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In this problem, we use the Method of Images and get the resulting properties of the charge distribution on the plane. At the end of this process, Griffiths, in Introduction to Electrodynamics, comments that we could have said that the charge on the plane was going to be $-q$ with some hindsight. How can this result have been obvious before arriving at the solution mathematically? I can't think of any thought process that would lead to the same conclusion.

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I think that the "hindsight" goes as follows?

Electric field lines start on a positive charge and end on a negative charge so where else can the electric field lines from the $q$ charge go but to the infinite grounded conducting plate?

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  • $\begingroup$ I understood the part that the grounded plate is going to be negatively charged and the field lines will terminate perpendicular to the surface. I specifically want to know how does one guess that the charge will be $-q$ exactly $\endgroup$ – user117913 Feb 12 '17 at 9:10
  • $\begingroup$ By assuming that there are a certain number of electric field lines leaving a $=q$ charge and the same number are arriving at a $-q$ charge. Gauss can be used by having a Gaussian surface around the $+q$ charge and another Gaussian surface above the infinite conducting plane. $\endgroup$ – Farcher Feb 12 '17 at 9:12
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Maybe if I say this less laconically it will be clearer: We are assuming there is no E field below the grounded plane. Look at how your charge density formula was derived. If you don't make an assumption like this you can't figure out the charge density. For instance if the image charge was a real charge there would be field below the plane and the charge on the plane would be zero.

There is no E field below so if we take a Gaussian surface with one side below the plane and the other sides going off to infinity, there is no flux, therefore no net charge, therefore the plane must contain -q.

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