Method of Images - Electromagnetics

[elexhobby]

Hello friends,

Could somebody please explain me how to prove the Method of Images so often used in Electromagnetics. I simply fail to understand how replacing an infinite conductor with an opposite charge configuration exactly below the conductor, creates an equivalent effect & leaves the system unchanged. I am referring to Engineering Electromagnetics by Hayt & Buck.

Thanks in advance,
Regards

 

[RadioRon]

If you found that you could build two entirely different antennas that produce exactly the same radiation pattern when viewed from far away, would you consider these antennas equivalent? I would, as long as the intensity and polarization and rotation of the electric and magnetic fields were identical in any direction. You get this when you replace an ideal ground plane with the "reflected" image of the original antenna over the ground plane. This is because the reflection of energy off the ground plane just happens to be identical to the radiation from the image antenna.

I presume you already know the properties of the image. That is, that charge polarity is reversed relative to the real antenna, and that the vertical components of the image are the same as those of the real antenna while the horizontal components are reversed in direction.

The classic way to help understand this is to use ray tracing of energy radiating from a point radiator above ground (the simplest case). You usually see this example in text books. If you trace the reflected ray you see that it is in the same line as a ray that would radiate from the image point. These rays are equivalent.

 

[elexhobby]

Thanks for your immediate reply. But my doubt still isn’t cleared. Sorry, I should have stated my doubt clearly.
With the conducting surface present & say a charge +Q above it, for any other test charge, the only force is due to the +Q charge. Now consider the conducting plane removed, & a charge –Q placed equally below. Now the force on any test charge will be that due to +Q as well as –Q. This is not the same as the force which was present with the conducting plane in place..
I understand that the Method of Images can’t be wrong; there is obviously some flaw in my reasoning. Kindly help me understand it..
Thanks a lot,
Regards.

 

[RadioRon]

Its fun to try and look into your mind to see where the roadblock is on this. Let's try this statement: you say that in the first case with the conducting plane the only force is due to the +Q charge. I dispute this. The existance of the +Q charge above the conducting plane influences the conducting plane and induces an arrangement of charge (or current) in that conducting plane. Charges will move and arrange themselves to satisfy the boundary condition that the tangential component of the E field and the normal component of the H field are zero at the surface of the conducting plane. The total electric and magnetic fields in the region will be due to not only the charge that you call +Q but also to these "induced" charges on the conducting plane.

The force seen by a third test charge will be due to this total field.

As far as the fields in the region above the conducting plane are concerned, the same results can indeed be obtained by replacing the conducting plane with the "image" charges.

(source: Electromagnetic Waves and Radiating Systems; Jordan and Balmain, Prentice Hall, 1968, pp 369-370. I never throw out technical books of any sort)

 

[elexhobby]

Hey

Thanks a lot.. Your explanation helps clear many false assumptions I had made.. Will definitely refer to your suggested material for the proof of why the two electric fields are equivalent.. Surprisingly, Hayt doesn't mention this.. Will write to you if I have any further doubts..

Thanks again,
Regards