This question seems to come up often, but I cannot find a satisfying explanation. There is a point charge +Q some distance above an infinite conducting plane. Supposedly, the electric field below the plane must be zero. I have trouble understanding why this is true. The total charge on the plane must be zero, since the plane is a neutral conductor, and conservation of charge implies that any induced negative charge on the upper surface must be cancelled by an induced positive charge on the lower surface. Thus, if you look at a point P *outside* the plane, but *below* it, the fields from the positive and negative charges should not cancel each other. Indeed, the field must clearly be different in the configuration +Q * - - - - - - - - - - - P * +++++++++++++ than in +Q * - - - - - - - - - - - +++++++++++++ P * because the field from the positive charges has now switched directions (but not the field of the negative charges). But since the field is zero inside the conductor, then it must be nonzero outside. Moreover, if you use Gauss' law on a box containing P and going across the plate surface (but not containing Q), the net charge must be 0 (the plate has no net charge), so the electric flux must be 0 too. But the electric field is nonzero above the plane, so it must be nonzero below it as well. Another thing which confuses me is the meaning of "grounded". The problem is supposedly different if the plane is grounded or ungrounded. But is there a distinction between "grounded conductor" and "ungrounded conductor"? Grounded supposedly means that the potential is 0. But I thought the actual value of the potential was irrelevant, so I could decide that the potential would be 117 V on the plate instead, without affecting the solution.