say there is an infinite grounded conducting plane and a point charge of charge q is above it. the induced charge on the plane is -q and i know how to find this using the method of images to find the potential and then finding and integrating the charge density, but my text says that it should be intuitively clear that this induced charge must be -q. i'm just not seeing why it has to be, can anyone give me a simple explanation? thanks.
Apply Gauss's law to a box that contains the entire region with z<=0 (strictly speaking, you should take the limit as the box gets infinitely large). Since the field at the box surface is the same in the case of the grounded conducting plane and the image charge, the total enclosed charges are the same in both cases.
how do you know that the field at the box surface is the same for the plane and the image charge? in the point charge case the field at z=0 is 2q^2/(4*pi*e*r^2) but i don't see how you know what the field is in the case of the plane.
My comment may be late, i register just now. I may be having an answer about what you are asking for. You may think about the problem in reverse. To be clear, if one takes two equal and sign opposite charges +q and -q. let r be the distance between them. Then from basic electrostatics, we know that each charge will create its own potential in the surrounding space. Equipotential surfaces will then be created around the given charge +q and -q respectively. Knowing that the expression of the potential created by a point charge Q at any point of the space is given by: U=Q/(4pi*epsilon0*R) with R being the distance between Q and the point of the space where the potential is evaluated. Then after the theorem of superposition (applicable to the electrostatic potential), at any point of the space around +q and -q, the total potential is the sum of U(+q) and U(+q). This lead to find a an equipotential surface where the total potential U=U(+q)+U(-q) is zero (especially for the particular case where +q and -q have identical magnitudes in Coulomb unit). It is rather clear that this equipotential will be placed at a distance r/2 from each one of the two charges. Let the equipotential surface where the total potential is zero be S. This mean that if one brings any metallic sheet and insure its potential to be zero (by grounding it for example), and fit it exactly to S, then this will change nothing of the electrostatics configuration of the two charges +q and -q. And here you got your answer. This means that solving the electrostatic interaction between a point charge +q and a grounded surface is exactly the same as if it was between two point charges +q and -q placed at equal distances from each part of the grounded surface. For more clear interpretation i recommand you to look to the First volume of Feynman lectures on electromagnetics, "the method of images" section. Its very well exposed.