工程电磁场第三章解读ppt课件.ppt

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1、3.1 Electric Flux Density2. Electric Flux:)(coulombsQ3. Electric Flux Density (coulombs/square meter):direction (the direction of the flux lines at that point) and magnitude (the number of flux lines crossing a surface normal to the lines divided by the S. area).D)(42inneraaQDrar)(42outerabQDrbr)(42

2、braarQDr4. Shown in the right figure1. Faradays experiment: a larger positive charge on the inner sphere induced a corresponding larger negative charge on the outer sphere.3.1 Electric Flux Density5. If the inner sphere becomes a point charge of Q, we still have24rQDar6. Compared with , we haverarQE

3、204)space freein (0ED7. For a general volume charge distribution:rvolrvolaRdDaRdE220448. For dielectrics, the relationship between and will be more complicatedDE3.1 Electric Flux Density9. Example 3.1: find D in the region about a uniform line charge of 8nC/m lying along the z axis in free space.10.

4、 Exercise: D3.1, D3.23.2 Gausss Law1. Gausss law: The electric flux passing through any closed surface is equal to the total charge enclosed by that surface. -the generalization of Faradays experiment2.A cloud of point charges(total charge Q) are shown in the following Figure. There is some value DS

5、 at every point on the surface. How to describe an incremental element of area?SS n The flux crossing is SsDS The total flux passing through the closed surface issurfaceclosedsSdDd3.Consider the nature of an incremental of the surface :S3.2 Gausss Law4.To a gaussian surface, the mathematical formula

6、tion of Gausss lawcharge closed()SnLSSDdSQQdLdSd 5. The last form is usually useddSdDSS2200sinsin4sin4rSSQdSad d aD dSd dQDdSd dQ 6. For example: placing a point charge Q at the origin of a spherical coordinate system and choose a sphere of radius a as the gaussian S.rarrEDraaQDarQDarQE22204440 7. E

7、xercise: D3.3(P61)3.3 Application of Gausss Law:some symmetrical charge.1. Determining if the charge distribution is knownSDChoose a closed surface in which is everywhere either normal or tangential to the closed surfaceSDOn that portion of the closed surface for which is not zero, =constantSdDSSD2.

8、 Example: a point charge Q at the origin of a spherical coordinate system, and the results agree with those of Chap.2.rSSSSsphSSSarQDrQDDrddrDdSDSdDQ2220022444sinChoose a closed surface centered at the origin: a sphere of radius r 3.3 Application of Gausss Law3. A second example: the uniform line ch

9、arge distribution lying along the z axis and extending from - to +Only the radial component of D is presentaDDSo we can choose a cylindrical surface to which is everywhere normal D20000 2222SScylsidesz LSSzLLSQDdSDdSDd dLDLLQDLL 4. If symmetry does not exist, we cannot use Gausss Law to obtain a sol

10、ution. The radial component is a function of only( )Df3.3 Application of Gausss Law5. A coaxial cable, the inner of radius a and the outer radius b, a charge distribution of on the outer surface of the inner conductorChoose a right circular cylindrical of length L and radius and , so we have S200002

11、SScylsidesz LSSzQDdSDdSDd dzDL ba2002LSSzQad dzaL SSSSaaDDa22LLSaDa 0( and )SDab3.3 Application of Gausss Law6. Example 3.2: Select a 50-cm length of coaxial cable having an inner radius of 1mm, and an outer radius of 4mm. The space between conductors is assumed to be filled with air. The total char

12、ge on the inner conductor is 30nC. We wish to know the charge density on each conductor, and E and D vector fields.7. Exercise: D3.5(P66)3.4 Application of Gausss Law:Differential Volume Element1. Apply Gausss Law to the problem without any symmetry.Choose a very small closed surface: D is almost co

13、nstant and the small change in D can be represented by the first two terms of Taylors-series expansion for D.2. The aim is to receive some information about the way D varies in the region of our small surface.3. Consider any point, shown in the following Figure. Choose the small rectangular box as t

14、he closed surface0000 xxyyzzDD aD aD aSSQDdS3.4 Application of Gausss Law:Differential Volume Element3. Consider any point, shown in the following Figure. Choose the small rectangular box as the closed surface0()2xxfrontDxDy zx SSfrontbackleftrighttopbottomQDdS.00 22frontfrontfrontxx frontfrontxx fr

15、ontxxxDSDy zaDy zDxxDDrate of change of D with xDx .0.0()()22backbackbackxx backbackxxbackxx backxDSDy z aDy zDxDy zDxxDDx xfrontbackDx y zx 3.4 Application of Gausss Law:Differential Volume ElementBy the same process, we can obtain, yrightleftDx y zy ztopbottomDx y zz These results may be collected

16、 to yield ()()yyxxzzSSDDDDDDQDdSx y zxyzxyz 4. Apply Gausss law to the closed surface surrounding the volume element and have an approximate resultCharge enclosed in volume ()volume yxzDDDxyz3.4 Application of Gausss Law:Differential Volume Element5. Example 3.3: find an approximate value for the to

17、tal charge enclosed in an incremental volume of 10-9 m3 located at the origin, if 2sincos2/xxxxzDeyaeyaza C m6. Exercise: D3.6(P70)3.5 Divergence1. This equation can be written as Or as a limit:()SyxSzDdSDDDQxyz00()limlimSyxSzDdSDDDQxyz 2. The last term is the volume charge density, hence0()limSyxSz

18、DdSDDDxyz 3. The methods could have been used on any vector A00()lim()limSyyxSxSzzDdSA dSDADADAxyzxyz 3.5 Divergence4. The divergence of the vector flux density A is : the outflow of flux from a small closed surface per unit volume as the volume shrinks to zero.5. A positive divergence indicates a s

19、ource; a negative divergence indicates a sink.div yxzDDDDxyz6. In cartesian coordinate system:0Divergence of div limSA dSAA ()11div (cylindrical)zDDDDz22(sin)()111div (spherical)sinsinrDDr DDrrrr7. Note:Divergence is performed on a vector; but the result is a scalar.3.5 Divergence8. Example 3.4: Fin

20、d div D at the origin if 9. Exercise: D3.7(P73)sincos2xxxyzDeyaeyaza3.6 Maxwells First Equation (Electrostatics)1. The first is the definition of divergence; The second is the result of applying the definition to a differential volume element.0div DlimSD dS div DyxzDDDxyz2. This is the first of Maxw

21、ells equations as they apply to electrostatics and steady magnetic fields. It states that the electric flux per unit volume leaving a vanishingly small volume unit is exactly equal to the volume charge density there.div Dthe point form of Gausss law3. Maxwells first equation is described as the diff

22、erential-equation form of Gausss law and Gausss law is recognized as the integral form of Maxwells first equation.div D3.6 Maxwells First Equation (Electrostatics)4. Consider the divergence of D in the region about a point charge Q located at the origin.24rQDar5. Exercise: D3.8(P74)222222(sin)()1111

23、div ()0(0)4sinsin4rDDr DQQDrrrrr rrrrr3.7 The vector operator and the divergence theorem1. Define the del operator as a vector operator,xyzaaaxyz () ()yxzxyzxxyyzzDDDdiv DDaaaD aD aD axyzxyz 2. The use of is much more prevalent than that of D div D3. Divergence theorem: The integral of the normal co

24、mponent of any vector field over a closed surface is equal to the integral of the divergence of this vector throughout the volume enclosed by the closed surfaceSvolSvolvolD dSQQdD dSQdDdD3.7 The vector operator and the divergence theorem4. The divergence theorem is true for any vector field5. Give an example to specify the divergence theorem.6. Example 3.5: Evaluate both sides of the divergence theorem for the field and the rectangular parellelepiped formed by the planes x=0 and 1, y=0 and 2, z=0 and 3.222(/)xyDxyax a C m7. D3.9(P78)