(272) |

for , not outside , and

(273) |

when lies on or on . The Green's function also has the symmetry property

(274) |

Let us try

Note that the above function is symmetric with respect to its arguments, because . It follows from Equation (25) that

(276) |

However, if , do not lie outside then the argument of the latter delta function cannot be zero. Hence, for , not outside , this function takes the value zero, and the above expression reduces to

(277) |

as required. Equation (276) can be written

where . When written in this form, it becomes clear that when lies on (i.e., when ) or on (i.e., when ). We conclude that expression (279) is the unique Green's function for the Dirichlet problem within the domain .

According to Equation (246), the electrostatic potential within the domain is written

(279) |

Here, is the charge distribution within (i.e., the region ), is the potential on (i.e., the surface ), and the potential at infinity (i.e., on the surface ) is assumed to be zero. Moreover, we have made use of the fact that on , because the unit vector points radially inward. Finally, it is easily demonstrated that

(280) |

Hence,

where , , and , , are standard spherical coordinates, in terms of which,

As a specific example, suppose that represents the surface of a grounded spherical conductor. It follows that . Suppose that there is a single charge, , in the domain , located on the -axis at , , and . It follows that

(283) |

Thus, Equations (282) and (283) yield

in the region . Of course, in the region . It follows from Equations (154) and (203) that there is a charge sheet on the surface of the conductor (because the normal electric field is non-zero just above the surface, but zero just below) whose density is given by

(285) |

Thus, the net charge induced on the surface of the conductor is

(286) |

In Equation (285), the first term inside the square brackets clearly represents the electric potential generated by the charge , which suggests that

(287) |

is the potential generated by the charges induced on the surface of the conductor. Hence, the attractive force acting on the charge due to the induced charges is

(288) |

Of course, an equal and opposite force acts on the conductor.

As a final example, suppose that there are no charges in the domain , but that is prescribed on the surface . It follows from Equation (282) that the electric potential on the -axis (i.e., , , and ) is

(289) |

for . Let correspond to the surface of two conducting hemispheres (separated by a thin insulator). Suppose that the upper (i.e., ) hemisphere is held at the potential , whereas the lower (i.e., ) hemisphere is held at the potential . It follows that

(290) |

where , which yields

(291) |

for .