Energy stored in an electric field I know the mathematical proof that $U=\frac{\epsilon_0}{2}\int\vec{E}^2dv$ is the energy stored in a particular volume in space due to an electric field, but I don't get what it actually means. I lack the physical intuition to this result. For example, if I want to calculate what work needs to be done in order to assemble a charged sphere with radius $R$, why is it required to integrate over the entire space from radius $r=\infty$ to radius $r=R$? Every insight will help, it is just a pure mathematical result for me now and I'm not even sure how to properly use it.
 A: One needs to perform work in order to construct a charge distribution. Doing work is an energy transfer and this energy is stored in the form of the electric field.
An easy example is a parallel plate capacitor. Its capacitance is $$C=\frac{\epsilon_0A}{d},$$ whereas the energy built by moving the charge from one plate to the other is $$U=\frac{Q^2}{2C}=\frac{CV^2}{2}=
\frac{\epsilon_0A}{d}\frac{E^2d^2}{2}=\frac{\epsilon_0E^2}{2}Ad,$$
i.e. exactly what we call the energy of the field in volume $Ad$.
A: Suppose you have a collection of positive charges. Now, you can intuitively feel that the closer they are, more is the electrostatic potential energy stored because they would want to fly apart. Suppose this collection is in the form of uniformly charged sphere, containing very tiny positive charges spread uniformly throughout the volume of the sphere with charge density $ρ$.
Now, the electrostatic potential energy is somewhat related to the electric field generated by this sphere in the space around and within it. In case of a spring which is compressed, we say that the potential energy is stored inside the spring as elastic energy. But, in case of electrostatic energy, we see that the effect is not local anymore and we may say that energy is stored inside the $field$ which is present everywhere.
To support this viewpoint, when you perform the integral to calculate electrostatic energy for the charged sphere ( integrating from $r=0$ to $r=∞$ ), you get the result as $$U=\frac35kQ^2/r$$
As the radius of sphere decreases, potential energy increases and so does the electric field created by the configuration increases, and hence the relation involves electric field.
I hope it makes things a bit more intuitive.
