Why Does the Mirror Charge Method Double the Potential Energy?

[Aroldo]

Hi everybody,
The situation is the classic one: a point charge q+ in a distance d above a conductor plane grounded:
The conductor is grounded so V = 0, for z = 0.
Also, far away from the system (x² + y² + z² >> d) V --> 0

The argument to replace it for a q- charge seems perfect to me.
What I can't understand why W (potential energy) of the two point charges is twice as much as the conductor plane + q+ charge system.
I mean, why E = 0 for z < 0 in the conductor plane + q+.
Thank you a lot!

 

[BvU]

Try Gauss law on the "underside" of the conductor: no charge, no field.

 

[Aroldo]

What do you mean by 'underside', the region immediately bellow the conductor, or a phill box that starts inside the conductor and ends in z < 0?

 

[BvU]

Both give E = 0.

The mirror charge method only says something about the region where the real charge is (because a solution to the field equation is unique, anything is allowed, but the solution is only for that region).