电磁场与电磁波第12讲焦耳定律边界条件电阻计算及第5章复习-课件.ppt（33页）

1、Field and Wave Electromagnetic电磁场与电磁波电磁场与电磁波2015.10.3121.Current Density and Ohms Law SIJ dsA2 (/)JuA m2 (A/m)JE12lIVIRIsGReview33.Equation of Continuity and Kirchhoffs Current Law2.Electromotive Force and Kirchhoffs Voltage Law1121212221dd=d-VVViEl=El El Outside the sourceInside the source (V)jkkjk

2、R I3 (A/m)Jt 0jjI EConducting mediumPNEImpressed sourceEi0J0SJ dsSdQJ dsdt 4We are now in a position to prove this statement and to calculate the time it takes to reach an equilibrium.tEJEJt E0twhere 0 is the initial charge density at t=0.The time constant is called the relaxation time(驰豫时间）驰豫时间）.铜，

3、铜，1.521.5210-19SAn initial charge density 0 will decay to 1/e or 36.8%of its value in a time equal to0ttCee5For a homogeneous conducting medium0JWe know that a curl-free vector field can be expressed as the gradient of a scalar potential field.Let us writeJ Substitution of this equation into 0Jyield

4、s a Laplaces equation in ;that is206Two fields are found to be very similar in source-free region.Steady Electric Current Field)0(E d0lJl d0SJS0J0JElectrostatic Field)0(d0lEl d0SES0E0E The electric current density J corresponds to the electric field intensity E,and the electric current lines to the

5、electric field lines.In some cases,since the steady electric current field is easy to be constructed and measured,the electrostatic field can be investigated based on the steady electric current field with the same boundary conditions,and this method is called electrostatic simulation.7Capacitancess

6、LLD dsE dsQCVE dlE dl JEResistanceLLssE dlE dlVRIJ dsE dsCRCGBased on the equations for two fields,we can find the resistance and conductance between two electrodes as8In certain situations,electrostatic and steady-current problems are not exactly analogous,even when the geometrical configurations a

7、re the same.This is because current flow can be confined strictly within a conductor(which has a very large in comparison to that of the surrounding medium),whereas electric flux usually cannot be contained within a dielectric slab of finite dimensions.The range of the dielectric constant of availab

8、le materials is very limited,and the fluxfringing around conductor edges makes the computation of capacitance less accurate.9Main topic 恒定电流恒定电流3.电阻的计算电阻的计算1.功率耗散和焦耳定律功率耗散和焦耳定律2.电流密度的边界条件电流密度的边界条件101.功率耗散和焦耳定律功率耗散和焦耳定律 宏观上，导体中的电子受电场的影响，发生漂移运动；在微观上，宏观上，导体中的电子受电场的影响，发生漂移运动；在微观上，这些电子与晶格上的原子发生碰撞。因此，能量从电

9、场传到作热振动的这些电子与晶格上的原子发生碰撞。因此，能量从电场传到作热振动的原子上。将电荷移动了一段距离原子上。将电荷移动了一段距离 ，电场，电场E 作的功为作的功为q E ，则其所对则其所对应的功率为：应的功率为：00limlimttWqE dlPqE utt 其中其中 u 为漂移速度。传递到体积为漂移速度。传递到体积 dv 内所有电荷载体的总功率为内所有电荷载体的总功率为:3d or (W/m)iiiiiiiiiiPpEN qudvdPdPE JdvE JdvJN qu 11The total electric power converted into heat in volume V:

10、(W)VPE Jdv This is known as Joules law.The point function EJ is a power density under steady-current conditions.In a conductor of a constant cross section,we can written as2 WPI R122.电流密度的边界条件电流密度的边界条件When current obliquely crosses an interface between two media with different conductivities(1 2),th

11、e current density vector changes both in direction and in magnitude.A set of boundary conditions can be derived for J in a way similar to that used in Section 3-9 for obtaining the boundary conditions for D and E.The governing equations for steady current density J in the absence of non-conservative

12、 energy sources are0J0J Differential form0SJ ds0CJdlIntegral formGoverning Equations for Steady Current Density13E2E1 2 1at w hacdban2hS 2 1an2D1D2 s D0E0CE dl SD dsQ2121221212()0 or )or nttnsnnsaEEEEaDDDD（121t2t1t2t d d d d d0 d d()0bcdalabcdbdacEwEwEE 1212ElElElElElElElEwEw21 1212222121n2nd()(sSbo

13、ttomsidetopnntopbottomnsDdSD dSDdSSDdSDdSDaSDaSaDDSDDSS DS140J 0J 0SJds0CJdlJ2J1 2 1at w hacdban2hS2 1an2J1J2 s122120nJJa1122ttJJ12212 -0nnnJJaJJthe normal component of current density vector J being continuous.the ratio of the tangential components of current density vector J at two sides of an int

14、erface is equal to the ratio of the conductivities.15 1,1an2E2,D2,J2 s 2,2E1,D1,J1When a steady-current flows across the boundary between two different lossy dielectrics:2122120nsnaDDdielectricsaEE1221221200nnJJaconductoraJJ212212212-00nsnnaDDaEEaJJ16For a homogeneous conducting medium0JWe know that

15、 a curl-free vector field can be expressed as the gradient of a scalar potential field.Let us writeJ Substitution of this equation into 0Jyields a Laplaces equation in ;that is2017Example 5-4 P214:An emf is applied across a parallel-plate capacitor of area S.The space between the conductive plates i

16、s 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 int

17、erface.x1d2d122na1E2E1JsiS12212J+-12yo181.ISdSdRRI)()(221121122121122121/ddSddSSIJ2121221 (A/m)yJadd x1d2d122na1E2E1JsiS12212J+-12yo2.JEMethod one211221121221-yyEaddEaddOr method two11221122E dE dEE193.2121122:nsysiBC aDDaEE211212211221yyysiaaadddd211221221 (C/m)sidd 21211121221112211221212222221122

18、12211221up surface:(C/m)lower surface:(C/m)nsyyynsyyyaDDaEaaddddaDDaEaadddd 201.Choose an appropriate coordinate system for the given geometry.2.Assume a potential difference V0 between conductor terminals.3.Find E from E=-V(2V=0),or other relations.4.Find the total current where S is the cross-sect

19、ional area over which I flows.5.Find resistance R by taking the ratio V0/I.SSIJ dsE dsThe procedure for computing the resistance of a piece of conducting material between specified equipotential surfaces(or terminals)is as follows:3.电阻计算电阻计算(Resistance Calculations)21 Example 5-6.一导电材料形状是四分之一的扁平圆垫圈一

20、导电材料形状是四分之一的扁平圆垫圈,其均匀其均匀厚度为厚度为 h,电导率为电导率为 ,垫圈内半径为垫圈内半径为 a,外半径外半径b,如下图所示如下图所示,求端面求端面之间的电阻之间的电阻.0lVdlRIs0 AssumeVE20 VEV IJJEIJ dsRIVR00 AssumeIJERV0IJSJE0LVE dl 00VRI思路思路V0yxhabr0(r,)0 22 Solution:The cylindrical coordinate system should be selected.Assume the electric potential difference between tw

21、o end faces is V0,and let Since the electric potential V is related to the angle ,it should satisfy the following equation22d0dVThe general solution is12VCCThe electric potential at 0V 0The electric potential at0VV2V0yxhabr0(r,)0 23 Based on the given boundary conditions,we find02VV02VVJEVaarr The c

22、urrent density J in the conducting medium is Then the current I flowing into the conducting medium across the end face at is202()bSaVIJ dSaa hdrr 0022dlnbaV hV hrbraConsequently,the resistance R between two end faces is2lnURbIha24summary1.Power Dissipation and Joules Law (W)VPE Jdv 2 WPI R3.Resistan

23、ce CalculationsCRCGelectrostatic simulation252.Boundary Conditions for Current Density0J 0J Differential form0SJds0CJdlIntegral formGoverning Equations for Steady Current Density212212212-00nsnnaDDaEEaJJWhen a steady-current flows across the boundary between two different lossy dielectrics:26homewor

24、kThank you!Bye-bye!Thank you!Bye-bye!答疑安排答疑安排时间：时间：地点：地点：1401，1403P.5-15271.Current Density and Ohms Law SIJ dsA2 (/)JuA m2 (A/m)JE12lIVIRIsGReview Steady Electric Currents283.Equation of Continuity and Kirchhoffs Current Law2.Electromotive Force and Kirchhoffs Voltage Law1121212221dd=d-VV VEl=El El

25、 Outside the sourceInside the source (V)jkkjkR IEConducting mediumPNEImpressed sourceE3 (A/m)Jt 0jjI 0J0SJ dsSdQJ dsdt 294.Power Dissipation and Joules Law (W)VPE Jdv 2 WPI R6.Resistance CalculationsCRCGelectrostatic simulation305.Boundary Conditions for Current Density0J 0J Differential form0SJds0C

26、JdlIntegral formGoverning Equations for Steady Current Density212212212-00nsnnaDDaEEaJJWhen a steady-current flows across the boundary between two different lossy dielectrics:31abc 1,1 2,2V0I111222012001212211ln221ln222()()lnlncabcdrcRrLLadrbRrLLcVLVVRR IIcbRRac 3212012021212(lnln)2(lnln)LVVIJcbcbSrLracac abc 1,1 2,2V0I2122-nsnraDDaa12011111212102222221(lnln)(lnln)VJDEcbracVJDEcbrac 3312012121022121120212(2:;1:1)(lnln)(1:;2:2)(lnln)(1:2;2:1)()(sarsbrscrraconductorsdielectricsVaDcbaacrbconductorsdielectricsVaDcbbacrcdielectricsdielectricsVaDD 1lnln)cbcacabc 1,1 2,2V0I2122-nsnraDDaa