Charge above and infinite grounded plane

[aaaa202]

The classic image problem is this: Suppose we have a charge q over an infinite grounded plane. What will be the potential at a given point for z>0. It's not just the potential of the charge q since it will induce a certain charge on the conductor.
By looking at the mirror problem of a charges of opposite sign and displaced the same length relative to z=0 but in different directions, one finds from the uniqueness theorem that this solution also holds for the problem above.
Now: According to my book, even though the fields will the same, then the electrostatic energy of the charge is different in the two problems.
In the problem with the conductor the energy is namely only half the energy of the image problem. I have some trouble understanding this. The electrostatic energy is negative because we have to do work to separate the system in both cases. In the first case a negative charge and positive charge are brought in. In the second case a charge q is brought in while the induced charge runs into sustain a zero potential in the conductor. But why would separation require half the energy?
In the first case you would need to pull away the positive charge away while keeping the negative fixed. In the conductor case, what would then happen?

 

[gabbagabbahey]

The classic image problem is this: Suppose we have a charge q over an infinite grounded plane. What will be the potential at a given point for z>0. It's not just the potential of the charge q since it will induce a certain charge on the conductor.
By looking at the mirror problem of a charges of opposite sign and displaced the same length relative to z=0 but in different directions, one finds from the uniqueness theorem that this solution also holds for the problem above.
Now: According to my book, even though the fields will the same, then the electrostatic energy of the charge is different in the two problems.
In the problem with the conductor the energy is namely only half the energy of the image problem. I have some trouble understanding this. The electrostatic energy is negative because we have to do work to separate the system in both cases. In the first case a negative charge and positive charge are brought in. In the second case a charge q is brought in while the induced charge runs into sustain a zero potential in the conductor. But why would separation require half the energy?
In the first case you would need to pull away the positive charge away while keeping the negative fixed. In the conductor case, what would then happen?

This easiest way to see this is to look at the expression for electrostatic energy in term of the electric field $$W=\frac{\epsilon_0}{2}\int_{\text{all space}} E^2 d^3 x$$

The fields for $z\geq0$ are the same for both cases, but what about for $z<0$?