[Physics] Is the electric force a vector or a vector field

coulombs-lawelectric-fieldselectrostaticsVector Fieldsvectors

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 $\mathbf{F}_{12}=\mathbf{F}_{12}(\mathbf{r}_{12})$?

Best Answer

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 $\vec r_2$.