Page 138 - Engineering Electromagnetics, 8th Edition
P. 138

120                ENGINEERING ELECTROMAGNETICS





                                                                       n








                                         Figure 5.4 An appropriate closed path and gaussian surface are used to
                                         determine boundary conditions at a boundary between a conductor and free
                                         space; E t = 0 and D N = ρ S.


                                     and normal components of D and E on the free-space side of the boundary. Both
                                     fields are zero in the conductor. The tangential field may be determined by applying
                                     Section 4.5, Eq. (21),

                                                                   E · dL = 0
                                     around the small closed path abcda. The integral must be broken up into four parts
                                                               b     c     d     a
                                                                              = 0
                                                               +    +    +
                                                             a    b    c    d
                                     Remembering that E = 0 within the conductor, we let the length from a to b or c to
                                     d be  w and from b to c or d to a be  h, and obtain

                                                                   1
                                                                               1
                                                       E t  w − E N,at b  h + E N,at a  h = 0
                                                                   2           2
                                        As we allow  h to approach zero, keeping  w small but finite, it makes no
                                     difference whether or not the normal fields are equal at a and b, for  h causes these
                                     products to become negligibly small. Hence, E t  w = 0 and, therefore, E t = 0.
                                        The condition on the normal field is found most readily by considering D N rather
                                     than E N and choosing a small cylinder as the gaussian surface. Let the height be  h
                                     and the area of the top and bottom faces be  S.Again, we let  h approach zero.
                                     Using Gauss’s law,


                                                                   D · dS = Q
                                                                  S
                                     we integrate over the three distinct surfaces

                                                               +       +     = Q
                                                             top  bottom  sides
                                     and find that the last two are zero (for different reasons). Then

                                                              D N  S = Q = ρ S  S
   133   134   135   136   137   138   139   140   141   142   143