Is Coulomb's law accurate for moving charges?
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.
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.