Talk:Method of image charges

Valid OUTSIDE the sphere
The resulting potential when the imaginary charge is INSIDE the sphere is valid OUTSIDE the sphere. The article says it's only valid INSIDE the sphere. If you agree with this, change the article. Also; It's quite simple to derive the magnitude and position of the imaginary charge. If there is interest, I can provide the derivation in form of a picture and some of you guys can add it to the article.

Edit: nevermind, I saw now that the image charged is outside the sphere and the "real" charge is inside. I was confused because Griffiths got it the other way around.

— Preceding unsigned comment added by 188.113.82.187 (talk) 18:29, 9 July 2012 (UTC)

Uniqueness Theorem
I wrote an article on uniqueness theorem as the one provided here is rather sad... Perhaps the section in this article can be eliminated (or else should be rewritten)? IlyaV (talk) 21:32, 3 April 2010 (UTC) --- The x=0 in the pictures at the side should be y=0. If the author or someone could fix this that would be great.

Thanx Noigmn
 * I agree JabberWok 02:44, 28 September 2006 (UTC)

I'd really appreciate a description of the case in which we have more than one conductive surface

p.s. monday I have an electromagnetism exam and I can't find a description of this case :P

--Paracetamolo 17:30, 5 January 2007 (UTC)

For more than one conductive surface, see Foundations of Electromagnetic Theory, 3rd/e, by Reitz, Milford and Christy, p65.Kissnmakeup (talk) 18:22, 31 July 2008 (UTC)

Further discussion about the method of images with respect to perfect dipoles would be appreciated. For instance, the orientation of the image dipole is pretty confusing to me.

fixed
I have fixed the pictures so that they all now all say y=0.

the formula for the potential is in cylindrical coordinates where the distance from the infinite plane is measured along the z-axis. This would imply that the infinite plane is in the x-y plane but the picture has it in the x-z plane. This would be very confusing to someone who did not already understand the material.

rho
What is $$\rho\,$$ in the formula   $$V\left(\rho,y\right) = \frac{1}{4 \pi \epsilon_0} \left( \frac{q}{\sqrt{\rho^2 + \left(y-a \right)^2}} + \frac{-q}{\sqrt{\rho^2 + \left(y+a \right)^2}} \right) \, $$? --Saippuakauppias ⇄ 12:33, 23 January 2011 (UTC)


 * It's the radial coordinate in cylindrical coordinates, that is, the distance from the axis of the cylinder (which in this case is the line through the charge and its image). —Keenan Pepper 18:26, 24 January 2011 (UTC)

Generalization to conductors of arbitrary shape?
The article so far includes the case of a linear or spherical grounded conductor. It would be of interest to know whether the method can be generalized to conductors of arbitrary shape. The page on the Schwarz Integral Formula contains the intriguing statement that the method can be generalized from a unit disk to any open set using a conformal map. I wonder whether this also applies to image-charges?

Relation to Green's Functions?
Since Green's Functions are also used to solve the Poisson equation with boundary conditions, it would be of interest to know of any connection between the Image Charges method and the Green's Function method.

Conducting Plane Reflection for Electric Dipole Moments
The article says the image moment is rotated by pi radians azimuthally, but I believe this should be a polar rotation (about the z-axis). That is the rotation given in the equation, and in the figure. Please change if you agree. — Preceding unsigned comment added by Jacobbaron (talk • contribs) 16:11, 10 April 2014 (UTC)

Merge with another article
I think the Method of Images page should be merged with this page.  Pratyush Sarkar ( talk ) 14:10, 28 September 2014 (UTC)