Can we use Coulomb's law to calculate the force between two charges which are not at rest? If not, what formula should be used to calculate the force? I searched it, but I couldn't find a clear answer.


Coulomb's law is not precisely true when charges are moving-the electrical forces depend also on the motions of the charges in a complicated way. One part of the force between moving charges we call the magnetic force. It is really one aspect of an electrical effect. That is why we call the subject "electromagnetism."

There is an important general principle that makes it possible to treat electromagnetic forces in a relatively simple way. We find, from experiment, that the force that acts on a particular charge-no matter how many charges there are or how they are moving-depends only on the position of that particular charge, on the velocity of the charge, and on the amount of the charge. We can write the force $\mathbf{F}$ on a charge $q$ moving with a velocity $\mathbf{v}$ as $$\mathbf{F}=q\left(\mathbf{E}+\mathbf{v}\times \mathbf{B}\right)\,.$$
We call $\mathbf{E}$ the electric field and $\mathbf{B}$ the magnetic field at the location of the charge.1

Credits: 1Feynman lectures on Physics-II.

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    $\begingroup$ You can search for the PDF version of Feynman's lectures on Physics. You will get good answers for your question in depth. $\endgroup$
    – Sensebe
    Jul 14 '14 at 22:04
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    $\begingroup$ In the frame where the source charge is stationary, there is no $B$ field, so shouldn't Coulomb's Law be ok? If the source charge is moving, then I can see that there will be a magnetic field, plus one needs to account for the finite speed of the transmission of the field (speed of light). In small systems, which means almost everything that we see on a day to day basis, the retardation can be ignored. $\endgroup$
    – garyp
    Jul 14 '14 at 22:30
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    $\begingroup$ After reflection ... my comment above only applies if there exists an inertial frame in which the source is at rest. If not the answer becomes complicated. The source will radiate, and we have near-field and far-field ... so the short answer to the OP is "it depends". $\endgroup$
    – garyp
    Jul 14 '14 at 22:40
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    $\begingroup$ So the magnetic field that would be generated by a moving charge with ,for example, 2 m/s velocity is negligible. Right? $\endgroup$
    – Starior
    Jul 15 '14 at 3:31

Suppose in some reference frame S, we have two stationary charged particles, $q_1$ and $q_2$. The force experienced by the first particle due to the second is given by

$$\textbf{F}_{12} = k \frac{q_1 q_2}{r^2_{12}} \hat{r}_{12}\,,$$

where $k$ is Coulomb's constant, $q_i$ is the charge of the respective particle, $r_{12}$ is the distance between the two particles, and $\hat{r}_{12}$ is the unit radial vector point from particle 2 to particle 1.

As is commonly done in most electromagnetism textbooks, it then makes sense to introduce the electric field vector, defined as the "force per unit charge." In the same frame S, the stationary particle (at position $\textbf{r} = \textbf{0}$) with charge $q$ creates the electric field

$$\textbf{E}(\textbf{r}) = k \frac{q}{r^2} \hat{r} \, ,$$

where $r = ||\textbf{r}||$ and $\hat{r} = \frac{\textbf{r}}{||\textbf{r}||}$.

With a bit of magic (just kidding, all math), it is possible to show that this statement is equivalent to Gauss's law:

$$\oint_\Sigma \textbf{E} \cdot d\textbf{A} = \frac{Q_{\text{enc}}}{\epsilon_0} \, .$$

It is understood that the field is integrated over a closed surface $\Sigma$ and that $Q_{\text{enc}}$ is the amount of charge located within said surface.

As it turns out, the first so-called "definition" of the electric field is only true for the frame(s) where the particle is stationary. For other frames, it is necessary to apply the standard transformations of special relativity. The most convenient approach is to recognize that charge is an invariant quantity (i.e., it does not change under a Lorentz transform). This implies that Gauss's law is true for all reference frames. The only quantities that transform are the electric field and the surface of integration:

$$\oint_{\Sigma'} \textbf{E}' \cdot d\textbf{A} = \oint_\Sigma \textbf{E} \cdot d\textbf{A}\,.$$

Thanks again to special relativity, we can show that a moving particle must generate a magnetic field in some frames of reference. The final expressions for the general electric and magnetic fields for a frame of reference where the particle is moving with an arbitrary velocity and acceleration are complicated, but nevertheless can be calculated.

Applying these fields along with the Lorentz force law,

$$\textbf{F} = q(\textbf{E} + \textbf{v} \times \textbf{B})\,,$$

we can find the trajectories of both particles. However, these trajectories are quite complicated, and need to be simulated on a computer.

TL;DR - No, in a sense Coulomb's law cannot be applied to the motion of two interacting charged particles.


As many people have already said in their answer, if you ask yourself what is the force between any two charged particles which have some velocities at some time in the frame of reference of the lab, then what you find is the Lorentz force: $\mathbf{F} = q(\mathbf{E} + \mathbf{v} \times \mathbf{B})$.

Now, the problem is when you try to resolve the dynamics of such a system because it turns out that the Lorentz force is not reciprocal that is to say, it doesn't satisfy Newton's third law ($\mathbf{F}_{12} \neq - \mathbf{F}_{21}$).

This is very well explained here (with animations and all).

This was already figured out by Poincare and it puzzled him for a while until he realized that the reason for this violation was because we were not accounting for variations of the momentum-energy tensor of the electromagnetic field itself.

Hence, if you want to solve for real this dynamical problem, this is a very complex one where the electromagnetic field needs to be resolved at the same time as the positions and momenta of your two particles.

Now, it turns out that if you are only interested in statistical properties of an (classical) assembly of charges (at thermodynamic equilibrium), then one can show that it is sufficient to consider only the Coulomb interaction between charges; so in a way, depending on what you want to do, yes Coulomb can be enough otherwise it becomes very complicated.


Lorentz's force is acting on the charge $$F = q(E+v\times B)$$

If the charge is moving in an uniform electric field $E$, there will be no $B$ and the force is $F = qE$. In the case of a non-uniform electric field (e.g. a point charge), the electric field at the charge will change in time and thus, by the Ampere's law, a $B$ will be induced. But usually (in non-relativistic cases) the induced $B$ will be negligible.

  • $\begingroup$ Risking a display of ignorance: Is it not true that in the case of a stationary point charge, there is no magnetic field, period. So Coulomb's Law is valid in that frame. In the frame that moves with the test particle, there is a magnetic field, so Coulomb's Law needs help. So isn't the answer the same as the the Benchley's reply to Dorothy Parker's quip about girls and glasses: "It depends on the frame." $\endgroup$
    – garyp
    Jul 15 '14 at 0:44

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