工程电磁场第八版课后答案第7章

工程电磁场第八版课后答案第7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 neglig。</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>