Page 52 - Engineering Electromagnetics, 8th Edition
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34                 ENGINEERING ELECTROMAGNETICS


                   EXAMPLE 2.3
                                     Asanexampleoftheevaluationofavolumeintegral,wefindthetotalchargecontained
                                     in a 2-cm length of the electron beam shown in Figure 2.5.
                                     Solution. From the illustration, we see that the charge density is
                                                                      −6 −10 ρz
                                                          ρ ν =−5 × 10 e   5  C/m 2
                                     The volume differential in cylindrical coordinates is given in Section 1.8; therefore,

                                                         0.04     2π     0.01   5
                                                                           −6 −10 ρz
                                                  Q =               −5 × 10 e     ρ dρ dφ dz
                                                       0.02  0  0
                                     We integrate first with respect to φ because it is so easy,
                                                             0.04     0.01
                                                                              5
                                                                        −5
                                                                    −10 πe  −10 ρz ρ dρ dz
                                                      Q =
                                                           0.02  0
                                     and then with respect to z, because this will simplify the last integration with respect
                                     to ρ,
                                                                    −5
                                                             0.01    −10 π  5      z=0.04
                                                      Q =            5  e −10 ρz ρ dρ
                                                           0     −10 ρ            z=0.02
                                                             0.01
                                                                  −5
                                                        =      −10 π(e −2000ρ  − e −4000ρ )dρ
                                                           0



























                                                      Figure 2.5 The total charge contained
                                                      within the right circular cylinder may be

                                                      obtained by evaluatingQ =  vol  ρ ν dν.
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