Page 72 - Engineering Electromagnetics, 8th Edition
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54                 ENGINEERING ELECTROMAGNETICS

                                     or a surface charge,

                                                  Q =   ρ S dS  (not necessarily a closed surface)
                                                       S
                                     or a volume charge distribution,

                                                                Q =     ρ ν dv
                                                                      vol
                                        The last form is usually used, and we should agree now that it represents any or
                                     all of the other forms. With this understanding, Gauss’s law may be written in terms
                                     of the charge distribution as


                                                                           ρ ν dv                     (6)
                                                                D S · dS =
                                                              S          vol
                                     a mathematical statement meaning simply that the total electric flux through any
                                     closed surface is equal to the charge enclosed.


                   EXAMPLE 3.1
                                     To illustrate the application of Gauss’s law, let us check the results of Faraday’s
                                     experiment by placing a point charge Q at the origin of a spherical coordinate system
                                     (Figure 3.3) and by choosing our closed surface as a sphere of radius a.
                                     Solution. We have, as before,
                                                                       Q
                                                                 D =      a r
                                                                     4πr 2
                                     At the surface of the sphere,
                                                                       Q
                                                                 D S =    a r
                                                                      4πa 2
                                     The differential element of area on a spherical surface is, in spherical coordinates
                                     from Chapter 1,

                                                                           2
                                                             2
                                                       dS = r sin θ dθ dφ = a sin θ dθ dφ
                                     or
                                                                   2
                                                             dS = a sin θ dθ dφ a r
                                     The integrand is
                                                           Q                     Q
                                                               2
                                                D S · dS =    a sin θ dθ dφa r · a r =  sin θ dθ dφ
                                                         4πa 2                   4π
                                     leading to the closed surface integral

                                                              φ=2π     θ=π  Q
                                                                         sin θ dθ dφ
                                                            φ=0   θ=φ 4π
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