Suppose we have two moving particles, both with velocity $v$ (relative to the observer) and a distance of $r$ apart (perpendicular to the velocity of both particles). The charges are $q_1$ and $q_2$. The magnetic field from charge $1$ at the position of charge $2$ is then $$B=\frac{\mu_0}{4\pi}\frac{q_1v\times \hat{r}}{r^2}.$$
Moreover the force acting on charge $2$ by this magnetic field is $$F=q_2v\times B.$$
Since $v$, $\hat{r}$ and $B$ are all perpendicular we have that the magnitudes of $B$ and $F$ are $$B=\frac{\mu_0}{4\pi}\frac{q_1v}{r^2}$$ and $$F=q_2v\frac{\mu_0}{4\pi}\frac{q_1v}{r^2}=\frac{\mu_0}{4\pi}\frac{q_1q_2v^2}{r^2}.$$
However if we view the situation from the perspective of one of the charges, they are both at rest, there will be no magnetic field and only an electric field. The electric field of charge $1$ is now $$E=\frac{1}{4\pi \varepsilon_0}\frac{q_1}{r^2}.$$ The force on particle $2$ must now be $$F=q_2E=\frac{1}{4\pi \varepsilon_0}\frac{q_1q_2}{r^2}.$$ These two forces on charge $2$ must be equal and so $$\frac{1}{4\pi \varepsilon_0}\frac{q_1q_2}{r^2}=\frac{\mu_0}{4\pi}\frac{q_1q_2v^2}{r^2}$$ which implies $$v^2=\frac{1}{\varepsilon_0\mu_0}.$$ But this implies that $v=c$ since $$c^2=\frac{1}{\varepsilon_0\mu_0}.$$ Where is the fault in my reasoning? Do the distances, velocities or charges somehow change according to relativity? What would the correct equations then be?
