# Why can we ignore self energy?

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.

• It breaks my heart that someone is reading Purcell and using a bunch of $\epsilon_0$ everywhere. – Mark Eichenlaub Mar 31 '15 at 18:21