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The electric force (Coulomb's law) on a point charge $Q_2$ due to $Q_2$: \begin{gather*} \mathbf{F}_{12}=\frac{Q_1Q_2}{4\pi\epsilon_0}\frac{\mathbf{r}_2-\mathbf{r}_1}{|\mathbf{r}_2-\mathbf{r}_1|^3} \end{gather*}

The seperation vector between the charges is $\mathbf{r}_{12}=\mathbf{r}_2-\mathbf{r}_1=(x_2-x_1)\hat{\textbf{i}}+(y_2-y_1)\hat{\textbf{j}}=(z_2-z_1)\hat{\textbf{k}} $.

Does this mean that the force is a vector field or just a vector?

Is the force a function of $\mathbf{r}_{12}$, so $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?

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The electric force is just a vector, not a vector field.

The Coulomb force formula you wrote depends on "some" position vectors $\vec r_1,\vec r_2$, much like fields $\Phi(\vec r)$ depend on the position vector $\vec r$, but it's a different kind of a dependence.

First, a field must depend on a single position in space, $\vec r$, while the Coulomb force depends on two.

Second, different values of the vector $\vec r$ that a field $\Phi(\vec r)$ depends upon must be equally "true" and equally "exist" at the same moment. So both $\Phi(\vec r_\textrm{Boston},t)$ and $\Phi(\vec r_\textrm{Paris},t)$ exist at a given moment $t$. On the other hand, the Coulomb force is a force between two particular objects that sit at particular places $\vec r_1,\vec r_2$, so one choice of values of $\vec r_1,\vec r_2$ is "right" while all others are "wrong". The electric force you mention only depends on $t$ because the locations $\vec r_1,\vec r_2$ of the two charged objects are functions of $t$ themselves.

So the dependence of the force is $\vec F(\vec r_1(t),\vec r_2(t))$. Because time $t$ is the only independent variable, there are no independent spatial variables and we don't talk about a field.

Above, I assumed that we talk about "particular" charged objects. That was justified because you wrote "particular" charges $Q_1,Q_2$ which indicate that we talk about well-defined objects. If you omitted $Q_2$, you would get a formula for the electric field $\vec E(\vec r_2)$ which is a field. It's the electric force (from the $Q_1$ at $\vec r_1$ source) per unit charge that you would get if you placed the charge at the point $\vec r_2$.

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  • $\begingroup$ I’m not sure I fully understand. If we fix one of the two particles $j$ than the force only depends on the other $\textbf{r}_i$; if we choose to fix neither, than the displacement $\textbf{r}$ can be chosen at hand as variable, can it not? Dividing by $Q_2$ we certainly get a field, but that would conceptually be equivalent to fix charge and position of one of the two on the right hand side of the force and only consider the remaining as variable, or am I mistaken? $\endgroup$ – gented Jul 8 '16 at 10:29
  • $\begingroup$ Dear Gennaro, whether something is a field isn't given just by its "dependence" on some variables, as I argued, so you are just repeating the same misc0nception and completely ignoring everything I wrote. A field is something that objectively exists at each point of space. The transition from mechanics to field theory was a transition in physics in which the number of dynamical variables increased from a "few" at each moment, $\vec x(t)$, to "infinitely many" at each moment, $\Phi(\vec r,t)$. There's some information at each point of space. $\endgroup$ – Luboš Motl Jul 8 '16 at 10:37
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    $\begingroup$ Now, the force $\vec F$ is a concept from mechanics, by $\vec F = m \vec a$, it determines acceleration of objects. It is not a concept that requires field theory, so it's always wrong to say that a particular force is a "field". Saying that a force is a "field" is exactly as wrong as saying that the position of the Earth $\vec r_{\rm Earth}$ is a field. It also depends on some position vector - it's equal to that vector - but that's not enough for something to be a "field". A field is a collection of infinitely many variables that may have independent values from each other at a moment $t$. $\endgroup$ – Luboš Motl Jul 8 '16 at 10:39
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    $\begingroup$ A field is something that assigns a value or values to each point in space and time. So it has to be a collection of more or less independent numbers $\Phi(x,y,z,t)$ - so that the value of $\Phi$ for one value of $x,y,z,t$ may be chosen independently from the values of the fields at all other points. That's clearly not satisfied by the force between two particular charges. $\endgroup$ – Luboš Motl Jul 8 '16 at 10:47
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    $\begingroup$ @LubošMotl Well, you could try using standard terminology, for example; I have never encountered the word objectively in physics. Moreover "the latter is a function of one variable $t$, and is therefore not a field, while the former is a function of 4 spacetime variables, and is a field": would this imply that only functions depending on more than one variable can be considered fields? Either you want to express that we want $x$ not to depend on $t$ but be independent by itself (I agree, but you're using way overterminology for it) or I still don't see it. $\endgroup$ – gented Jul 8 '16 at 12:50
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If an electric force was acting on charge that was spread out continuously over a region, then then each point in that region would have an force vector associated with it, and so in that case there would be a vector function of coordinates describing the forces involved. (There are some issues with each point would have an infinitesimal amount of charge, but I think that can be left for later.)

The the case of a point charge in an electric field, there is really just one point in space of interest, the point where that charge is located. I guess in a trivial way, you could call the forces a vector field that was zero almost everywhere, except where the charge was located.

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