Why is energy associated with the electric field given as $U = \dfrac{\epsilon_0}{2} \int E^2 dv$?

Question

This is an excerpt from Edward M. Purcell's Electricity & magnetism:

Suppose a spherical shell of charge is compressed slightly, from an initial radius of $r_0$ to a smaller radius. This requires that work be done against the repulsive force $\dfrac{\sigma^2}{2\epsilon_0}$ for each square meter of the surface. The displacement, being $dr$ , the total work done is $\left(\dfrac{\sigma^2}{2\epsilon_0}\right)\left(4\pi r_0^2\right)$. This represents an increase in the energy required to assemble the system of charges, the energy $U$: $$dU =\dfrac{\sigma^2}{2\epsilon_0}r_0^2\sigma^2$$. Now, notice how the electric field $E$ has been changed. Within the shell of thickness $dr$, the field was $0$ & is now $4\pi\sigma ^1$. Beyond $r_0$, the field is unchanged. In effect, we have created a field of strength $E= \dfrac{\sigma}{\epsilon_0}$, filling a region of volume $4\pi r_0^2 dr$. We have done so by investigating an amount of energy given by the above equation which, if we substitute $\epsilon_0 E$ for $\sigma$, can be written as $$dU = \dfrac{E^2 \epsilon_0}{2} 4\pi r_0^2 dr$$. This is an instance of a general theorem: The potential energy $U$ of a system of charges, which is the total work required to assemble the system, can be calculated from the electric field itself simply by asssigning an amount of energy $\dfrac{E^2 \epsilon_0}{2}dv$ to every volume element $dv$ & integrating over all space where there is electric field.

$$U = \dfrac{\epsilon_0}{2} \int_\text{entire field} E^2 dv$$

My questions are:

1)What is actually meant by assembling of charges? To bring the charges to make the desired arrangement, right? In order to find the energy, wouldn't we need to find the work done from bringing the charges from infinity to the desired configuration?? But here, Mr. Purcell just squeezed the sphere & found the energy required to squeeze the shell against the repulsive force! How can the energy required for squeezing be the energy required for assembling the charges? I am not getting the intuition. Please explain.

2)$^1$ How can $4\pi\sigma$ be the field? How did he deduce it? Moreover, can anyone tell me what he is saying in the second para?

Answer 1

How can the energy required for squeezing be the energy required for assembling the charges?

The energy required for squeezing the sphere from initial radius $r_0 \to (r_0-dr)$ is not the energy required to assemble the charges. He merely calculated the change in energy (or the work required) each time you squeezed the sphere by a displacement of $dr$. To find the work required for assembling the charges you'll have to integrate over all space (As in you had a sphere of Radius $\to$ infinity and squeeze it to the desired radius).

Mr. Purcell just squeezed the sphere & found the energy required to squeeze the shell against the repulsive force! ...

He found how much work is required to squeeze a sphere a little bit, and again the entire energy required to assemble the charges is to sum over all this little bit of energy from infinity to the desired radius.

It's some how like recursion; Suppose you know the energy $U$ required to assemble the charges at their current state, and now you want to squeeze them a little bit closer so you must invest extra work: $$U(r_0-dr)=U(dr)+U(r)=2U(dr)+U(r+dr)= \cdots =\text{[integration over all} \qquad U(dr) \qquad \text{from infinity to the current radius]} + U\text{(infinity)} \text{[Which is 0]}$$

If you go backwards, and let the electrons move freely to infinity (If they can do that), you'll gain back energy which is basically equal ,in value, to the work required in assembling them in the first place.

How can $4\pi\sigma$ be the field? How did he deduce it? Moreover, can anyone tell me what he is saying in the second para?

I think you have an old version of the book; The one I have has $\dfrac{\sigma}{\epsilon_0}$ instead of that. By the way, there is a "1" near the sigma, there must be an explanation at the bottom of the page I believe.

Hope this helps!

Answer 2

What is actually meant by assembling of charges? To bring the charges to make the desired arrangement, right? In order to find the energy, wouldn't we need to find the work done from bringing the charges from infinity to the desired configuration?? But here, Mr. Purcell just squeezed the sphere & found the energy required to squeeze the shell against the repulsive force! How can the energy required for squeezing be the energy required for assembling the charges? I am not getting the intuition. Please explain.

Use the work-energy theorem and the conservation of electrostatic potential. Then you'll know that the work of taking the electron from infinity to its place is equal to the potential energy-difference.