Page 88 - Engineering Electromagnetics, 8th Edition
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70                 ENGINEERING ELECTROMAGNETICS

                                                         3     2            3     2
                                                  =−       (D x ) x=0 dy dz +  (D x ) x=1 dy dz
                                                       0  0               0  0
                                                         3     1            3     1
                                                    −      (D y ) y=0 dx dz +  (D y ) y=2 dx dz
                                                       0  0               0  0
                                        However, (D x ) x=0 = 0, and (D y ) y=0 = (D y ) y=2 , which leaves only

                                                              3     2             3     2
                                                   D · dS =      (D x ) x=1 dy dz =  2ydydz
                                                  S         0  0                0  0
                                                              3
                                                              4 dz = 12
                                                         =
                                                            0
                                     Since
                                                              ∂        ∂   2
                                                      ∇ · D =  (2xy) +   (x ) = 2y
                                                             ∂x        ∂y
                                     the volume integral becomes

                                                              3     2     1         3     2
                                                                                       2ydydz
                                                  ∇ · D dv =         2ydx dydz =
                                                vol         0   0  0              0  0
                                                              3
                                                         =    4 dz = 12
                                                            0
                                     and the check is accomplished. Remembering Gauss’s law, we see that we have also
                                     determined that a total charge of 12 C lies within this parallelepiped.


                                                                    1
                                                                                  1
                                                                                          2
                                        D3.9. Given the field D = 6ρ sin φ a ρ +1.5ρ cos φ a φ C/m , evaluate both
                                                                    2
                                                                                  2
                                        sides of the divergence theorem for the region bounded by ρ = 2, φ = 0,
                                        φ = π, z = 0, and z = 5.
                                        Ans. 225; 225
                                     REFERENCES

                                     1. Kraus, J. D., and D. A. Fleisch. Electromagnetics. 5th ed. New York: McGraw-Hill,
                                        1999. The static electric field in free space is introduced in Chapter 2.
                                     2. Plonsey, R., and R. E. Collin. Principles and Applications of Electromagnetic Fields.
                                        New York: McGraw-Hill, 1961. The level of this text is somewhat higher than the one we
                                        are reading now, but it is an excellent text to read next. Gauss’s law appears in the second
                                        chapter.
                                     3. Plonus, M. A. Applied Electromagnetics.New York: McGraw-Hill, 1978. This book
                                        contains rather detailed descriptions of many practical devices that illustrate
                                        electromagnetic applications. For example, see the discussion of xerography on
                                        pp. 95–98 as an electrostatics application.
                                     4. Skilling, H. H. Fundamentals of Electric Waves.2d ed. New York: John Wiley & Sons,
                                        1948. The operations of vector calculus are well illustrated. Divergence is discussed on
                                        pp. 22 and 38. Chapter 1 is interesting reading.
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