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.ppt

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1、Field and Wave Electromagnetic电磁场与电磁波2015.10.3121.Current Density and Ohms LawReview33.Equation of Continuity and Kirchhoffs Current Law2.Electromotive Force and Kirchhoffs Voltage LawOutside the sourceInside the sourceEConducting mediumP N E Impressed sourceEi4We are now in a position to prove this

2、 statement and to calculate the time it takes to reach an equilibrium.where 0 is the initial charge density at t=0.The time constant is called the relaxation time(驰豫时间）.铜，=1.5210-19SAn initial charge density 0 will decay to 1/e or 36.8%of its value in a time equal to5For a homogeneous conducting med

3、iumWe know that a curl-free vector field can be expressed as the gradient of a scalar potential field.Let us writeSubstitution of this equation into yields a Laplaces equation in;that is6Two fields are found to be very similar in source-free region.Steady Electric Current Field Electrostatic Field T

4、he electric current density J corresponds to the electric field intensity E,and the electric current lines to the 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

5、 current field with the same boundary conditions,and this method is called electrostatic simulation.7CapacitanceJEResistanceBased 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 n

6、ot exactly analogous,even when the geometrical configurations are 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 fi

7、nite dimensions.The range of the dielectric constant of available materials is very limited,and the fluxfringing around conductor edges makes the computation of capacitance less accurate.9Main topic 恒定电流3.电阻的计算1.功率耗散和焦耳定律2.电流密度的边界条件101.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，这些电子与晶格上的原子发生碰撞。因此，能量从电场传

8、到作热振动的原子上。将电荷移动了一段距离，电场E 作的功为q E，则其所对应的功率为：其中 u 为漂移速度。传递到体积 dv 内所有电荷载体的总功率为:1 1The total electric power converted into heat in volume V: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 as122

9、.电流密度的边界条件When current obliquely crosses an interface between two media with different conductivities(1 2),the 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 c

10、onditions for D and E.The governing equations for steady current density J in the absence of non-conservative energy sources areDifferential formIntegral formGoverning Equations for Steady Current Density13E2E1 2 1atwhacdban2hS 2 1an2D1D2s14J2J1 2 1atwhacdban2hS2 1an2J1J2sthe normal component of cur

11、rent density vector J being continuous.the ratio of the tangential components of current density vector J at two sides of an interface is equal to the ratio of the conductivities.15 1,1an2E2,D2,J2s 2,2E1,D1,J1When a steady-current flows across the boundary between two different lossy dielectrics:16F

12、or a homogeneous conducting mediumWe know that a curl-free vector field can be expressed as the gradient of a scalar potential field.Let us writeSubstitution of this equation into yields a Laplaces equation in;that is17Example 5-4 P214:An emf is applied across a parallel-plate capacitor of area S.Th

13、e 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

14、 densities on the plates and at the interface.x12+-yo181.x12+-yo2.Method oneOr method two193.201.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

15、is the cross-sectional area over which I flows.5.Find resistance R by taking the ratio V0/I.The 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(Resistance Calculations)21 Example 5-6.一导电材料形状是四分之一的扁平圆垫

16、圈,其均匀厚度为 h,电导率为,垫圈内半径为 a,外半径b,如下图所示,求端面之间的电阻.思路V0yxhabr0(r,)022 Solution:The cylindrical coordinate system should be selected.Assume the electric potential difference between two end faces is V0,and let Since the electric potential V is related to the angle,it should satisfy the following equationTh

17、e general solution isThe electric potential at The electric potential atV0yxhabr0(r,)023 Based on the given boundary conditions,we findThe current density J in the conducting medium is Then the current I flowing into the conducting medium across the end face at isConsequently,the resistance R betwee

18、n two end faces is24summary1.Power Dissipation and Joules Law3.Resistance Calculationselectrostatic simulation252.Boundary Conditions for Current DensityDifferential formIntegral formGoverning Equations for Steady Current DensityWhen a steady-current flows across the boundary between two different l

19、ossy dielectrics:26homeworkThank you!Bye-bye!答疑安排时间：地点：1401，1403P.5-15271.Current Density and Ohms LawReview Steady Electric Currents283.Equation of Continuity and Kirchhoffs Current Law2.Electromotive Force and Kirchhoffs Voltage LawOutside the sourceInside the sourceEConducting mediumP N E Impress

20、ed sourceE294.Power Dissipation and Joules Law6.Resistance Calculationselectrostatic simulation305.Boundary Conditions for Current DensityDifferential formIntegral formGoverning Equations for Steady Current DensityWhen a steady-current flows across the boundary between two different lossy dielectrics:31abc1,12,2V0I32abc1,12,2V0I33abc1,12,2V0I