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# Dirichlet Green's Function for Spherical Surface

As an example of a boundary value problem, suppose that we wish to solve Poisson's equation, subject to Dirichlet boundary conditions, in some domain that lies between the spherical surfaces and , where is a radial spherical coordinate. Let and denote the former and latter surfaces, respectively. The Green's function for the problem, , must satisfy

 (272)

for , not outside , and

 (273)

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

 (274)

Let us try

 (275)

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

 (278)

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,

 (281)

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

 (282)

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

 (284)

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 .

Next: Exercises Up: Electrostatic Fields Previous: Boundary Value Problems
Richard Fitzpatrick 2014-06-27