電磁場(chǎng)與電磁波第12講焦耳定律邊界條件電阻計(jì)算及第5章復(fù)習(xí)(編輯修改稿)
2025-06-22 13:27 本頁(yè)面
【文章內(nèi)容簡(jiǎn)介】 ???? ? ?????1 122ttJJ??? ? ?1 2 2 1 2? 0n n nJ J a J J? ? ?the normal ponent of current density vector J being continuous. the ratio of the tangential ponents of current density vector J at two sides of an interface is equal to the ratio of the conductivities. 15 ? 1 , ?1 an2 E2 , D2 , J2 ?s ? 2 , ?2 E1 , D1 , J1 When a steadycurrent flows across the boundary between two different lossy dielectrics: ? ?? ?2 1 22 1 2?? 0nsna D Dd ie le c tric sa E E?? ? ? ???? ? ???? ?122122 1 2? 0? 0nnJJac o n d u c to ra J J??? ??? ? ?? ?????? ? ? ??? ?? ?? ?2 1 22 1 22 1 2? ? 0? 0nsnna D Da E Ea J J?? ????? ? ???? ? ???16 For a homogeneous conducting medium 0J? ? ?We know that a curlfree vector field can be expressed as the gradient of a scalar potential field. Let us write J ?? ??Substitution of this equation into 0J? ? ?yields a Laplace’s equation in ? 。 that is 2 0???17 Example 54 P214: An emf ? is applied across a parallelplate capacitor of area S. The space between the conductive plates is filled with two different lossy dielectrics of thickness d1 and d2 , permittivities ?1 and ?2 ,and conductivities ?1 and ?2 respectively. Determine (a)the current density between the plates, (b)the electric field intensities in both dielectrics, and (c)the surface charge densities on the plates and at the interface. x 1d2d1 2 2na?1E2E1Jsi??S1?2?2?1?2J+ + + + + + + + + + + + 1?2?y o 18 1. ISdSdRRI )()(221121 ??? ????122121122121 /??????????ddSddSSIJ?????? ? 2121 2 2 1? ( A / m )yJadd? ? ??????x 1d2d1 2 2na?1E2E1Jsi??S1?2?2?1?2J+ + + + + + + + + + + + 1?2?y o 2. JE??Method one ? ?? ?211 2 2 1121 2 2 1??yyEaddEadd????????????Or method two 1 1 2 21 1 2 2E d E dEE??????? ??19 3. ? ? ? ?2 1 2 1 1 2 2? ?: n s y s iB C a D D a E E? ? ? ?? ? ? ? ? ? ?? ? ? ?21121 2 2 1 1 2 2 1? ? ?y y y s ia a ad d d d? ? ? ?? ? ?? ? ? ???? ? ? ? ???????? ?2 1 1 2 21 2 2 1 ( C/m )si dd? ? ? ? ?? ????? ?? ? ? ?? ?? ? ? ?? ?2 1 2 1 1 121 2 2 111 2 2 1 1 2 2 12 1 2 2 2 222 1 1 221 2 2 1 1 2 2 1? ?up
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