQW )( 22112 21 VQVQW (1) Two charges 1Q 2Q 12R 1P 2P
17、where V2 is the potential at P2 established by charge Q1, chose the reference point for the potential at infinity; This work is stored in the assembly of the two charges as potential energy. Another form where V1 is the potential at P1 established by charge Q2 . 22 Bring another charge Q3 from infin
18、ity to a point that is R13 to charge Q1 and R23 from charge Q2 in free space ,an additional work is required that equals where V3 is the potential at P3 established by charges Q1 and Q2 , W3 , which is stored in the assembly of the three charges Q1 , Q2 , and Q3 , is 230 2 130 1 333 44 R Q R QQVQW 2
19、3 32 13 31 12 21 0 23 4 1 R QQ R QQ R QQWWW (2) Three charges 1Q 2Q 12R 1P 2P 3Q 3P 13R 23R 23 We can rewrite W3 in the following form )( 332211 230 2 130 1 3 230 3 120 1 2 130 3 120 2 13 2 1 44 44442 1 VQVQVQ R Q R Q Q R Q R Q Q R Q R Q QW 23 32 13 31 12 21 0 23 4 1 R QQ R QQ R QQWWW where V1 is th
20、e potential at Q1 established by charges Q2 and Q3 , similarly, V2 and V3 are the potentials at Q2 and Q3 , respectively, in the three-charge assembly. 3 1 1 2 2 3 3 1 () 2W Q V Q V Q V 24 Extending this procedure of bringing in additional charges, we arrive at the following general expression for t
21、he potential energy of a group of N discrete point charges at rest. (The purpose of the subscript e on We is to denote that the energy is of an electric nature.) We have 1 1 2 N e k k k W Q V 10 1 4 N j k j jk jk Q V R where Vk , the electric potential at Qk, is caused by all the other charges and h
22、as the following expression: (3) A group of N discrete point charges at rest 25 1 1 2 N e k k k W Q V Two remarks are in order here. First, We can be negative. In that case, work is done by the field (not against the field). Second, We in this equation represents only the interaction energy (mutal e
23、nergy) and does not include the work required to assemble the individual point charges themselves (self-energy). 10 1 4 N j k j jk jk QV R (4) a continuous charge distribution 1 2e VW V dv We replace Qk by dv and the summation by an integration and obtain: Note that We in this equation includes the
24、work (self-energy) required to assemble the distribution of macroscopic charges, because it is the energy of interaction of every infinitesimal charge element with all other infinitesimal charge elements. 26 Example 3-23 P136 Find the energy required to assemble a uniform sphere of charge of radius
25、b and volume charge density . R b dR 1 2e VW V dv 27 (4) Electrostatic Energy in Terms at Field Quantities 11 () 22e VVW Vdv D Vdv Recalling the vector identity, ( ) ,V D V D D V 1 1 1 ( ) ( ) 2 2 2 11 22 e V V V nSV W V D D V dv V D dv D V dv V D a ds D E dv We can write as 28 11 22en SVW V D a ds
26、D E dv Since V can be any volume that includes all the charges, we may choose it to be a very large sphere with radius R (R ). 2 2 11, , w he n the f ir st inte gr a l w il l v a nishV D S R R RR ， . 1 ( J ) 2e VW D Edv For a linear medium, we have D=E, 21 ( J) 2e VW E d v 21 ( J )2e V DW d v 29 11
27、22eeV V VW D E dv D E dv w dv We can always define an electrostatic energy density we mathematically, 2 2311 ( J /c m ) 2 2 2e Dw D E E However, this definition of energy density is artificial because a physical justification has not been found to localize energy with an electric field; all we know
28、is that the volume integrals in Equation give the correct total electrostatic energy. 30 Example 3-24 P138 QSC Vd y x 2211 2 2 2 Q e CW Q V C V 電容器儲存的總能量： 31 summary 1. Boundary Conditions for Electrostatic Fields 2 1 2 1 2 2 1 2 1 2 ( ) 0 o r ) o r n t t n s n n s a E E E E a D D D D （ 2. Capacitan
29、ce and Capacitors ( c /v or F )QC V 12 (c / v o r F )QC V 32 3. Electrostatic Energy 1 1 2 N e k k k W Q V 10 1 4 N j k j jk jk QV R 1 2e VW V dv 11 22eeV V VW D E dv D E dv w dv 2 2311 ( J /c m ) 2 2 2e Dw D E E 2211 2 2 2 Q e CW Q V C V 33 homework Thank you! Bye-bye! 答疑安排 時(shí)間： 地點(diǎn)： 1401， 1403 P.3-2
30、5, 3-44 34 In addition, we can find the relationship between the bound charges and the normal component of the electric field intensity as 12 n n sPP 122 ( ) nsa P P dSq PS 21 dS top s ide bottomP d S P d S P d S PS 12 1 2 1 22 2 2 1n 2n d ( ) ( S top bo tt om n n n s P d S P d S P a S P a S a P P S
31、 P P S S PS 1 n 2 n sP S P S S h S 2 1 an2 D 1 D2 s 35 2) Capacitances In Multiconductor Systems 3 2 N 1 1 11 1 12 2 1 2 21 1 22 2 2 1 1 2 2 , , . NN NN N N N NN N V p Q p Q p Q V p Q p Q p Q V p Q p Q p Q 1 1 1 1 1 2 2 1 2 2 1 1 2 2 2 2 1 1 2 2 , , . NN NN N N N NN N Q c V c V c V Q c V c V c V Q c V c V c V
電磁場與電磁波第8講邊界條件電容能量
