Method of Images problem

  • Thread starter vjraghavan
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Dear all,
I have some difficulties in understanding to use the method of images to solve Poisson's/Laplace's equation with boundary values.

I understand that the Uniqueness theorem(s) enables this method. In this method we use image charges and setup a different configuration without affecting the given boundary values and rho(the charge density/distribution) where it is easy to calculate the potential in the region of interest. Uniqueness theorem guarantees that the potential we thus determine is same as the potential due to the original configuration. This would mean that we shall NOT put any of the image charges in the region where we want to determine the potential. For, it would then change rho and we would be dealing with a different problem.

The difficulty arises when we are to determine the potential in all of space with the boundary conditions and rho given. This would mean that we cannot put the image charges anywhere in space. How then would be able to solve this problem using the method of images?:uhh:

http://en.wikipedia.org/wiki/Method_of_image_charges" [Broken] is Wikipedia's article on Method of Images if you want to have a look
 
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  • #2
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The difficulty arises when we are to determine the potential in all of space with the boundary conditions and rho given. This would mean that we cannot put the image charges anywhere in space. How then would be able to solve this problem using the method of images?
I don't think you can use it in this case, since there is nowhere to put the image charge. The method of images only works for certain problem setups where you can place an image charge without affecting the field in the region of interest.
 
  • #3
ZapperZ
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Dear all,
I have some difficulties in understanding to use the method of images to solve Poisson's/Laplace's equation with boundary values.

I understand that the Uniqueness theorem(s) enables this method. In this method we use image charges and setup a different configuration without affecting the given boundary values and rho(the charge density/distribution) where it is easy to calculate the potential in the region of interest. Uniqueness theorem guarantees that the potential we thus determine is same as the potential due to the original configuration. This would mean that we shall NOT put any of the image charges in the region where we want to determine the potential. For, it would then change rho and we would be dealing with a different problem.

The difficulty arises when we are to determine the potential in all of space with the boundary conditions and rho given. This would mean that we cannot put the image charges anywhere in space. How then would be able to solve this problem using the method of images?:uhh:

http://en.wikipedia.org/wiki/Method_of_image_charges" [Broken] is Wikipedia's article on Method of Images if you want to have a look
Why would you want to use method of images for such a problem, especially when there's no "image"? Why not either directly use Coulomb's Law, or directly solve the Poisson's equation?

Method of images is there for convenience when it is applicable. You don't use Gauss's Law, for example, unless you have a highly symmetric problem, or else it becomes rather unmanagable. The same applies here.

Zz.
 
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  • #4
vanesch
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The difficulty arises when we are to determine the potential in all of space with the boundary conditions and rho given. This would mean that we cannot put the image charges anywhere in space. How then would be able to solve this problem using the method of images?:uhh:
The method of images serves mostly to allow one to dispose of a boundary condition. The reason is that the boundary condition is the "hard" part in the solution of a Poisson equation ; for the rho distribution in free space, the Green function is known and the solution is simply a convolution of the Green function with rho.

For instance, the method of images can mimick an infinite plane with constant potential, or a spherical surface at constant potential. Conformal transformations (especially useful in 2D) can then transform these surfaces into other ones, extending the applicability of the method to these other surfaces.

But if you have charges in all of space, there are no boundary conditions to be mimicked with the method of images !
 

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