Hello guys,(adsbygoogle = window.adsbygoogle || []).push({});

I'm having some trouble getting my head around the method of images, partly due to confusing notes and things.

If I have a charge +q at a fixed point (0,0,a) above a conducting plane that is held at zero potential, it is said that the plane can be replaced with a charge of -q at (0,0,-a) as this solution satisfies the same boundary conditions (as per the uniqueness theorem).

How is it that the potential on the surface of the plane can be 0 but the electric field be described as q/(2pi x espilon_0 x a^2)?

I would have thought the potential be non-zero in the case of the mirror charge. Can the two charges not be thought of as a dipole? As in their field lines are both going in the same direction. Only if both charges were the +q or both -q would the potential, in my mind at least, be zero midway between them.

Any help would be greatly appreciated,

watty

**Physics Forums - The Fusion of Science and Community**

# Method of images with infinite, earthed, conducting plane

Know someone interested in this topic? Share a link to this question via email,
Google+,
Twitter, or
Facebook

Have something to add?

- Similar discussions for: Method of images with infinite, earthed, conducting plane

Loading...

**Physics Forums - The Fusion of Science and Community**