What formula is this and what does it signify? (Electric Field and Potential) I probably skipped the useful part of the lecture, but while we were being taught about electric potential energy, my professor mentioned an equation, which he said we will seldom use, but which is significant. The equation is for the potential energy per unit volume. The equation was:
$$
\frac{dU}{dV}=\frac{1}{2}\varepsilon_{0}E^{2}.
$$
He then rearranged and integrated to give
$$
U=\frac{1}{2}\int\varepsilon_{0}E^{2}\, dV,
$$
where $\varepsilon_{0}$ is the permittivity of free space, $E$ is the magnitude of electric field at the point of focus, and $dV$ is a volume element of the space in focus.
He then ended the lecture with saying, this shows that electric potential energy is stored as electric field in free space. So, I want more insight into this equation, what is the significance of this and if possible a name of any of these two.
 A: 
So, I want more insight into this equation, what is the significance
of this and if possible a name of any of these two.

Perhaps showing you the connection between the formula and the potential energy stored in the volume containing the electric field between the plates of a parallel plate capacitor may give you some insight.
First we start with the potential energy stored in the capacitor as a function of its voltage $V$ and capacitance $C$.
$$U=\frac{1}{2}CV^2$$
Next we relate the voltage $V$ across the capacitor to the electric field $E$ and the plate separation distance $d$
$$V=Ed$$
Finally, a parallel plate capacitor's capacitance in terms of the area $A$ of the plates and the electrical permittivity $\epsilon$ of the dielectric material between the plates is given by
$$C=\frac{\epsilon A}{d}$$
Substituting the last two equations into the first
$$U=\frac{1}{2}\frac {\epsilon A E^{2}d^{2}}{d}=\frac{1}{2}\epsilon E^{2}V$$
Where $V=Ad$ is now the volume of the space containing the electric field of the capacitor.
For air or vacuum, $\epsilon=\epsilon_{o}$
Since the electric field $E$ is considered constant in a parallel plate capacitor, it would come out of the integral you gave, making the last equation identical to your second equation.
Hope this helps.
