What is the force between two charges moving at 0.5c relative to each other? [closed]

Question

Suppose two charges, $q_{1}$ and $q_{2}$ are in uniform translational motion relative to each other. If the relativity velocity is relativistic, e.g. $0.5c$, how can the force between the two charges be computed? This probably would be frame-dependent, so let's just say $q_{1}$ is in the laboratory frame and $q_{2}$ is in uniform translational motion at $0.5c$ relative to $q_{1}$. Would the force exerted on $q_{1}$, as measured in the laboratory reference frame, still be computed from the Lorentz force $\vec{F} = q(\vec{E} + \vec{v} \times \vec{B})$? I know that in this case $q_{2}$'s motion would induce a magnetic field, $\vec{B}$ in addition to the electric field, $\vec{E}$, but would these fields need to be corrected because of the relativistic speed of $0.5c$? If so, how? How would $\vec{E} = \frac{kq}{r^{2}} \vec{r}$ and $ \vec{B} = \frac{\mu_{0}}{4\pi} \frac{q\vec{v} \times \vec{r}}{r^{2}}$ change because of the relativistic speed $v$? Note that here I am using $\vec{r}$ to denote the radial vector directed from the origin outward and $r$ to denote the magnitude of $\vec{r}$.

Answer 1

An approach would be to calculate the electric field of $q_2$ in its own stationary frame of reference; then transform this to find the electric and magnetic fields present in the laboratory frame (here are the transformation equations); then work out what force $q_1$ experiences due to those transformed electric and magnetic fields. This force will depend on the relative displacement (not just the scalar separation) of $q_1$ and $q_2$.

If $q_1$ is initially at rest in the lab frame, the initial force is only due to the electric field from $q_2$; then once it is moving there will also be an additional force due to the magnetic field.