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By Newton's laws, the acceleration of an object depends on the force acting on it and its mass by $$\frac{\vec{F}}{m} = \vec{a}$$ and the gravitational acceleration is defined as $$\frac{\vec{F}_{\text{grav}}}{m} = \vec{g}$$ so that the gravitational field can be interpreted as the acceleration of a massive particle.

I also learned that an electric field can be defined by the force (from the field) acting on a charge: $$\frac{\vec{F}_{\text{elec}}}{q} = \vec{E}$$ This equation looks similar; can electric field can be thought of as the "acceleration of charge" acting on point charges (only, because the field ignores neutral particles)?

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Kind of! In the case of gravity, where $F \propto m$, the quantity $F/m$ is constant, so all objects fall with the same acceleration. This is a hint that gravity is really a geometric phenomenon, as shown in general relativity.

However, in the case of electromagnetism, we instead have $F \propto q$, so the acceleration is instead proportional to the charge to mass ratio $q/m$. Since this quantity is different for different particles, it precludes a geometric description of electromagnetism.

If you only consider particles with a constant charge to mass ratio, you can indeed interpret electric fields geometrically as generating accelerations, and magnetic fields as generating rotations, as I explain here.

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  • $\begingroup$ great, thanks for the distinction of $q/m$ $\endgroup$ – khaverim Jan 27 '17 at 2:21

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