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Sep 10, 2022 at 7:33 history edited user330563 CC BY-SA 4.0
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Sep 9, 2022 at 11:00 comment added user330563 He doesnt have to allow for an isolated point charge into his energy integral (conf. that he starts out from $\frac{q_1q_2}{4\pi \epsilon_0 r_{12}}$. The E- field form of the integral could be a numerical coincidence so to speak), until he interprets it to actually mean that energy is physically located in the fields
Sep 9, 2022 at 10:09 comment added kricheli Generally, for a charge distribution $\rho$ the self-energy is $U=\frac{1}{2}\int \rho \phi dV = \frac{\varepsilon_0}{2}\int E^2 dV$ (the equality is shown by partial integration and using Gauß' law). It doesn't really matter which of the two integrals you use for the point charge, both are divergent/do not exist.
Sep 9, 2022 at 8:56 history answered user330563 CC BY-SA 4.0