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Figure 3.5: Computation of potential from known sources and values on bounding sur-
                        faces.


                        plates of a capacitor, which may be charged and on which the potential is assumed
                        to be known. To solve for  (r) within V we must know the potential produced by a

                        point source. This potential, called the Green’s function, is denoted G(r|r ); it has two

                        arguments because it satisfies Poisson’s equation at r when the source is located at r :
                                                      2


                                                    ∇ G(r|r ) =−δ(r − r ).                     (3.52)
                        Later we shall demonstrate that in all cases of interest to us the Green’s function is
                        symmetric in its arguments:


                                                       G(r |r) = G(r|r ).                      (3.53)
                        This property of G is known as reciprocity.
                          Our development rests on the mathematical result (B.30)known as Green’s second
                        identity. We can derive this by subtracting the identities

                                             ∇· (φ∇ψ) = φ∇· (∇ψ) + (∇φ) · (∇ψ),
                                             ∇· (ψ∇φ) = ψ∇· (∇φ) + (∇ψ) · (∇φ),
                        to obtain

                                                                           2
                                                                   2
                                              ∇· (φ∇ψ − ψ∇φ) = φ∇ ψ − ψ∇ φ.
                        Integrating this over a volume region V with respect to the dummy variable r and using

                        the divergence theorem, we obtain

                                       2             2
                              [φ(r )∇ ψ(r ) − ψ(r )∇ φ(r )] dV =−  [φ(r )∇ ψ(r ) − ψ(r )∇ φ(r )] · dS .
                             V                                    S
                        The negative sign on the right-hand side occurs because ˆ n is an inward normal to V .
                        Finally, since ∂ψ(r )/∂n = ˆ n ·∇ ψ(r ),wehave








                                                                         ∂ψ(r )       ∂φ(r )
                                       2             2
                               [φ(r )∇ ψ(r ) − ψ(r )∇ φ(r )] dV =−   φ(r )     − ψ(r )       dS
                             V                                    S       ∂n           ∂n
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