Page 74 - Engineering Electromagnetics, 8th Edition
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56                 ENGINEERING ELECTROMAGNETICS

                                     3.3 APPLICATION OF GAUSS’S LAW: SOME
                                            SYMMETRICAL CHARGE DISTRIBUTIONS
                                     We now consider how we may use Gauss’s law,


                                                                Q =    D S · dS
                                                                      S
                                     to determine D S if the charge distribution is known. This is an example of an integral
                                     equation in which the unknown quantity to be determined appears inside the integral.
                                        The solution is easy if we are able to choose a closed surface which satisfies two
                                     conditions:

                                     1. D S is everywhere either normal or tangential to the closed surface, so that
                                        D S · dS becomes either D S dS or zero, respectively.
                                     2. On that portion of the closed surface for which D S · dS is not zero, D S =
                                        constant.
                                        This allows us to replace the dot product with the product of the scalars D S and
                                     dS and then to bring D S outside the integral sign. The remaining integral is then
                                       dS over that portion of the closed surface which D S crosses normally, and this is

                                      S
                                     simply the area of this section of that surface. Only a knowledge of the symmetry of
                                     the problem enables us to choose such a closed surface.
                                        Let us again consider a point charge Q at the origin of a spherical coordinate
                                     system and decide on a suitable closed surface which will meet the two requirements
                                     previously listed. The surface in question is obviously a spherical surface, centered
                                     at the origin and of any radius r. D S is everywhere normal to the surface; D S has the
                                     same value at all points on the surface.
                                        Then we have, in order,

                                                  Q =    D S · dS =  D S dS
                                                       S          sph
                                                                       φ=2π     θ=π
                                                                                 2
                                                    = D S    dS = D S           r sin θ dθ dφ
                                                          sph         φ=0   θ=0
                                                         2
                                                    = 4πr D S
                                     and hence
                                                        Q
                                                 D S =
                                                       4πr 2
                                     Because r may have any value and because D S is directed radially outward,


                                                               Q              Q
                                                         D =      a r   E =       a r
                                                              4πr 2         4π  0 r 2
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