Method of images with infinite, earthed, conducting plane


by watty
Tags: electric field, equipotential, method of images
watty
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#1
Dec24-08, 08:49 AM
P: 9
Hello guys,

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
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Doc Al
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Dec24-08, 09:32 AM
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Quote Quote by watty View Post
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)?
Note that this is the field immediately above the plane, due to the induced surface charge.

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.
Yes, the charge and the mirror charge form a dipole, and their field lines do go in the same direction at the plane.
Only if both charges were the +q or both -q would the potential, in my mind at least, be zero midway between them.
The potential midway between two like charges is not zero. Recall that potential is a scalar--the potential from each charge is the same, thus the potential midway is non-zero. (Only for unlike charges would the potential contributions cancel at the midway point.)

Here's a clear discussion of the method of image charges that might help you: Method of Images


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