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where r < = min{r, a} and r > = max{r, a}. Using orthogonality of the exponentials we
                        find that only the m = 0 terms contribute:

                                                ∞  Y n0 (θ, φ) r n     π

                                                                        ∗

                                     F(r) = 2πa 2            <     f (θ )Y (θ ,φ ) sin θ dθ .




                                                            n+1         n0
                                                n=0  2n + 1 r >  0
                        Finally, since

                                                           2n + 1
                                                    Y n0 =       P n (cos θ)
                                                             4π
                        we have
                                                ∞
                                            1  2  
        r < n     π




                                     F(r) =  a     P n (cos θ)     f (θ )P n (cos θ ) sin θ dθ .  (3.101)
                                            2              r > n+1  0
                                                n=0
                          As an example, suppose f (θ) = cos θ = P 1 (cos θ). Then
                                              ∞
                                          1  2  
         r < n     π




                                    F(r) =  a    P n (cos θ)  n+1  P 1 (cos θ )P n (cos θ ) sin θ dθ .
                                          2                   0
                                              n=0        r >
                        The orthogonality of the Legendre polynomials can be used to show that
                                                π                          2

                                                P 1 (cos θ )P n (cos θ ) sin θ dθ =  δ 1n ,
                                              0                            3
                        hence
                                                             a 2   r <
                                                      F(r) =   cos θ  .                       (3.102)
                                                                    2
                                                             3     r >
                        3.2.7   Field produced by a permanently polarized body
                          Certain materials, called electrets, exhibit polarization in the absence of an external
                        electric field. A permanently polarized material produces an electric field both internal
                        and external to the material, hence there must be a charge distribution to support the
                        fields. We can interpret this charge as being caused by the permanent separation of
                        atomic charge within the material, but if we are only interested in the macroscopic field
                        then we need not worry about the microscopic implications of such materials. Instead, we
                        can use the Maxwell–Boffi equations and find the potential produced by the material by
                        using (3.99). Thus, the field of an electret with known polarization P occupying volume
                        region V in free space is dipolar in nature and is given by

                                               1     −∇ · P(r )       1     ˆ n · P(r )




                                       (r) =                  dV +                 dS


                                             4π	 0  V  |r − r |     4π	 0  S |r − r |
                        where ˆ n points out of the volume region V .
                          As an example, consider a material sphere of radius a, permanently polarized along
                        its axis with uniform polarization P(r) = ˆ zP 0 . We have the equivalent source densities
                                       ρ p =−∇ · P = 0,   ρ Ps = ˆ n · P = ˆ r · ˆ zP 0 = P 0 cos θ.
                        Then

                                                1     ρ Ps (r )     1     P 0 cos θ


                                         (r) =              dS =                 dS .


                                               4π	 0  S |r − r |  4π	 0  S |r − r |
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