Method of Images and Green's Functions

We have seen how we can obtain the electric pontials and fields for some problems involving simple geometries. We shall now show how the solutions thus obtained can be used to construct the Green's function for these problems. As an illustration we shall consider the case of a point charge in front of an infinite conducting plane. We have seen that the solution for the electric potential is $\phi(\vec r) = \frac{q}{\vert\vec r - \vec r'\vert} +\frac{-q}{\vert\vec r - \vec r''\vert}$ where $q$ and $\vec r'$ are the strength and position of the charge and $\vec r''$ is the position of the image charge. If we choose the conducting plane along x-y plane, the coordinates of $\vec r''$ are $(x',y',-z')$ if $(x',y',z')$ are the coordinates of $\vec r'$. The solution is valid for z-component of $\vec r$ positive. Further, the charge is also to be placed on the same side of the plane where the potential is computed, or the z-component of $\vec r'$ is positive.

For the problem we are considering, the potential $\phi(x)$ vanishes on the plane $z=0$. That is we have Dirichlet boundary condition. So, we define the Green's function to be a solution of equation

$$\nabla^2 G(\vec r , \vec r') = - 4 \pi \delta(\vec r - \vec r')$$

with $G(\vec r, \vec r') =0$ when $\vec r$ or $\vec r'$ is on the plane $z=0$. If we choose

$$G(\vec r , \vec r') = \frac{1}{\vert\vec r - \vec r'\vert} - \frac{1}{\vert\vec r - \vec r' + 2 z' \hat k\vert}$$ (54)

where $z'$ is the z-component of $\vec r'$. Then by definition, the electric potential due to an arbitrary charge distribution $\rho(\vec r)$ is

$$\phi(\vec r) = \int d^3r'\left ( \frac{\rho(\vec r'}{\vert\vec r - \vec r'\vert}... ...c{1} {4 \pi} \int dS \phi(\vec r') \hat k \cdot \vec \nabla G(\vec r , \vec r')$$

$$= \int d^3r' \frac{\rho(\vec r')}{\vert\vec r - \vec r'\vert} - \int d^3r' \frac{ \rho(\vec r')}{\vert\vec r - \vec r' + 2 z' \hat k\vert}$$ (55)

because the the potential on the conducting plane vanishes. This is exactly the potential we have obtained using the method of images.

For the case of the conducting sphere of radius $a$, the potential using the method of images is

$$\phi(\vec r) = \frac{q}{\vert\vec r - \vec r'\vert} - \frac{aq/r'}{\vert\vec r - \vec r''\vert}$$ (56)

where $\vec r'' = a^2 / r' \hat r'$ and $r'$ is the length of vector $r'$ and $\hat r'$ is the unit vector in the direction of vector $r'$. We can therefore define the Green's function

$$G(\vec r, \vec r') = \frac{1}{\vert\vec r - \vec r'\vert} - \frac{a/r'} {\vert\vec r - (a/r')^2 \vec r'\vert}$$

$$= \frac{1}{\sqrt{r^2+r'^2-2rr'\cos\gamma}} - \frac{1}{\sqrt{(rr'/a)^2 +a^2-2rr'\cos\gamma}}.$$ (57)

where $\gamma = \hat r \cdot \hat r'$. From the equation above we see that the functional dependence pf Green's function on $\vec r$ and $\vec r'$ is same. Also, we find that when $\vec r$ or $\vec r'$ is on the conducting sphere ( $r=a$ or $r'=a$ ), the Green's function vanishes. In order to compute the electric potential we need the normal derivative of the Green's function on the surface of the sphere. The electric potential for a given charge distribution $\rho(\vec r)$ is

$$\phi(\vec r) = \int d^3r' G(\vec r, \vec r') \rho(\vec r') + \frac{1}{4\pi} \int d\Omega a^2 \phi(\vec r') \hat r' \cdot \nabla_{\vec r'} G(\vec r, \vec r')$$

$$= \int d^3r' \left ( \frac{1}{\vert\vec r - \vec r'\vert} - \frac{a/r'}{\vert\vec r - (a/r')^2 \vec r'\vert} \right ) \rho(\vec r')$$

$$+ \frac{1}{4\pi}\int d\Omega \phi(a,\theta,\phi) \frac{a(r^2 - a^2)} {(r^2+a^2-2ar\cos(\hat a \cdot \hat r))^{3/2}}$$ (58)

If in a given problem the potential on the surface of the sphere is specified, we can use this Green's function to compute the potential. As an example, consider that the spherical surface is maintained at constant potential $V_0$. Then the electric potential for this problm is

$$\phi(\vec r) = \frac{V_0}{4 \pi} \int_0^{2\pi} d\phi' \int_{-1}^1 d(\cos \theta') \frac{a(r^2 - a^2)}{(r^2+a^2-2ar \cos \theta')^{3/2}}$$ (59)

In the right hand side, the z axis has been chosen along $\vec r$. After completing the integrations, we get $\phi(\vec r) = \frac{V_0 a}{r}$, which is same as the potential due to a charge $V_0 a$ at the origin.

For general functional forms of the potential on the surface of the sphere, analytic expression for the potential is not possible and numerical integration should be done.