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

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?

• Like last time, you can gain some understanding by thinking about two electrons. Pushing them closer together is like compressing a spring. Then imagine pushing a circle of electrons closer together, then a sphere of electrons closer together. But note that again, we have a force versus field issue here. E isn't really a field, Fµν is the field. E is the linear force that results from Fµν field interactions. It takes two to tango. – John Duffield Jun 23 '15 at 13:00
• $F_{\mu\nu}$ decomposes to $E$ and $B$ when restricting from Lorentz transformations to rotations, so there is nothing wrong with considering the $E$ and $B$ vectors alone! – Sebastian Riese Jun 23 '15 at 13:14
• @John Duffield: I want to know sir, how this squeezing energy is equal to the energy required to assemble the charges. I'm not getting the sense. – user36790 Jun 23 '15 at 13:18
• @Sebastian Riese : I'm afraid there is. User36790 : see what Dylan said. I sympathise with you, it isn't always easy to understand some given textbook, and there's often a reticence to criticise a textbook that is in any way unclear or misleading. – John Duffield Jun 23 '15 at 19:16
• I disagree. As long as you do not consider boosts, you loose nothing and gain notation you are more used to (it is a completely different thing, that it might be better to teach electromagnetism in covariant formalism or in terms of differential forms from the beginning ...). But if you now what you are doing, working with $E$ and $B$ fields is not harmful. – Sebastian Riese Jun 25 '15 at 14:04

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!

• Note that this site has MathJax enabled, which means you can use Latex-like syntax to add in equations for readability. – Kyle Kanos Jun 23 '15 at 15:09
• Sir, I have given $1$ in order to highlight my problem. This is disgusting. I've spent two days only to understand this single piece & now this is a typo! – user36790 Jun 23 '15 at 15:35
• Ahh sorry, my fault then! And yeah I know the feeling.. It's better if you acquire the latest edition (the 3rd one). Good luck! – Dylan132 Jun 23 '15 at 15:41
• +1. Really thanks for the enlightenment! My book is "special Indian edition" for Indian students at cheap price; I thought it doesn't contain any errors however I was wrong:( Which edition do you have, sir? – user36790 Jun 23 '15 at 15:57
• @user36790 It isn't a random typo, it depends on whether you measure surface charge density in statcoulombs per square centimeter like in cgs Gaussian units or in coulombs per square meter like in mks SI, so it is correct for a particular choice of system of units. – Timaeus Jun 23 '15 at 20:14

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.

• What about question 2? – Kyle Kanos Jun 23 '15 at 14:36
• What does sigma stand for? – err513 Jun 23 '15 at 15:05
• Looks like surface charge density to me. – Kyle Kanos Jun 23 '15 at 15:10
• @Kyle Kanos: Yes, it is. – user36790 Jun 23 '15 at 15:24