Let us say we have a uniform electric field, like between two charged plates separated by a distance $d$.
The formula for the voltage between the plates is $\Delta V=Ed$.
But what is the value of this uniform field $E$? When I look for the formula, I see $E=\Delta V/d$, but this circular reference doesn't help. Well, when the field is not uniform, the formula is $E=kQ/r^2$, but what is $r$ here?
With gravity, it is easier. For example, one measures that the gravitational field close to the surface of the Earth (ultimately acceleration) is $g$ = $GM_{Earth}/R^2$ and you assume that this is not going to change significantly if you add to that whatever height, which is negligible compared to the radius of the Earth $R$. Then the gravitational potential difference, sometimes called "liftage", is the same thing over an additional distance $h$, i.e. $\Delta V=gh$. In conclusion, liftage is proportional to the constant $G$, the source mass $M_{Earth}$ and the height $h$ and inversely proportional to $R^2$.
My problem with the electric potential is that I only see the additional distance $d$, not the original distance $r$. So what is electric potential or voltage dependent on, is it only proportional to the constant $k$, the source charge $Q$ and the distance $d$?