电磁场与电磁波课件(电子科大)-FWEJ-Ch04解读.ppt

1、Chapter 4 Steady Electric CurrentsElectric current,Electromotive forcePrinciple of current continuity,Energy dissipation.1.Current&Current Density2.Electromotive Force3.Principle of Current Continuity 4.Boundary Conditions for Steady Electric Currents 5.Energy Dissipation in Steady Electric Current

2、Fields6.Electrostatic Simulation1.Current&Current Density Classification:Conduction current and convention current.The conduction current is formed by the free electrons(or holes)in a conductor or the ions in an electrolyte.The convection current is resulting from the motion of the electron,the ions

3、,or the other charged particles in vacuum,a solid,a liquid or a gas.The amount of charge flowing across a given area per unit time is called the electric current intensity or electric current,and it is denoted by I.The unit of electric current is ampere(A)The relationship between electric current I

4、and electric charge q istqIdd The current density is a vector,and it is denoted as J.The direction of the current density is the same as the flowing direction of the positive charges,and the magnitude is the amount of charge through unit cross-sectional area per unit time.The relationship between th

5、e current element dI across a directed surface element dS and the current density J isSJ ddIThe electric current across the area S isSI d SJWhich states that the electric current across an area is the flux of the current density through the area.In most conducting media,the conduction current densit

6、y J at a point is proportional to the electric field intensity E at that point so thatEJwhere is called the conductivity,and its unit is S/m.A large means that the conducting ability of the medium is stronger.The above equation is called the differential form of the following Ohms lawIRU A conductor

7、 with infinite is called a perfect electric conductor,or p.e.c.A medium without any conductivity is called a perfect dielectric or an insulator.In a perfect electric conductor,electric current can be produced without the influence of an electric field.There is no steady electric field in a perfect e

8、lectric conductor.Otherwise,an infinite current will be generated,and it results in an infinite energy.In nature there exists no any p.e.c.or perfect dielectric.The conductivities of several mediaunit in S/mMediaConductivitiesMediaConductivitiesSilverSea waterCopperPure waterGoldDry soilAluminumTran

9、sformer oilBrassGlassIronRubber71017.671080.531071010.451071054.3111071057.1121071015104 The magnitude of the current density of the convection current is not proportional to the electric field intensity,and the direction may be different from that of electric field intensity.vJ As the polarization

10、of dielectrics,the conducting properties of a medium can be homogeneous or inhomogeneous,linear or nonlinear,and isotropic or anisotropic,with same meanings as before.If the charge density is ,and the moving velocity is v,and thenThe above equations are valid only for a linear isotropic medium.ECond

11、ucting medium2.Electromotive Force We first discuss the chemical action inside the impressed source under open-circuit condition.In the impressed source,under the influence of non-electrostatic force the positive charges will be moved continuously to the positive electrode plate P,while the negative

12、 charges to the negative electrode plate N.These charges on the plates produce an electric field E,with the direction pointed to the plate N from the plate P,and the electric field E will be stronger with the increase of the charges on the plates.PNEImpressed sourceE The electric force caused by the

13、 charges on the plates will resist the movement of the charges in the source.When the electric force is equal to the non-electrostatic force,the charges are stopped,and the charges on the plates will be constant.This impressed electric field intensity is still defined as the force acting on unit pos

14、itive charge,but it is denoted as E.Since the non-electrostatic force behaves as the force acting on the charge,the non-electrostatic force is usually considered as that produced by an impressed electric field.The impressed electric field E pushes the positive charges to the positive electrode plate

15、,and the negative charges to the negative electrode plate,and the direction of is opposite to that of the electric field E produced by the charges on the plates.If the conducting medium is connected,the positive charges on the positive electric plate will be moved to the negative electric plate thro

16、ugh the conducting medium,while the negative charges on the negative electric plate to the positive electric plate.In this way,the charges on the plates will be decreased,and E E.The charges in the source will be moved again.when the impressed electric field is equal but opposite to the electric fie

17、ld produced by the charges on the plates,and the charges will be at rest.The impressed source will continuously provide the positive charges to the positive electric plate,whereas the negative charges to the negative electric plate,and in view of this a continuous current is formed.When it is in dyn

18、amical balance,the charges on the plates will be constant,and they produce a steady electric field in the impressed source and in the conducting medium.In the impressed source,and there is a steady current in the circuit consisting of the impressed source and the conducting medium.EE Consequently,in

19、 order to generate the continuous current in the conducting medium,it must rely on an impressed source.Although the distribution of the charges on the plates is unchanged,the charges are not at rest.These charges are replaced without interruption.Hence,they are called sustained charges.The steady el

20、ectric field in conducting medium is produced just by the sustained charge.Once the impressed source is disconnected,the supply of sustained charge to the conducting medium will vanish.The line integral of the impressed electric field along the path from the negative electric plate N to the positive

21、 electric plate P is defined as the electromotive force of the impressed source,and it is denoted as e,i.e.lEd PNe When it is in dynamical balance,in the impressed source.Therefore,the above equation can be rewritten asEElE d PNe The steady electrode field caused by the sustained charges on the plat

22、es is also a conservative field,and the line integral of it around a closed circuit should be zero,i.e.l 0d lEFor homogeneous media,the above equation becomesl 0d lJConsider that in the conducting medium,we haveEJl 0d lJUsing Stokes theorem,we have0 J0 J In homogeneous conducting media the steady el

23、ectric current field is irrotational.3.Principle of Current Continuity Assume the density of the sustained charges in the volume V bound by the closed surface S is,thenVVq dVttqSVdd SJthen Because the distribution of the charges in the steady electric current field is independent of time,i.e.,we fin

24、d0tS 0d SJwhich states that in the steady electric current field the flux of the current density through any closed surface will be zero.In this way,the electric current lines must be closed,with no beginning or end.This result is called the principle of current continuity.If we use a set of curves

25、to describe the current field and let the tangential direction at a point on the curves be the direction of the current density at the point,these curves can then be called the electric current lines.By using the divergence theorem,we obtaint Jwhich is called the charge conservation principle in dif

26、ferential form.Hence,for a steady electric current field,we have0 Jwhich states that the steady electric current field is solenoidal.4.Boundary Conditions for Steady Electric Currents The integral forms of the equations for steady electric current field are as follows:l 0d lJS 0d SJ0 J0 JAnd the cor

27、responding differential forms are From the equations in integral form we can find the boundary condition for the tangential components of the current densities to be 2t21t 1JJAnd the normal components are2n1nJJ The tangential components of the current densities are discontinuous,while the normal com

28、ponents are continuous.Since ,we find the boundary conditions for the steady electric field can be obtained as follows:EJn221n12tt 1EEEE2t21t 1JJ2n1nJJ Since there is no the electric field cannot in a perfect electric conductor,the tangential components of a steady electric current cannot exist on t

29、he surface.Therefore,when an electric current flows into or out of a perfect electric conductor,the electric current lines are always perpendicular to the surface.5.Energy Dissipation in Steady Electric Current Fields In a conducting medium,the collision of free electrons with the atomic lattice wil

30、l generate thermal energy,and this is an irreversible energy conversion process.The impressed source has to compensate the energy dissipation in order to maintain the steady electric current.In a steady electric current field,we construct a small cylinder of length dl and end face area dS,and assume

31、 the two end faces of the cylinder are equipotential surfaces.dlUJdS Under the influence of the electric field,electric charge dq is moved to the right end face from the left end face in dt,with the corresponding work done by the electric force aslqEqWdddddlEThe power dissipation P islSEJlEIltqEtWPd

32、dddddddThen the power dissipation per unit volume as22JEEJpl If the direction of J is different from that of E,the above equation can be written in the following general formJE lpWhich is called the differential form of Joules law,and it states that the power dissipation at a point is equal to the p

33、roduct of the electric field intensity and the current density at the point.Suppose the electric potential difference between two end faces is U,then .And we know that .Hence,the power dissipation per unit volume can be expressed aslUEdSIJdVUIlSUIpldddThe total power dissipation in the cylinder isUI

34、VpPldwhich is Joules law.Example 1.A parallel plate capacitor consists of two imperfect dielectrics in series.Their permittivities are 1 and 2,the conductivities are 1 and 2,and the thickness are d1 and d2,respectively.If the impressed voltage is U,find the electric field intensities,the electric en

35、ergies per unit volume,and the power dissipations per unit volume in two dielectrics.Solution:Since no current exists outside the capacitor,the electric current lines in the capacitor can be considered to be perpendicular to the boundaries.Then we have2211EEUdEdE2211In view of this we find UddE12212

36、1UddE122112 1 1 2 2d1d2UThe electric energies per unit volume in two dielectrics,respectively,are 2222e2111e21 ,21EwEwThe power dissipations per unit volume in two dielectrics,respectively,are 22222111 ,EpEpllTwo special cases are worth noting:If ,then ,.0201E0e1w01lp22dUE If ,then ,.0111dUE 02E02ew

37、02lpd1d2 1=0E 2=0UE 1=0 2=0U Example 2.A quarter of a flat circular conducting washer is shown in the figure.Calculate the resistance between two end faces.Uyxtabr0(r,)0 Solution:The cylindrical coordinate system should be selected.Assume the electric potential difference between two end faces is U,

38、and let Since the electric potential is related to the angle,it should satisfy the following equation0dd22The general solution is21CCThe electric potential at 010The electric potential atU22 Based on the given boundary conditions,we find2UrUr2eeEJThe current density J in the conducting medium is The

39、n the current I flowing into the conducting medium across the end face at is2)d(2drtrUISSeeSJ abUtrrUtbaln2d2Consequently,the resistance R between two end faces isabtIURln 26.Electrostatic Simulation Two fields are found to be very similar in source-free region.Steady Electric Current Field)0(EElect

40、rostatic Field)0(0d llJ0d llE0d SSJ0d SSE0J0E0 J0 E The electric current density J corresponds to the electric field intensity E,and the electric current lines to the electric field lines.If the steady electric current field has the same boundary conditions as that for the electrostatic field,the di

41、stribution of the current density will be the same as that of the electric field intensity.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 condit

42、ions,and this method is called electrostatic simulation.Based on this similarity,the solution of the steady electric current field can be found directly from the results of the electrostatic field.The electrostatic field and the steady electric current field between two electrodes as follows:PNStead

43、y electric current fieldPNElectrostatic field The calculation of the resistance of conducting media can be determined based on the results of the corresponding electrostatic field.CR CG Based on the equations for two fields,we can find the resistance and conductance between two electrodes as If the

44、capacitance between two electrodes is known,from the above equations the resistance and the conductance between two electrodes can be found out.The capacitance of a coaxial line per unit length is ,where b is the inner radius of the outer conductor,and a is the radius of the internal conductor.If th

45、e conductivity of the filled dielectric is,the leakage conductance per unit length G1 is)/ln(21abC )/ln(21abGdSdSG The capacitance of a parallel plate capacitor of plate area S and separation d is .If the conductivity of the imperfect dielectric is,the leakage conductance G between two electric plates of the parallel plate capacitor isdSCCR CG