工程电磁场第八版课后答案第08章

工程电磁场第八版课后答案第08章Tag内容描述：<p>1、CHAPTER 4 4.1. The value of E at P( = 2, = 40, z = 3) is given as E = 100a 200a+ 300azV/m. Determine the incremental work required to move a 20C charge a distance of 6 m: a) in the direction of a: The incremental work is given by dW = q E dL, where in this case, dL = da= 6 106a. Thus dW = (20 106C)(100V/m)(6 106m) = 12 109J = 12nJ b) in the direction of a: In this case dL = 2da= 6 106a, and so dW = (20 106)(2。</p><p>2、CHAPTER 11 11.1. Show that Exs= Aejk0z+is a solution to the vector Helmholtz equation, Sec. 11.1, Eq. (30), for k0= 00and any and A: We take d2 dz2 Aejk0z+= (jk0)2Aejk0z+= k2 0Exs 11.2. A 100-MHz uniform plane wave propagates in a lossless medium for which r= 5 and r= 1. Find: a) vp: vp= c/r= 3 108/5 = 1.34 108m/s. b) : = /vp= (2 108)/(1.34 108) = 4.69 m1. c) : = 2/ = 1.34 m. d) Es: Assume real amplitude E0, forward z travel, and x。</p><p>3、CHAPTER 3 3 1 Suppose that the Faraday concentric sphere experiment is performed in free space using a central charge at the origin Q1 and with hemispheres of radius a A second charge Q2 this time a p。</p><p>4、CHAPTER 9 9.1. In Fig. 9.4, let B = 0.2cos120t T, and assume that the conductor joining the two ends of the resistor is perfect. It may be assumed that the magnetic fi eld produced by I(t) is negligible. Find: a) Vab (t): Since B is constant over the loop area, the fl ux is ? = (0.15)2B = 1.41 10?2cos120t Wb.Now, emf = Vba(t) = ?d?/dt = (120)(1.41 10?2)sin120t. Then Vab(t) = ?Vba(t) = ?5.33sin120t V. b) I(t) = Vba(t)/R = 5.33sin(120t)/250 = 21.3sin(120t) mA 9.2. In the example descri。</p><p>5、CHAPTER 1 1.1. Given the vectors M = ?10ax+ 4ay? 8azand N = 8ax+ 7ay? 2az , fi nd: a) a unit vector in the direction of ?M + 2N. ?M + 2N = 10ax? 4ay+ 8az+ 16ax+ 14ay? 4az= (26,10,4) Thus a = (26,10,4) |(26,10,4)| = (0.92,0.36,0.14) b) the magnitude of 5ax+ N ? 3M: (5,0,0) + (8,7,?2) ? (?30,12,?24) = (43,?5,22), and |(43,?5,22)| = 48.6. c) |M|2N|(M + N): |(?10,4,?8)|(16,14,?4)|(?2,11,?10) = (13.4)(21.6)(?2,11,?10) = (?580.5,3193,?2902) 1.2. Vector A extends from the origin to (1,2,3) and vector。</p><p>6、CHAPTER 2 2.1. Three point charges are positioned in the x-y plane as follows: 5nC at y = 5 cm, -10 nC at y = ?5 cm, 15 nC at x = ?5 cm. Find the required x-y coordinates of a 20-nC fourth charge that will produce a zero electric fi eld at the origin. With the charges thus confi gured, the electric fi eld at the origin will be the superposition of the individual charge fi elds: E0= 1 40 15 (5)2 ax? 5 (5)2 ay? 10 (5)2 ay ? = 1 40 3 5 ax? aynC/m The fi eld, E20, associated with the 20。</p>