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

 

[safwatonline]

simply they have the same field at boundry conditions ,
which makes the field solution true for the area above the conductor and not true below it.
when we replace the conductor (perfect ,and hence no fields on it) by an oppiste charge configuration ,this will impose the same field configuration above the conductor (u can think that opposite charges so the fields at half the distance must equal zero), but this will introduce fields under tha conductor due to the opp. charge (this filed doesnt exist naturally cause tangential field ends on the perfect conductor surface), so when applying the image theory take the part of the field above the conductor only.
note this is a good aproximation since we was considering an infinte conducting surface ,which is no true.
regards,
a.safwat

 

[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.

 

[safwatonline]

ok, i am not sure, but here what i understand , consider
the one charge u stated +Q , this charge have field lines going outward of the charge and terminating perpendicular on the perfect conductor material (due to boundary conditions on perfect conductor surface) , now if we put an extra -ve chage -Q at the same distance field lines with opposite direction (coming into the -Q) will exist ,which will also terminate at the perfect conducting surface .
untill now we didnt change the field lines due to the +Q , then if we removed the conductor the field lines will be the same (i.e. the field lines will be connected to each other without changing the field).
so since we have the same field , we have a representation for the same setup. i can say but i am not quite sure that if the field is the same then we have the same force. refer to coulombs law F=-qE ; E is the electric field , so if the field is the same then the force is the same.
regards,
a.safwat

 

[elexhobby]

You have considered the case with conductor present. With the conductor absent, there is no reason for the field lines from the negative charge to terminate, & hence these field lines from the negative charge will start affecting the behavior of any test charge kept above the plane. In the presence of the conducting plane, the test charge was subjected to field lines only from the +Q charge, in the absence of the conducting plane, it is subjected to field lines from the positive as well as negative charges. How can these two seemingly different setups be equivalent to each other?
Help...

 

[safwatonline]

hello,
actually when two equal and opposite charges are seperate their field lines will be perpendicular to each plane (p2,p3,p4) containing the two points"this is at equal distance from the two charges".
P1 : plane of the perfect conductor
P2,P3,P4: planes conatining the two charges and all are perpendicular on P1
field lines are Perpendicular to P1
regards,
a.safwat