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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 $ V$ that lies between the spherical surfaces $ r=a$ and $ r=\infty$ , where $ r$ is a radial spherical coordinate. Let $ S$ and $ S'$ denote the former and latter surfaces, respectively. The Green's function for the problem, $ G_D({\bf r},{\bf r}')$ , must satisfy

$\displaystyle \nabla^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$ (272)

for $ {\bf r}$ , $ {\bf r}'$ not outside $ V$ , and

$\displaystyle G_D({\bf r},{\bf r}')=0$ (273)

when $ {\bf r}'$ lies on $ S$ or on $ S'$ . The Green's function also has the symmetry property

$\displaystyle G_D({\bf r}',{\bf r})=G_D({\bf r},{\bf r}').$ (274)

Let us try

$\displaystyle G_D({\bf r},{\bf r}') = -\frac{1}{4\pi\,\vert{\bf r}'-{\bf r}\vert}+ \frac{a}{4\pi\,r'\,\vert a^{\,2}\,{\bf r}'/r'^{\,2}-{\bf r}\vert}.$ (275)

Note that the above function is symmetric with respect to its arguments, because $ r'\,\vert a^{\,2}\,{\bf r}'/r'^{\,2}-{\bf r}\vert=
r\,\vert a^{\,2}\,{\bf r}/r^{\,2}-{\bf r}'\vert$ . It follows from Equation (25) that

$\displaystyle \nabla^{\,2} G_D({\bf r},{\bf r}') = \delta({\bf r}-{\bf r}')- \frac{a}{r'}\,\delta({\bf r}-a^{\,2}\,{\bf r}'/r'^{\,2}).$ (276)

However, if $ {\bf r}$ , $ {\bf r}'$ do not lie outside $ V$ then the argument of the latter delta function cannot be zero. Hence, for $ {\bf r}$ , $ {\bf r}'$ not outside $ V$ , this function takes the value zero, and the above expression reduces to

$\displaystyle \nabla^{\,2}G_D({\bf r},{\bf r}')=\delta({\bf r}-{\bf r}'),$ (277)

as required. Equation (276) can be written

$\displaystyle G_D({\bf r},{\bf r}') = -\frac{1}{4\pi\,(r^{\,2}-2\,r\,r'\,\cos\g...
...\frac{1}{4\pi\,(r^{\,2}\,r'^{\,2}/a^{\,2}-2\,r\,r'\,\cos\gamma+a^{\,2})^{1/2}},$ (278)

where $ \cos\gamma={\bf r}\cdot{\bf r}'/(r\,r')$ . When written in this form, it becomes clear that $ G_D({\bf r},{\bf r}')=0$ when $ {\bf r}'$ lies on $ S$ (i.e., when $ r'=a$ ) or on $ S'$ (i.e., when $ r'=\infty$ ). We conclude that expression (279) is the unique Green's function for the Dirichlet problem within the domain $ V$ .

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

$\displaystyle \phi({\bf r})=-\frac{1}{\epsilon_0} \int_V G_D({\bf r},{\bf r}')\...
..._S \phi_S({\bf r}')\,\frac{\partial G_D({\bf r},{\bf r}')}{\partial r'}\, d S'.$ (279)

Here, $ \rho({\bf r})$ is the charge distribution within $ V$ (i.e., the region $ r>a$ ), $ \phi_S({\bf r})$ is the potential on $ S$ (i.e., the surface $ r=a$ ), and the potential at infinity (i.e., on the surface $ S'$ ) is assumed to be zero. Moreover, we have made use of the fact that $ \partial/\partial n'=-\partial/\partial r'$ on $ S$ , because the unit vector $ {\bf n}'$ points radially inward. Finally, it is easily demonstrated that

$\displaystyle \left.\frac{\partial G_D}{\partial r'}\right\vert _{r'=a}= \frac{a-r^{\,2}/a}{4\pi\,(r^{\,2}-2\,r\,a\,\cos\gamma+a^{\,2})^{3/2}}.$ (280)


$\displaystyle \phi(r,\theta,\varphi)$ $\displaystyle =\frac{1}{4\pi\,\epsilon_0}\int_a^\infty\int_0^\pi\int_0^{2\pi}\f...
  $\displaystyle \phantom{=}-\frac{1}{4\pi\,\epsilon_0}\int_a^\infty\int_0^\pi\int...
  $\displaystyle \phantom{=}+\frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\frac{\phi_S(\th...
...\sin\theta'\,d\theta'\,d\varphi'}{(r^{\,2}-2\,r\,a\,\cos\gamma+a^{\,2})^{3/2}},$ (281)

where $ r$ , $ \theta$ , $ \varphi$ and $ r'$ , $ \theta'$ , $ \varphi'$ are standard spherical coordinates, in terms of which,

$\displaystyle \cos\gamma = \cos\theta\,\cos\theta'+\sin\theta\,\sin\theta'\,\cos(\varphi-\varphi').$ (282)

As a specific example, suppose that $ S$ represents the surface of a grounded spherical conductor. It follows that $ \phi_S=0$ . Suppose that there is a single charge, $ q$ , in the domain $ V$ , located on the $ z$ -axis at $ r=b>a$ , $ \theta=0$ , and $ \varphi=0$ . It follows that

$\displaystyle \rho({\bf r}')\,dV' \rightarrow q\,\delta(r'-b)\,\delta(\theta')\,\delta(\varphi')\,dr'\,d\theta'\,d\varphi'.$ (283)

Thus, Equations (282) and (283) yield

$\displaystyle \phi(r,\theta) = \frac{q}{4\pi\,\epsilon_0}\left[\frac{1}{(r^{\,2...
...- \frac{1}{(r^{\,2}\,b^{\,2}/a^{\,2}-2\,r\,b\,\cos\theta+a^{\,2})^{1/2}}\right]$ (284)

in the region $ r>a$ . Of course, $ \phi=0$ in the region $ r<a$ . 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

$\displaystyle \sigma(\theta)= -\epsilon_0\left.\frac{\partial\phi}{\partial r}\...
...\pi\,a}\,\frac{(b^{\,2}-a^{\,2})}{(a^{\,2}-2\,a\,b\,\cos\theta+b^{\,2})^{3/2}}.$ (285)

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

$\displaystyle q' = \int\sigma\,dS= 2\pi\,a^{\,2}\int_0^\pi\sigma(\theta)\,d\theta = - q\,\frac{a}{b}.$ (286)

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

$\displaystyle \phi'(r,\theta) = -\frac{q}{4\pi\,\epsilon_0} \frac{1}{(r^{\,2}\,b^{\,2}/a^{\,2}-2\,r\,b\,\cos\theta+a^{\,2})^{1/2}}$ (287)

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

$\displaystyle {\bf F}' = -\left.q\,\nabla\phi'\right\vert _{r=b,\theta=0} =- \frac{q^2}{4\pi\,\epsilon_0}\,\frac{b\,a}{(b^{\,2}-a^{\,2})^2}\,{\bf e}_z.$ (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 $ V$ , but that $ \phi_S$ is prescribed on the surface $ S$ . It follows from Equation (282) that the electric potential on the $ z$ -axis (i.e., $ r=z$ , $ \theta=0$ , and $ \varphi=0$ ) is

$\displaystyle \phi(z) = \frac{1}{4\pi}\int_0^\pi\int_0^{2\pi}\frac{\phi_S(\thet...
...\sin\theta'\,d\theta'\,d\varphi'}{(z^{\,2}-2\,z\,a\,\cos\theta'+a^{\,2})^{3/2}}$ (289)

for $ \vert z\vert\geq a$ . Let $ S$ correspond to the surface of two conducting hemispheres (separated by a thin insulator). Suppose that the upper (i.e., $ 0\leq \theta<\pi/2$ ) hemisphere is held at the potential $ +V$ , whereas the lower (i.e., $ \pi/2<\theta\leq \pi$ ) hemisphere is held at the potential $ -V$ . It follows that

$\displaystyle \phi(z) = \frac{V}{2}\,(z^{\,2}-a^{\,2})\,a\left[\int_0^1 \frac{d...
...2})^{3/2}}-\int_{-1}^0\frac{d\mu}{(z^{\,2}-2\,z\,a\,\mu+a^{\,2})^{3/2}}\right],$ (290)

where $ \mu=\cos\theta$ , which yields

$\displaystyle \phi(z) = {\rm sgn}(z)\,V\left[1-\frac{z^{\,2}-a^{\,2}}{\vert z\vert\,(z^{\,2}+a^{\,2})^{1/2}}\right]$ (291)

for $ \vert z\vert\geq a$ .

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