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I'm interested in determining the total gravitational and electric field of a charged particle. At reasonable distances the value of each field at a point is given by:

$$g = G\frac{m}{r^2}$$

$$E = \frac{q}{(4ε_0πr^2)}$$

The total of each field at every point is therefore just the integral of the relevant equation from 0 to infinity, but that obviously doesn't converge at $r = 0$. How do I calculate the total value for these fields?

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    $\begingroup$ What do you mean by "the total of each field at every point," and why do you think that should be a meaningful thing to talk about? $\endgroup$
    – J. Murray
    Commented Dec 16, 2022 at 17:57
  • $\begingroup$ The language might be a bit flowery, but I just meant the integral. I'm not sure how to address the meaningful nature of the question. Knowing the total gravitational field in principle gives you the total curvature of space imposed by that mass, while the total electric field relates similarly to the charge. $\endgroup$
    – WaveInPlace
    Commented Dec 16, 2022 at 18:11
  • $\begingroup$ It's not a matter of flowery language, I genuinely don't know what you mean by "the integral." Are you saying you want to compute $\int|\vec E| \mathrm d^3x$, integrated over all space? The reason I ask about the meaningfulness of the question is that simply adding up the magnitude of the electric field at different points doesn't correspond to anything physical or meaningful. The "total electric field" isn't, for example, the total energy contained in the electric field, which is proportional to the integral of $|\vec E|^2$ in electrostatics. $\endgroup$
    – J. Murray
    Commented Dec 17, 2022 at 18:28
  • $\begingroup$ I think we're on the same page here. Yes, I wish to sum the magnitude of the electric and gravitational fields at all points, from immediately adjacent to the particle to infinitely far away. In particular, I need to know how to integrate these fields at distances significantly less than 1 m, where the standard equations given in the question don't work (I can integrate the standard equations myself without any trouble). $\endgroup$
    – WaveInPlace
    Commented Dec 17, 2022 at 19:32
  • $\begingroup$ I still don't understand the question. If you integrate $1/r^2$ over all space, you obtain $\int \frac{1}{r^2} \mathrm d^3r = \int_0^\infty \frac{1}{r^2}4\pi r^2 \mathrm dr = 4\pi \int_0^\infty \mathrm dr \rightarrow \infty$. The integral doesn't have a finite value. You could impose a cutoff (only integrating out to a distance $R$, for example) to make it finite, of course. $\endgroup$
    – J. Murray
    Commented Dec 17, 2022 at 21:08

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Points are mathematical, not physical, entities. The physics answer is that you cannot do this. In any real physics problem, when there are convergence difficulties at small scales, there will be an inner scale, where a working mathematical formulation changes to allow convergence (and thus capture reality). Similarly, there may also be an outer scale.

Never take the math to be physically sensible automatically. A large part of physics is understanding the limits of your mathematical models.

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