What did Feynman want to mean by 'artificial energy'? I was reading Mechanical and electrical energies by Feynman when I came before this:

We have seen an analogous situation in electrostatics. We showed that the energy of a capacitor is equal to $Q^2/2C$. When we use the principle of virtual work to find the force between the plates of the capacitor, the change in energy is equal to $Q^2/2$ times the change in $1/C$. That is,
  $$\Delta U=\frac{Q^2}{2}\,\Delta\left(\frac1C\right)=−\frac{Q^2}2 \frac{\Delta C}{C^2}.\tag{15.14}$$
Now suppose that we were to calculate the work done in moving two conductors subject to the different condition that the voltage between them is held constant. Then we can get the right answers for force from the principle of virtual work if we do something artificial. Since $Q=CV$, the real energy is $\frac12 CV^2$. But if we define an artificial energy equal to $−\frac12 CV^2$, then the principle of virtual work can be used to get forces by setting the change in the artificial energy equal to the mechanical work, provided that we insist that the voltage $V$ be held constant. Then
  $$\Delta U_\textrm{mech}=\Delta\left(−\frac{CV^2}{2}\right)=−\frac{V^2}{2}\Delta C,\tag{15.15}$$
which is the same as Eq. $(15.14)$. We get the correct result even though we are neglecting the work done by the electrical system to keep the voltage constant. Again, this electrical energy is just twice as big as the mechanical energy and of the opposite sign.

I couldn't really comprehend what he wanted to tell in the above excerpt especially by the bold line. 
What did Feynman wanted to mean by 'artificial energy'? Meant to say, what is the need of introducing it here; what did he want to say in this para?
 A: He meant that the "energy" function
$$
E_\textrm{art}(C) = - \frac{1}{2}CV^2
$$
is introduced solely for the purpose of getting the right result with the "principle of virtual work". It is not really EM energy in the usual sense as energy stored in the capacitor, available for use. If there is potential difference $V$, the latter energy $E$ is actually always positive and equal to
$$
E = \frac{1}{2}CV^2.
$$
A: As we know the work done by a battery is $W=QV$ which is equal to $CV^2$, and as we know energy available to use in a capacitor is half of $W$, which means that half of the work done by battery cannot be used or some other source does the work $-CV^2/2$ (note we can consider this artificial energy to be work done by some virtual force and can be used to solve problems), so the energy $-CV^2/2$ cannot be used so he pins it as artificial energy.
Most of the time the artificial energy is emitted by the conductor as radiations.
My professor said that for a capacitor to be charged "artificial" energy has to be burned.
A: He demonstrated this analogy for the magnetostatics' case.
Magnetostatics : if Current $I $  is held constant , you can use the "fake" energy,
$ U = -\vec{\mu} .\vec{B}$
to find forces
Electrostatics: if Voltage $V$ is held constant, you can use the "fake" energy,
$U = -\frac{1}{2}CV^2 $
to find forces between parallel plates of a capacitor.
In both cases one thing is held constant and the expression for energy is false as it does not accommodate for energy provided by a source (like battery) to keep current constant(in Magnetostatics) and to keep voltage constant (in Electrostatics)
This was the analogy he was taking our attention to using the capacitor example.
