電磁場(chǎng)與電磁波課件(電子科大)FWEJCh4

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 Elec

2、tric Current 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 motio

3、n of the electron, the ions, 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 relat

4、ionship between electric current I 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

5、 unit time. The relationship between the 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 conduc

6、ting media, the conduction current density 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 f

7、orm of the following Ohms lawIRU A conductor 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.

8、 There is no steady electric field in a perfect electric 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/mMediaConductivitiesMediaConductivitiesSi

9、lverSea waterCopperPure waterGoldDry soilAluminumTransformer 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 fr

10、om that of electric field intensity. vJ As the polarization 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

11、 above equations are valid only for a linear isotropic medium. EConducting 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

12、 moved continuously to the positive electrode plate P, while the negative 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

13、 charges on the plates.PNEImpressed sourceE The electric force caused by the 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 imp

14、ressed electric field intensity is still defined as the force acting on unit positive 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 im

15、pressed electric field E pushes the positive charges to the positive electrode plate, 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

16、charges on the positive electric plate will be moved to the negative electric plate through 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 w

17、ill be moved again. when the impressed electric field is equal but opposite to the electric field 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 t

18、o the negative electric plate, and in view of this a continuous current is formed. When it is in dynamical 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 ste

19、ady current in the circuit consisting of the impressed source and the conducting medium. EE Consequently, in 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 no

20、t at rest. These charges are replaced without interruption. Hence, they are called sustained charges. The steady electric 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 vanis

21、h. The line integral of the impressed electric field along the path from the negative electric plate N to the positive 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. Theref

22、ore, the above equation can be rewritten asEElE d PNe The steady electrode field caused by the sustained charges on the plates 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 lJConside

23、r that in the conducting medium, , we haveEJl 0d lJUsing Stokes theorem, we have0 J0 J In homogeneous conducting media the steady electric 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 , t

24、henVVq dVttqSVdd SJthen Because the distribution of the charges in the steady electric current field is independent of time, i.e. , we find0tS 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 electri

25、c 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 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 ca

26、n then be called the electric current lines.By using the divergence theorem, we obtaint Jwhich is called the charge conservation principle in differential form. Hence, for a steady electric current field, we have0 Jwhich states that the steady electric current field is solenoidal.4. Boundary Conditi

27、ons 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 corresponding differential forms are From the equations in integral form we can find the boundary condition for the tangential components of the current d

28、ensities to be 2t21t 1JJAnd the normal components are2n1nJJ The tangential components of the current densities are discontinuous, while the normal components are continuous. Since , we find the boundary conditions for the steady electric field can be obtained as follows:EJn221n12tt 1EEEE2t21t 1JJ2n1

29、nJJ Since there is no the electric field cannot in a perfect electric conductor, the tangential components of a steady electric current cannot exist on the surface. Therefore, when an electric current flows into or out of a perfect electric conductor, the electric current lines are always perpendicu

30、lar to the surface.5. Energy Dissipation in Steady Electric Current Fields In a conducting medium, the collision of free electrons with the atomic lattice will generate thermal energy, and this is an irreversible energy conversion process. The impressed source has to compensate the energy dissipatio

31、n 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 the two end faces of the cylinder are equipotential surfaces. dlUJdS Under the influence of the electric field, electric charge dq is mo

32、ved 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 islSEJlEIltqEtWPddddddddThen the power dissipation per unit volume as22JEEJpl If the direction of J is different from that of E, the above equation ca

33、n 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 product of the electric field intensity and the current density at the point. Suppose the electric potential difference between two

34、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 isUIVpPldwhich is Joules law. Example 1. A parallel plate capacitor consists of two imperfect dielectrics in series. Their permitt

35、ivities 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 energies per unit volume, and the power dissipations per unit volume in two dielectrics. Solution: Since no current e

36、xists 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 UddE122121UddE122112 1 1 2 2d1d2UThe electric energies per unit volume in two dielectrics, respectively, are 2222e2111e

37、21 ,21EwEwThe power dissipations per unit volume in two dielectrics, respectively, are 22222111 ,EpEpllTwo special cases are worth noting:If , then , , , .0201E0e1w01lp22dUE If , then , , , .0111dUE 02E02ew02lpd1d2 1= 0E 2= 0UE 1= 0 2= 0U Example 2. A quarter of a flat circular conducting washer is

38、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, and let Since the electric potential is related to the angle , it should satisfy the

39、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 Then the current I flowing into the conducting medium across the end face at is2)d (2

40、drtrUISSeeSJ 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(EElectrostatic Field )0(0d llJ0d llE0d SSJ0d SSE0J0E0 J0 E The electric current den

41、sity 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 distribution 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,