工程电磁场第八版课后答案第03章.pdf
《工程电磁场第八版课后答案第03章.pdf》由会员分享,可在线阅读,更多相关《工程电磁场第八版课后答案第03章.pdf(15页珍藏版)》请在装配图网上搜索。
CHAPTER 3 3.1. Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin, Q 1 , and with hemispheres of radius a. A second charge Q 2 (this time a point charge) is located at distance R from Q 1 , where R a. a) What is the force on the point charge before the hemispheres are assembled around Q 1 ? This will be simply the force beween two point charges, or F = Q 1 Q 2 4 0 R 2 a r b) What is the force on the point charge after the hemispheres are assembled but before they are discharged? The answer will be the same as in part a because induced charge Q 1 now resides as a surface charge layer on the sphere exterior. This produces the same electric field at the Q 2 location as before, and so the force acting on Q 2 is the same. c) What is the force on the point charge after the hemispheres are assembled and after they are discharged? Discharging the hemispheres (connecting them to ground) neutralizes the positive outside surface charge layer, thus zeroing the net field outside the sphere. The force on Q 2 is now zero. d) Qualitatively, describe what happens as Q 2 is moved toward the sphere assembly to the extent that the condition R a is no longer valid. Q 2 itself begins to induce negative surface charge on the sphere. An attractive force thus begins to strengthen as the charge moves closer. The point charge field approximation used in parts a through c is no longer valid. 3.2. An electric field in free space is E = (5z 2 / 0 )a z V/m. Find the total charge contained within a cube, centered at the origin, of 4-m side length, in which all sides are parallel to coordinate axes (and therefore each side intersects an axis at 2. The flux density is D = 0 E = 5z 2 a z . As D is z-directed only, it will intersect only the top and bottom surfaces (both parallel to the x-y plane). From Gauss law, the charge in the cube is equal to the net outward flux of D, which in this case is Q encl = I D nda = Z 2 2 Z 2 2 5(2) 2 a z a z dxdy + Z 2 2 Z 2 2 5(2) 2 a z (a z )dxdy = 0 where the first and second integrals on the far right are over the top and bottom surfaces respectively. 25 3.3. The cylindrical surface = 8 cm contains the surface charge density, s = 5e 20|z| nC/m 2 . a) What is the total amount of charge present? We integrate over the surface to find: Q = 2 Z 0 Z 2 0 5e 20z (.08)ddz nC = 20(.08) 1 20 e 20z 0 = 0.25nC b) How much flux leaves the surface = 8cm, 1cm z 5cm, 30 90 ? We just integrate the charge density on that surface to find the flux that leaves it. = Q 0 = Z .05 .01 Z 90 30 5e 20z (.08)ddz nC = 9030 360 2(5)(.08) 1 20 e 20z .05 .01 = 9.4510 3 nC = 9.45pC 3.4. An electric field in free space is E = (5z 3 / 0 )a z V/m. Find the total charge contained within a sphere of 3-m radius, centered at the origin. Using Gauss law, we set up the integral in free space over the sphere surface, whose outward unit normal is a r : Q = I 0 E nda = Z 2 0 Z 0 5z 3 a z a r (3) 2 sindd where in this case z = 3cos and (in all cases) a z a r = cos. These are substituted to yield Q = 2 Z 0 5(3) 5 cos 4 sind =2(5)(3) 5 1 5 cos 5 2 0 = 972 3.5. Let D = 4xya x + 2(x 2 +z 2 )a y + 4yza z C/m 2 and evaluate surface integrals to find the total charge enclosed in the rectangular parallelepiped 0 x 2, 0 y 3, 0 z 5 m: Of the 6 surfaces to consider, only 2 will contribute to the net outward flux. Why? First consider the planes at y = 0 and 3. The y component of D will penetrate those surfaces, but will be inward at y = 0 and outward at y = 3, while having the same magnitude in both cases. These fluxes will thus cancel. At the x = 0 plane, D x = 0 and at the z = 0 plane, D z = 0, so there will be no flux contributions from these surfaces. This leaves the 2 remaining surfaces at x = 2 and z = 5. The net outward flux becomes: = Z 5 0 Z 3 0 D x=2 a x dydz + Z 3 0 Z 2 0 D z=5 a z dxdy = 5 Z 3 0 4(2)ydy + 2 Z 3 0 4(5)ydy = 360 C 26 3.6. In free space, volume charge of constant density v = 0 exists within the region x , y , and d/2 z d/2. Find D and E everywhere. From the symmetry of the configuration, we surmise that the field will be everywhere z-directed, and will be uniform with x and y at fixed z. For finding the field inside the charge, an appropriate Gaussian surface will be that which encloses a rectangular region defined by 1 x 1, 1 y 1, and |z| d/2. The outward flux from this surface will be limited to that through the two parallel surfaces at z: in = I D dS = 2 Z 1 1 Z 1 1 D z dxdy = Q encl = Z z z Z 1 1 Z 1 1 0 dxdydz 0 where the factor of 2 in the second integral account for the equal fluxes through the two surfaces. The above readily simplifies, as both D z and 0 are constants, leading to D in = 0 za z C/m 2 (|z| d/2) ( 0 d/2)a z (z d/2) ( 0 d/2 0 )a z (z d/2) V/m 3.7. Volume charge density is located in free space as v = 2e 1000r nC/m 3 for 0 r 1 mm, and v = 0 elsewhere. a) Find the total charge enclosed by the spherical surface r = 1 mm: To find the charge we integrate: Q = Z 2 0 Z 0 Z .001 0 2e 1000r r 2 sindrdd Integration over the angles gives a factor of 4. The radial integration we evaluate using tables; we obtain Q = 8 r 2 e 1000r 1000 .001 0 + 2 1000 e 1000r (1000) 2 (1000r1) .001 0 = 4.010 9 nC b) By using Gausss law, calculate the value of D r on the surface r = 1 mm: The gaussian surface is a spherical shell of radius 1 mm. The enclosed charge is the result of part a. We thus write 4r 2 D r = Q, or D r = Q 4r 2 = 4.010 9 4(.001) 2 = 3.210 4 nC/m 2 27 3.8. Use Gausss law in integral form to show that an inverse distance field in spherical coordinates, D = Aa r /r, where A is a constant, requires every spherical shell of 1 m thickness to contain 4A coulombs of charge. Does this indicate a continuous charge distribution? If so, find the charge density variation with r. The net outward flux of this field through a spherical surface of radius r is = I D dS = Z 2 0 Z 0 A r a r a r r 2 sindd = 4Ar = Q encl We see from this that with every increase in r by one m, the enclosed charge increases by 4A (done). It is evident that the charge density is continuous, and we can find the density indirectly by constructing the integral for the enclosed charge, in which we already found the latter from Gausss law: Q encl = 4Ar = Z 2 0 Z 0 Z r 0 (r 0 )(r 0 ) 2 sindr 0 dd = 4 Z r 0 (r 0 )(r 0 ) 2 dr 0 To obtain the correct enclosed charge, the integrand must be (r) = A/r 2 . 3.9. A uniform volume charge density of 80C/m 3 is present throughout the region 8mm r 10mm. Let v = 0 for 0 r 10 mm, find D r at r = 20 mm: This will be the same computation as in part b, except the gaussian surface now lies at 20 mm. Thus D r (20mm) = 16410 12 4(.02) 2 = 3.2510 8 C/m 2 = 32.5nC/m 2 28 3.10. An infinitely long cylindrical dielectric of radius b contains charge within its volume of density v = a 2 , where a is a constant. Find the electric field strength, E, both inside and outside the cylinder. Inside, we note from symmetry that D will be radially-directed, in the manner of a line charge field. So we apply Gauss law to a cylindrical surface of radius , concentric with the charge distribution, having unit length in z, and where b) 3.11. In cylindrical coordinates, let v = 0 for 1 mm, v = 2sin(2000) nC/m 3 for 1mm 1.5mm. Find D everywhere: Since the charge varies only with radius, and is in the form of a cylinder, symmetry tells us that the flux density will be radially-directed and will be constant over a cylindrical surface of a fixed radius. Gauss law applied to such a surface of unit length in z gives: a) for 1 mm, D = 0, since no charge is enclosed by a cylindrical surface whose radius lies within this range. b) for 1mm 1.5mm, we have 2D = 2 Z .001 210 9 sin(2000 0 ) 0 d 0 = 410 9 1 (2000) 2 sin(2000) 2000 cos(2000) .001 or finally, D = 10 15 2 2 h sin(2000) + 2 110 3 cos(2000) i C/m 2 (1mm 1.5mm) 3.12. The sun radiates a total power of about 3.8610 26 watts (W). If we imagine the suns surface to be marked o in latitude and longitude and assume uniform radiation, a) What power is radiated by the region lying between latitude 50 N and 60 N and longi- tude 12 W and 27 W? 50 N lattitude and 60 N lattitude correspond respectively to = 40 and = 30 . 12 and 27 correspond directly to the limits on . Since the sun for our purposes is spherically-symmetric, the flux density emitted by it is I = 3.8610 26 /(4r 2 )a r W/m 2 . The required power is now found through P 1 = Z 27 12 Z 40 30 3.8610 26 4r 2 a r a r r 2 sindd = 3.8610 26 4 cos(30 )cos(40 )(27 12 ) 2 360 = 8.110 23 W b) What is the power density on a spherical surface 93,000,000 miles from the sun in W/m 2 ? First, 93,000,000 miles = 155,000,000 km = 1.5510 11 m. Use this distance in the flux density expression above to obtain I = 3.8610 26 4(1.5510 11 ) 2 a r = 1200a r W/m 2 3.13. Spherical surfaces at r = 2, 4, and6 m carry uniform surface charge densities of 20 nC/m 2 , 4nC/m 2 , and s0 , respectively. a) Find D at r = 1, 3and5 m: Noting that the charges are spherically-symmetric, we ascertain that D will be radially-directed and will vary only with radius. Thus, we apply Gauss law to spherical shells in the following regions: r 2: Here, no charge is enclosed, and so D r = 0. 2 r 4 : 4r 2 D r = 4(2) 2 (2010 9 ) D r = 8010 9 r 2 C/m 2 So D r (r = 3) = 8.910 9 C/m 2 . 4 6) = 1610 9 r 2 + s0 (6) 2 r 2 Requiring this to be zero, we find s0 =(4/9)10 9 C/m 2 . 30 3.14. A certain light-emitting diode (LED) is centered at the origin with its surface in the xy plane. At far distances, the LED appears as a point, but the glowing surface geometry produces a far-field radiation pattern that follows a raised cosine law: That is, the optical power (flux) density in Watts/m 2 is given in spherical coordinates by P d = P 0 cos 2 2r 2 a r Watts/m 2 where is the angle measured with respect to the normal to the LED surface (in this case, the z axis), and r is the radial distance from the origin at which the power is detected. a) Find, in terms of P 0 , the total power in Watts emitted in the upper half-space by the LED: We evaluate the surface integral of the power density over a hemispherical surface of radius r: P t = Z 2 0 Z /2 0 P 0 cos 2 2r 2 a r a r r 2 sindd = P 0 3 cos 3 /2 0 = P 0 3 b) Find the cone angle, 1 , within which half the total power is radiated; i.e., within the range 0 1 : We perform the same integral as in part a except the upper limit for is now 1 . The result must be one-half that of part a, so we write: P t 2 = P 0 6 = P 0 3 cos 3 1 0 = P 0 3 1cos 3 1 1 = cos 1 1 2 1/3 = 37.5 c) An optical detector, having a 1 mm 2 cross-sectional area, is positioned at r = 1 m and at = 45 , such that it faces the LED. If one nanowatt (stated in error as 1mW) is measured by the detector, what (to a very good estimate) is the value of P 0 ? Start with P d (45 ) = P 0 cos 2 (45 ) 2r 2 a r = P 0 4r 2 a r Then the detected power in a 1-mm 2 area at r = 1 m approximates as PW . = P 0 4 10 6 = 10 9 P 0 . = 410 3 W If the originally stated 1mW value is used for the detected power, the answer would have been 4 kW (!). 3.15. Volume charge density is located as follows: v = 0 for 2 mm, v = 4 C/m 3 for 1 2 mm. a) Calculate the total charge in the region 0 1 , 0 z L, where 1 1 2 mm: We find, Q = Z L 0 Z 2 0 Z 1 .001 4dddz = 8L 3 3 1 10 9 C b) Use Gauss law to determine D at = 1 : Gauss law states that 2 1 LD = Q, where Q is the result of part a. So, with 1 in meters, D ( 1 ) = 4( 3 1 10 9 ) 3 1 C/m 2 31 3.15c) Evaluate D at = 0.8mm, 1.6mm, and 2.4mm: At = 0.8mm, no charge is enclosed by a cylindrical gaussian surface of that radius, so D (0.8mm) = 0. At = 1.6mm, we evaluate the part b result at 1 = 1.6 to obtain: D (1.6mm) = 4(.0016) 3 (.0010) 3 3(.0016) = 3.610 6 C/m 2 At = 2.4, we evaluate the charge integral of part a from .001 to .002, and Gauss law is written as 2LD = 8L 3 (.002) 2 (.001) 2 C from which D (2.4mm) = 3.910 6 C/m 2 . 3.16. An electric flux density is given by D = D 0 a , where D 0 is a given constant. a) What charge density generates this field? Charge density is found by taking the diver- gence: With radial D only, we have v = D = 1 d d (D 0 ) = D 0 C/m 3 b) For the specified field, what total charge is contained within a cylinder of radius a and height b, where the cylinder axis is the z axis? We can either integrate the charge density over the specified volume, or integrate D over the surface that contains the specified volume: Q = Z b 0 Z 2 0 Z a 0 D 0 dddz = Z b 0 Z 2 0 D 0 a a addz = 2abD 0 C 3.17. A cube is defined by 1 x,y,z b. We integrate the charge densities (piecewise) over the spherical volume of radius b: Q = Z 2 0 Z 0 Z a 0 0 r 2 sindrdd Z 2 0 Z 0 Z b a 0 r 2 sindrdd = 4 3 2a 3 b 3 0 3.21. Calculate the divergence of D at the point specified if a) D = (1/z 2 ) 10 xyza x + 5x 2 za y + (2z 3 5x 2 y)a z at P(2,3,5): We find D = 10y z + 0 + 2 + 10 x 2 y z 3 (2,3,5) = 8.96 b) D = 5z 2 a + 10za z at P(3,45 ,5): In cylindrical coordinates, we have D = 1 (D ) + 1 D + D z z = 5z 2 + 10 (3,45 ,5) = 71.67 c) D = 2rsinsina r + rcossina + rcosa at P(3,45 ,45 ): In spherical coordi- nates, we have D = 1 r 2 r (r 2 D r ) + 1 rsin (sinD ) + 1 rsin D = 6sinsin + cos2sin sin sin sin (3,45 ,45 ) =2 3.22. (a) A flux density field is given as F 1 = 5a z . Evaluate the outward flux of F 1 through the hemispherical surface, r = a, 0 /2, 0 0. When r = 0, we have a singularity in D, so its divergence is not defined. b) Replace the point charge with a uniform volume charge density v0 for 0 r a. Relate v0 to Q and a so that the total charge is the same. Find div D everywhere: To achieve the same net charge, we require that (4/3)a 3 v0 = Q, so v0 = 3Q/(4a 3 ) C/m 3 . Gauss law tells us that inside the charged sphere 4r 2 D r = 4 3 r 3 v0 = Qr 3 a 3 Thus D r = Qr 4a 3 C/m 2 and D = 1 r 2 d dr Qr 3 4a 3 = 3Q 4a 3 as expected. Outside the charged sphere, D = Q/(4r 2 )a r as before, and the divergence is zero. 3.24. In a region in free space, electric flux density is found to be: D = 0 (z + 2d)a z C/m 2 (2dz 0) 0 (z2d)a z C/m 2 (0z 2d) Everywhere else, D = 0. a) Using D = v , find the volume charge density as a function of position everywhere: Use v = D = dD z dz = 0 (2dz 0) 0 (0z 2d) b) determine the electric flux that passes through the surface defined by z = 0, axa, by b: In the x-y plane, D evaluates as the constant D(0) = 2d 0 a z . Therefore the flux passing through the given area will be = Z a a Z b b 2d 0 dxdy = 8abd 0 C c) determine the total charge contained within the region a x a, b y b, d z d: From part a, we have equal and opposite charge densities above and below the x-y plane. This means that within a region having equal volumes above and below the plane, the net charge is zero. d) determine the total charge contained within the region a x a, b y b, 0z 2d. In this case, Q = 0 (2a)(2b)(2d) =8abd 0 C This is equivalent to the net inward flux of D into the volume, as was found in part b. 35 3.25. Within the spherical shell, 3 r 0.08 m? The total surface charge should be equal and opposite to the total volume charge. The latter is Q = Z 2 0 Z 0 Z .08 0 20r(mC/m 3 )r 2 sindrdd = 2.5710 3 mC = 2.57C So now s = 2.57 4(.08) 2 =32C/m 2 3.28. Repeat Problem 3.8, but use D = v and take an appropriate volume integral. We begin by finding the charge density directly through v = D = 1 r 2 d dr r 2 A r = A r 2 Then, within each spherical shell of unit thickness, the contained charge is Q(1) = 4 Z r+1 r A (r 0 ) 2 (r 0 ) 2 dr 0 = 4A(r + 1r) = 4A 3.29. In the region of free space that includes the volume 2 x,y,z 3, D = 2 z 2 (yza x +xza y 2xya z ) C/m 2 a) Evaluate the volume integral side of the divergence theorem for the volume defined above: In cartesian, we find D = 8xy/z 3 . The volume integral side is now Z vol Ddv = Z 3 2 Z 3 2 Z 3 2 8xy z 3 dxdydz = (94)(94) 1 4 1 9 = 3.47 C b. Evaluate the surface integral side for the corresponding closed surface: We call the surfaces at x = 3 and x = 2 the front and back surfaces respectively, those at y = 3 and y = 2 the right and left surfaces, and those at z = 3 and z = 2 the top and bottom surfaces. To evaluate the surface integral side, we integrate Dn over all six surfaces and sum the results. Note that since the x component of D does not vary with x, the outward fluxes from the front and back surfaces will cancel each other. The same is true for the left and right surfaces, since D y does not vary with y. This leaves only the top and bottom surfaces, where the fluxes are: I D dS = Z 3 2 Z 3 2 4xy 3 2 dxdy | z top Z 3 2 Z 3 2 4xy 2 2 dxdy | z bottom = (94)(94) 1 4 1 9 = 3.47 C 37 3.30 a) Use Maxwells first equation, D = v , to describe the variation of the electric field intensity with x in a region in which no charge density exists and in which a non-homogeneous dielectric has a permittivity that increases exponentially with x. The field has an x component only: The permittivity can be written as (x) = 1 exp( 1 x), where 1 and 1 are constants. Then D = (x)E(x) = d dx 1 e 1 x E x (x) = 1 1 e 1 x E x +e 1 x dE x dx = 0 This reduces to dE x dx + 1 E x = 0 E x (x) = E 0 e 1 x where E 0 is a constant. b) Repeat part a, but with a radially-directed electric field (spherical coordinates), in which again v = 0, but in which the permittivity decreases exponentially with r. In this case, the permittivity can be written as (r) = 2 exp( 2 r), where 2 and 2 are constants. Then- 配套讲稿:
如PPT文件的首页显示word图标,表示该PPT已包含配套word讲稿。双击word图标可打开word文档。
- 特殊限制:
部分文档作品中含有的国旗、国徽等图片,仅作为作品整体效果示例展示,禁止商用。设计者仅对作品中独创性部分享有著作权。
- 关 键 词:
- 工程 电磁场 第八 课后 答案 03
装配图网所有资源均是用户自行上传分享,仅供网友学习交流,未经上传用户书面授权,请勿作他用。
关于本文