Deriving Solutions with Method of Images

[OliviaB]

Apply the method of images to derive the solution

$$\displaystyle \phi(x,y,z) = \frac{z}{2 \pi} \int_{- \infty}^{\infty} \int_{- \infty}^{\infty} \frac{f(x_0, y_0)}{((x - x_0)^2 + (y - y_0)^2 + z^2)^{\frac{3}{2}}} dx_0 dy_0$$

from

$$\displaystyle \bigtriangledown^2 \phi (x,y,z) = 0$$

$$\phi(x,y,0) = f(x,y)$$

from the region $$- \infty < x < \infty, - \infty < y < \infty, 0 < z < \infty$$

Its hard for me to start because I don't really understand the method of images.

 

[jvicens]

Can you explain it to me more clearly?

Sure, the method of images is a technique used in electrostatics and potential theory to solve problems involving point charges or conducting surfaces. It involves creating a mirror image of the original system and using the principle of superposition to find the solution.

In this case, we are trying to find a solution for the Laplace equation, \bigtriangledown^2 \phi (x,y,z) = 0, which describes the potential in a region where there are no charges present. We are given the boundary condition \phi(x,y,0) = f(x,y), which means that the potential at the surface z=0 is equal to a given function f(x,y).

To apply the method of images, we first imagine a point charge located at (x_0, y_0, z_0). This charge will create a potential \phi_1(x,y,z) given by the formula:

\phi_1(x,y,z) = \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2}}

where \epsilon_0 is the permittivity of free space and q is the charge of the point charge.

Next, we create a mirror image of this charge at (x_0, y_0, -z_0). This charge will also create a potential \phi_2(x,y,z) given by:

\phi_2(x,y,z) = \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{(x-x_0)^2 + (y-y_0)^2 + (z+z_0)^2}}

Now, we can use the principle of superposition to find the total potential \phi(x,y,z) in our region:

\phi(x,y,z) = \phi_1(x,y,z) + \phi_2(x,y,z)

Substituting the expressions for \phi_1 and \phi_2, we get:

\phi(x,y,z) = \frac{1}{4 \pi \epsilon_0} \frac{q}{\sqrt{(x-x_0)^2 + (y-y_0)^2 + (z-z_0)^2}} + \frac{1}{4 \pi \epsilon_0} \frac