电磁场与电磁波课件(电子科大)FWEJCh2

1、Chapter 2 Static Electric Fields Electric Field Intensity, Electric PotentialPolarization of Dielectric, Field EquationsBoundary Conditions, Energy and Force1. Field Intensity, Flux, and Field Lines2. Equations for Electrostatic Fields in Free Space 3. Electric Potential and Equipotential Surfaces4.

2、 Polarization of Dielectrics5. Equations for Electrostatic Fields in Dielectric6. Boundary Conditions for Dielectric Interfaces 7. Boundary Conditions for Dielectric-conductor Interface 8. Capacitance9. Energy in Electrostatic Field10. Electric Forces1. Field Intensity, Flux, and Field Lines The int

3、ensity of the electric field at a point is defined as the force produced by the electric field on a unit positive charge at that point, and is denoted by EqFE Where q is the test charge, and F is the force acting on the charge. The flux of the electric field intensity through a surface is called ele

4、ctric flux, and it is denoted as , given by, ,i.e. SSE d0d lEThe vector equation for electric field linesElectric field tubeTwo parallel charge plates A negativepoint charge A positive point chargeDistributions of electric field lines2. Equations for Electrostatic Fields in Free Space Physical exper

5、iments show that the intensity of an electrostatic field in free space satisfies the following two equations in integral formSq0d SEl 0d lEWhere 0 is the permittivity (or dielectric constant) of free space and its value is The left equation is called Gausss law, and it shows that the outward flux of

6、 the electric field intensity of an electrostatic field over any closed surface in free space is equal to the ratio of the charge in the closed surface to the permittivity of free space. F/m10361m/F10854187817. 89120 The right equation states that the circulation of an electrostatic field around any

7、 closed curve is equal to zero.0 E0Ewhich shows that the curl of an electrostatic field in free space is zero everywhere. Using the divergence theorem, from Gauss law we haveIt is called the differential form of Gausss law. It shows that the divergence of the electric field intensity of an electrost

8、atic field at a point in free space is equal to the ratio of the density of the charge at the point to the permittivity of free space.From Stokes theorem and the above equation, we have The electrostatic field in free space is a lamellar or irrotational one. AEVVVV d)( 41)(d)( 41)(|rr |rErA|rr |rErw

9、here After knowing the divergence and the rotation of the electric field intensity, one may write, with the aid of the Helmholtzs theorem xPzyrOVd)(rrrrVVV 0d)(41)(|rr |rr0)(rA Substituting the electric field equations into the above results givesEHence The scalar function is called the electric pot

10、ential, and the electric field intensity at a point in free space is equal to the negative gradient of the electric potential at that point. E According to the National Standard of China, the electric potential is denoted by the Greek small character , i.e. If the electric charge is distributed on a

11、 surface S or on a curve l, we haveSSS 0d|)(41)(|rrrrSSS 30d|)(41)(|rrrrrrEll d)(41)(0|rr|rrllll 30d|)(41)(|rrrrrrE Substituting the electric potential expression into this equation, we haveVVd4)()(30rrrrrrE It is easy to see that the electric field intensity can be determined directly from the abov

12、e equations if the distribution of the charge is known.（a）The charge q in the Gauss law should be the sum of all positive and negative charges in the closed surface S. Summary（b）The electric field lines cannot be closed and intersect each other. （c）The line integral of the electric field intensity a

13、long a path between any two points is independent of the path, and electrostatic field is a conservative field as the gravitational field.（d）If the distribution of the charge is known, the electric field intensity can be found based on Gauss law, the electric potential, or the distribution of the ch

14、arge. Example 1 Calculate the electric field intensity produced by a point charge. Solution: The point charge is the charge whose volume is zero. Because of the symmetry of the point charge, if the point charge is placed at the origin of a spherical coordinate system, the electric field intensity mu

15、st be independent of the angles and . Construct a sphere of radius r and let the point charge be at the center of the sphere, then the magnitude of electric field intensity at all of the points on the surface of the sphere will be equal. If the point charge is positive, the direction of the electric

16、 field intensity is the same as that of the outward normal to the surface of the sphere. Sq 0dSEApplying Gauss lawThe left hand side of the above equation is SSSErSE 2n 4dd dSeESEAnd we have204rqErrqeE204orSq 0dSE We also can use the formula for electric potential or electric field intensity to calc

17、ulate the electric field intensity produced by the point charge. If the point charge is placed at the origin, then . We find the electric potential produced by the point charge as r|rrrq04)(rrrqrqeE200414The electric field intensity E as rVrrqVreerE20 204d4)( If the equation for electric field inten

18、sity is used directly, we find the electric field intensity E as Example 2 Calculate the electric field intensity produced by an electric dipole. Solution: For an electric dipole consists of two point electric charges, we can use the principle of superposition to calculate the electric field intensi

19、ty. Then the electric potential produced by an electric dipole asrrrrqrqrq000444 If the distance from the observer is much greater than the separation l, we can consider that and are parallel with , andrererecoslrr2cos2cos2rlrlrrrx-q+qzylrr-r+Owhere the direction of the vector l is defined as the di

20、rection pointing to the positive charge from the negative charge. The product ql is often called the electric moment of the dipole, and is denoted by p, so thatlpq)(4cos42020rrqlrqelWe have20204cos4rprrepThen the electric potential of the dipole can be written assin11rrrEreee30304sin2cosrprpree By u

21、sing the relation , we can find the electric field intensity of the electric dipole asE* * The electric potential of the dipole is inversely proportional to the square of the distance, and the magnitude of the electric field intensity is inversely proportional to the third power of the distance. The

22、se properties are very different from that of a point charge.* * In addition, the electric potential and the electric field intensity are both dependent of the azimuthal angle . Electric field linesElectric field lines and equipotential surfaces of an electric dipoleElectric field linesEquipotential

23、 surfacesExample 3 Assume that an infinitely long charged cylinder of radius a is placed in free space. The density of the charge is . Calculate the electric field intensities inside and outside the cylinder. xzyaLS1Solution: Choose a cylindrical coordinate system. Since the cylinder is infinitely l

24、ong, for any z the environment remain the same. Hence the field is independent of the coordinate variable z. In addition, the cylinder is rotationally symmetrical. Thus the field must be independent of the angle . Since it is symmetrical about any plane of z = constant, the electric field intensity

25、must be perpendicular to the z-axis, and coplanar with the radial coordinate r. Since the direction of the electric field intensity coincides with the outward normal of lateral surface S1 of the cylinder everywhere and is perpendicular to the normal to the upper or the lower end faces, the left side

26、 of the surface integral becomesrLESESESSS2ddd11 SE For r a, the charge , we have Laq2rraeE022 where can be considered as the charge per unit length, so that the electric field can be thought of as caused by a line charge with density . In view of this we can derive the electric field intensity caus

27、ed by an infinite line charge with line density as2al2alxzyr21rOrrzdzrzere),2,(zrP Example 4 Find the electric field intensity caused by a line charge with length L and charge density .l Solution: Let the z-axis of a cylindrical coordinate system coincide with the line charge and the mid point of th

28、e line be at the origin. It is impossible to apply Gauss law to calculate the field because the direction of the electric field intensity cannot be found out in advance. We have to evaluate directly the electric potential and the electric field intensity. Since it is rotationally symmetrical, the fi

29、eld is independent of the angle .lLLd42 2 30|rr|rrEl2 Since the field is independent of the angle , for convenience, let the field point P be placed in the yz-plane, i.e. . ThenConsideringd cscdcot)sincos(csccsc|2rzrzzrrrzeerr|rrd csccscsincos42 22021rraazreeElWe have)cos(cos)sin(sin412120rzlreerlrl

30、rreeE00224This result is the same as that of Example 3.)cos(cos)sin(sin412120rzlreeEIf the length L , then 1 0, 2 and3. Electric Potential and Equipotential Surfaces The physical meaning of the electric potential at a point is the work W by the electric field force to move a unit charge from the poi

31、nt to infinity, Hence The electric potential at a point is in fact the electric potential difference between the point and infinity. In principle, any point can be taken as the reference point for the electric potential. Obviously, the electric potentials for a given point will be different when dif

32、ference references are chosen. qW( q is the amount of the charge.) However, the electric potential difference between any two points is independent of the reference point. When charges are confined to a region, infinity is usually taken as the reference point of the electric potential because the el

33、ectric potential at infinity will be zero. The selection of reference point has no effect on the value of electric field intensity. The electric field lines are perpendicular to the equipotential surfaces everywhere. Equipotential surfaces are the surfaces on which the electric potential are equaliz

34、ed, and the equation isCzyx),(where the constant C is equal to the value of electric potential.Electric field linesEquipotential surfacesE If the electric potential differences between any adjacent equi-potential surfaces are constant, then where the equipotential surfaces are concentrated, there th

35、e change of electric potential will be faster, and the field intensity will be stronger. In this way, the distribution density of equipotential surfaces can indicate the intensity of the electric field. 4. Polarization of Dielectrics The electrons in a conductor are called free electrons, and their

36、charges are called free charges. the electrons in the dielectric cannot escape from the binding force and these charges are called bound charges. The bound charges in dielectrics will be displaced due to the electric fields, and this phenomenon is called polarization. Usually, the polarization of no

37、npolar molecules is called displacement polarization, while the polarization of polar molecules is called orientation polarization. Based on the distribution of the bound charges in dielectric, molecules can be classified into two kinds: polar and nonpolar molecules. Polar MoleculeNonpolar MoleculeN

38、onpolar MoleculePolar MoleculeEaPolarization of dielectrics The polarization is formed gradually, as shown in the following figure. DielectricTotal field Ea+ EsApplied field Ea Because the directions of electric dipoles are almost coincided with that of applied field, the direction of the strongest

39、secondary field in the middle of a dipole is always in opposite direction of applied field, that isasaEEESecondary field EsPolarization In order to measure the polarization intensity, we define the vector sum of the electric dipole moments per unit volume as the polarization, denoted by P ViiNpP1whe

40、re pi is the electric dipole moment of the i-th electric dipole in volume V , and N is the number of electric dipoles in volume V . Here, V is an infinitesimal volume. Most dielectrics are polarized by an applied electric field, their polarization P is proportional to the total electric field intens

41、ity E in the dielectrics, so thatEPe0where e is called the electric susceptibility, and it is usually a positive real number. In the above dielectrics, the direction of polarization is the same as that of the total electric field intensity. When the susceptibility e remains unchanged for all directi

42、ons of the applied field, the material is called an isotropic dielectric. zyxzyxEEEPPP 33e32e31e23e22e21e13e12e11e0 For certain dielectric materials, the susceptibility is dependant upon the direction of the applied field. The polarization is in a different direction from the total field. The relati

43、onship between the polarization and the electric field intensity will need to be described in terms of the following matrix:This kind of material is called an anisotropic dielectric. Dielectrics whose electric suspectibilities are uniform in space are called homogeneous dielectrics, otherwise they a

44、re called inhomogeneous dielectrics. If the value of the electric susceptibility is independent of the magnitude of the electric field, the dielectric is called a linear dielectric, otherwise it is a nonlinear dielectrics. If the electric susceptibility is independent of time, the dielectric is call

45、ed a static medium, otherwise, it is called a time-dependent medium. Homogeneity, linearity, and isotropy stand for three very different properties that are in general independent of each other. Properties of dielectrics After a dielectric is polarized, there exists as a result some bound surface ch

46、arge on the boundary of the dielectric. if the dielectric is inhomogeneous, there will be some volume bound charges in the dielectric. These surface and volume charges are also called polarization charges. We can show that the electric potential produced by these polarization charges asVVSd|)(41|d)(

47、41)( 0 0rrrPrrSrPr The net volume bound charge and the net surface bound charge are equal but opposite. The right equation can be rewritten asSq dSPn)()(erPrS)()(rPrand 5. Equations for Electrostatic Fields in Dielectric The electrostatic field in dielectrics can be considered as the electrostatic f

48、ields produced by both the free charges and the bound charges in free space. In this way, in dielectrics the electric flux over any closed surface S iswhere q the free charges ， the bound charges . q)(1d 0 qqSSEqS 0d)( SPEWe have known , thenSq dSPLet ，we havePED0qS d SD DWhere the vector D defined

49、is called electric flux density (or electric displacement). We can see that the outward flux of electric flux density over any closed surface is equal to the total net free charges in the closed surface, and independent of the bound charges. The above equation is called Gauss law for dielectrics. Us

50、ing the divergence theorem and considering , we can find VVq dwhere is the differential form of Gauss law for dielectrics, and it states that the divergence of the electric flux density at a point is equal to the density of the net free charge at the point. Electric flux density lines: The direction

51、 of tangent line at a point on the curve is the direction of electric flux density. Electric flux density lines start from positive free charges and end with negative free charges, and they are independent of bound charges. If the flux in all electric flux density tubes constructed by electric flux

52、density lines are equalized, and the axis of an electric flux density tube is taken as the electric flux density line, then the distribution density of the electric flux density lines can represent the magnitude of the electric flux density.The polarization of isotropic dielectrics , and EPe0EEED)1

53、(e0e00Let , then)1 (e0EDwhere is called the permittivity of the dielectrics (or called dielectric constant). Since the electric susceptibility e is usually a positive real number, the permittivities of most dielectrics are greater than that of free space. Relative permittivity r is defined ase0r1The

54、 relative permittivities of all dielectrics are usually greater than 1. Relative permittivities of several dielectrics Dielectrics rDielectrics rAir1.0Quartz3.3Transformer Oil2.3Mica6.0Paper1.34.0Ceramic5.36.5Plexiglass2.63.5Pure Water81Paraffin Wax2.1Resin3.3Polythene2.3Polystyrene2.6 For anisotrop

55、ic dielectrics, the relationship between the electric flux density and the electric field intensity can be written aszyxzyxEEEDDD 333231232221131211 The relationship between the electric flux density and the electric field intensity is dependent upon the direction of applied electric field. In addit

56、ion, the permittivity of a homogeneous dielectric is independent of the space coordinates. The permittivity of a static media is independent of time. The permittivity of linear dielectrics is independent of the magnitude of the electric field intensity. For homogeneous dielectrics, since the permitt

57、ivity is not a function of the coordinates, and we haveqS d SE E The previous relationship between the electric field, the electric potential and the free charges still hold as long as the permittivity of free space is replaced by the permittivity of dielectrics.6. Boundary Conditions for Dielectric

58、 Interfaces Since different properties of the media, the field quantities will be discontinuous at the interface between two media. This change is governed by the boundary conditions for the electrostatic field. Usually, the boundary conditions include tangential and normal components. To derive the

59、 relationships between the field quantities at the boundary, we may let h 0, then two of the linear integrals become 0d d 1 4 3 2 lElEE2E1 1 2etTangential Components1 4 4 3 3 2 2 1 d d d d d lElElElElEl Then the circulation of the electric field intensity around the rectangular path is given by Cons

60、tructing a rectangle about the point of consideration on the boundary, the length is l and the height is h. lh3124 In order to find the relationships between the field quantities at a point of the boundary, the length l should be small enough so that the field is constant over l, thenlElEd d dt21t4