Page 253 - Electrical Properties of Materials
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The effective field                       235



                                      Dielectric




                                                      – Charge         dA
                                                                 θ

             Void                                                  r

                                                                +            Fig. 10.10
                                                           Polarized charge  Calculation of effective internal field.


            giving a radial electric field,

                                          P dA cos θ
                                    dE r =        ,                  (10.42)
                                           4π  0 r 2
            in the middle of the sphere. Clearly, when we sum these components the net
            horizontal field in our drawing will be zero, and we have to consider only the
            vertical field. We get this field, previously called E , by integrating the vertical

            component and adding to it the applied field, that is

                                       £

                                 E =      dE r cos θ + E ,           (10.43)
                                     surface
            whence

                                        P cos θ dA
                                     £      2

                             E – E =
                                          4π  0 r 2
                                             2
                                        π  P cos θ
                                                2
                                   =          2  r 2π sin θ dθ
                                      0  4π  0 r
                                      P
                                   =    .                            (10.44)
                                     3  0

            Substituting for P from eqn (10.5) and solving for E we get
                                         1


                                     E = (  +2)E .                   (10.45)
                                         3
            This result is clearly acceptable for   = 1; and it is also consistent with our


            assumption that E is proportional to E .
               We can now derive an expression for the polarizability α as well, by
            combining our expression for the local field with eqns (10.7) and (10.8),  This is known as the Clausius–
            yielding                                                         Mossotti equation. It expresses the
                                                                             microscopically defined quantity

                                          –1 3  0                            α in terms of measurable macro-
                                     α =      ·   .                  (10.46)
                                                                             scopic quantities.
                                          +2 N m
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