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The energy of point charge configuration can be written as: $$W = \frac{1}{2}\sum_{i=1}^{n}q_{i}V(r_{i}) \, ,$$ which can take both positive and negative values. However, when we integrate the equation to get the energy of a continuous charge dustribution: $$W = \frac{1}{2}\int\rho Vd\tau \Rightarrow W = \frac{\epsilon_{0}}{2}\left [ \int E^{2} d\tau + \oint VE\cdot da\right ] \, .$$

Take the volume to integrate to be all space, then the second term vanishes: $$W = \frac{\epsilon_{0}}{2}\int E^{2}d\tau \, .$$ This formula can take only positive values.

So there is a discrepancy between the two formulas. What caused the discrepancy? According to Griffith's Introduction to Electrodynamics, it says in the former equation $V(r_{i})$ represent the potential due to all charges but $q_{i}$, whereas later $V(r_{i})$ is the full potential. But why would the original charge has any potential when there is no other charge already present?

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The difference is the zero point. When summing over charges, the reference is a state in which this charges are infinitely separated. Those are still distinct, localized charges, just separated from each other.

When integrating $E^2$ over all space, the reference state has all charge separated. Even the individual charges from the first method are broken up, so that $E=0$ everywhere.

The second method has its reference at an "absolute zero", so to speak. The first method has its reference with a lot of positive energy needed to gather the individual charges. That's why the first method can have negative values.

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Your formula for the first energy is incorrect. Instead use:

$$W = \frac{1}{2}\sum_{i\neq j}q_{i}V_j(\vec r_{i}) \, .$$

Or even:

$$W = \frac{1}{2}\sum_{i\neq j}\frac{q_iq_j}{4\pi\epsilon_0|\vec r_i-\vec r_j|} \, .$$

And now you see right away that you are avoiding the energy of the point charges themselves. Because naively it would be zero.

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