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I have been doing some practice questions in a text book [Electricity and Magnetism by Purcell and Morin]. So I know that the energy the potential energy of a system is the total work required to assemble the system from infinity and can be found from: $$U=\frac{\epsilon_0}{2}\int_{entire\ field }E^2dv$$ [This is stated in section 1.15 of this book]

In the problem that I am doing [question 1.33] it gets you to derive the total energy of two protons. If $\vec E_1$ is the field due to one and $\vec E_2$ that due to the other then the total potential is given by: $$U=\frac{\epsilon_0}{2}\int E_1^2dv+\frac{\epsilon_0}{2}\int E_2^2dv+\frac{\epsilon_0}{2}\int \vec E_1\cdot \vec E_2dv$$ But for this the third integral alone is equal to $$\frac{e^2}{4\pi \epsilon_0 b}$$ (where $b$ is the separation of the protons). But the first two integrals must be none zero. So this means one of two things:

  1. The total energy to build up an assembly of protons from infinity (where the total energy is taken to be 0) is not: $$\frac{e^2}{4\pi \epsilon_0 b}$$
  2. Or: $$U=\frac{\epsilon_0}{2}\int_{entire\ field }E^2dv$$ does not represent the total work done to assemble the charges from infinity.

One of the above statements must be true else we have a contradiction. I think it is the first one but have no idea why, please can someone explain.

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    $\begingroup$ It breaks my heart that someone is reading Purcell and using a bunch of $\epsilon_0$ everywhere. $\endgroup$ – Mark Eichenlaub Mar 31 '15 at 18:21
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The problem is about the energy needed to assemble the charges, assuming that they already exist.

So, imagine that the charges are separated from each other by a very large distance. Then your third expression is effectively zero.

As you bring the charges together, the first two expressions don't change at all, and the third expression does. So the work done, which is the change in energy, is completely represented by the third integral.

The answer to your question is explanation 2, that integral does not represent the work done. It's the change in that integral as we move the charges in from infinity that we're interested in.

However, there is still a problem in that the first two integrals diverge, indicating infinite energy for a point charge. So when we say that the first two integrals don't change, we are speaking loosely. You can perform a coordinate transformation to show that the integrals themselves are the same integrals, but not that the values are the same because the integrals don't have a value. For more on that, see Infinite Energy of Point Charges (in the context of classical field theories)

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  • $\begingroup$ thanks for your answer. Are there any conditions that if they are met then the whole integral stated above is indeed the total work done, rather then the change. I know it works for a charged sphere, so will it be the total work done if the charge distribution is continuous (i.e. of finite rather then point size)? $\endgroup$ – Quantum spaghettification Mar 31 '15 at 18:55
  • $\begingroup$ Right; as long as it's continuous that integral works. $\endgroup$ – Mark Eichenlaub Mar 31 '15 at 19:39

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