
#1
Feb2512, 08:33 PM

P: 3

Consider the classic image problem with a conducting plane and a point charge. After finding out the charge density on the plane we integrate to find out the total charge induced. It comes out to be "q",where q is the charge outside
My book says "It comes out to be q, as you can convince yourself with the benefit of hindsight". It isn't that obvious why it should come out equal to the image charge. Can somebody explain? 



#2
Feb2612, 12:19 AM

P: 404





#3
Nov712, 12:54 PM

P: 11

I would describe it a little bit different.
For me, the main point is that the tangential component of the electric field on the plate has to vanish. Otherwise, there would be movement of the electrons at the surface of the metal. So, the metal's surface is an equipotential surface of the electrostatic potential. Now you can think of an imaginary charge distribution that could be found such that the potential vanishes on the surface of the metal if the metal was not there. Of course, you find that it is a symmetrically placed charge q. You can go even further and generalize this procedure and ask for metallic corners subject to a point charge etc. The bottom line is that image charges are not real, just something that gives us some intuition about boundary value problems :) 



#4
Nov712, 04:03 PM

Sci Advisor
HW Helper
P: 1,937

Method of Images
Use Gauss's law. If the field is that from q'=q, then that must also be the total surface charge.



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