Method of images with infinite, earthed, conducting plane

[watty]

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

 

[Doc Al]

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: http://farside.ph.utexas.edu/teaching/em/lectures/node64.html"

 

[charng]

Hello watty,

The method of images is a mathematical technique used to solve problems involving conductors and charges. In this particular case, the method of images allows us to simplify the problem by replacing the conducting plane with a charge that satisfies the same boundary conditions. This is possible because the presence of the conducting plane creates a mirror image of the original charge, which cancels out the electric field on the surface of the plane.

Regarding your question about the potential being zero but the electric field being non-zero, this is due to the fact that the potential is a scalar quantity while the electric field is a vector quantity. The potential represents the amount of work needed to move a unit charge from one point to another, while the electric field represents the force experienced by a unit charge at a given point. In this case, the potential at the surface of the plane is zero because it is at the same potential as the charge on the plane. However, the electric field is non-zero because it is influenced by both the original charge and the image charge, resulting in a non-zero net field.

In terms of thinking about the two charges as a dipole, this is a valid way to conceptualize the situation. However, the method of images allows us to simplify the problem and find a solution that satisfies the boundary conditions without having to explicitly consider the dipole. This is why it is a useful technique in solving problems involving conductors and charges.

I hope this helps clarify the concept of the method of images for you. If you have any further questions, please don't hesitate to ask. Good luck with your studies!