In other words if I have a static point mass and a static point charge, we model them as having a scalar potential field surrounding them: $$V(r)\propto\frac{1}{r}$$ and their vector fields are defined as the negative gradient of the potential: $$\vec{E}(r)=-\nabla V(r)\propto\frac{1}{r}$$ and they exert a force on other masses and charged particles: $$\vec{F}(r)\propto\frac{1}{r^2}$$ What I'm effectively asking is are these mathematically the same model just using M/G or Q/K depending on whether we're talking about gravitational or electric fields?

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    $\begingroup$ Yes, except Q can be negative. $\endgroup$ Commented Nov 24, 2019 at 0:42
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    $\begingroup$ And that is a big difference. Also, like gravitational charges attract $\endgroup$
    – Dale
    Commented Nov 24, 2019 at 0:49
  • $\begingroup$ "Mathematically equivalent" usually means one can be derived from another, thus the two are essentially the same thing. What do you mean by mathematically equivalent here? and how can $M/G$ be the same as $Q/K$ ? I don't see how your question is meaningfully posted $\endgroup$
    – Graviton
    Commented Nov 24, 2019 at 1:07
  • $\begingroup$ My use of the phrase "mathematically equivalent" might be wrong, I'm saying that the way the electric and gravitational fields are described for point charges and point masses are the same. They both are surrounded by a scalar field and a vector field and the two equations for potential V(r) behave in the exact same way, the only difference being that we have labelled one with "charges" and the other with "masses" and require that masses be positive. I hope that makes sense. $\endgroup$
    – Charlie
    Commented Nov 24, 2019 at 1:18
  • $\begingroup$ Also where I have said M/G I don't mean the fraction, I mean M and G as in F=GMm/r^2 and F=kQq/r^2 $\endgroup$
    – Charlie
    Commented Nov 24, 2019 at 1:21

2 Answers 2


They both have the same inverse square law, but the difference is electric charges can be positive and negative, so the force can be attractive or repulsive.

Mass can only be positive, and two masses attract whereas two positive charges repel.

Electric fields can also be circular if there are changing magnetic fields around. That does not happen with gravity.

Enjoy the similarity, but keep them separate in your head!


It's not immediately obvious that your question is posed in a meaningful way, because you assume that quantities like force and energy have meaning in a completely electrostatic classical universe, or in a classical universe where the only interactions are purely static gravitational ones. In such a universe you can't build a spring scale or a calorimeter, for example. Such a universe can't even self-consistently exist, as a special case of a dynamical universe, due to Earnshaw's theorem.

As a concrete example of what goes wrong here, the energy density of the gravitational field in Newtonian gravity is $-g^2$ (ignoring a positive constant of proportionality), while the energy density of the electric field is $E^2$. This is a huge difference, and yet we can't measure this difference in a universe where there are no energy-measuring devices.


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