# Is electric field, like gravitational field, an analog of acceleration?

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)?

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