According to Feynman (Lectures on Physics, Vol. II, 21–6), the potential of a moving point charge is:
$$\phi = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ \gamma^2 \left( x-vt \right)^2 + y^2 + z^2 }}$$
I tried the following relativistic argument to derive it:
Call $S'$ the frame of reference of the particle, and call $S$ our frame. In $S'$, the particle is at rest, and in $S$, it moves toward positive $x$ with velocity $v$.
Now, in $S'$ I have the static potential: $$\begin{align}\phi' & = \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ {x'}^2 + {y'}^2 + {z'}^2 }} \\ \mathbf {A'} & = \mathbf 0 \end{align}$$ Then I make the Lorentz transformations in $\left(t;x,y,z\right)$ and in $\left(\phi ; \mathbf A\right)$ to get them at $S$: $$\begin{align} x & =\gamma\left(x'+vt\right) \\ t & =\gamma\left(v/c^2\cdot x'+t'\right) \\ \phi & = \gamma\phi' \\ \mathbf A & = \gamma \mathbf v \phi' \end{align}$$
Finally, substituting the potential $\phi'$: $$\begin{align}\phi & = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ {x'}^2 + {y'}^2 + {z'}^2 }} \\ & = \gamma \cdot \frac q {4\pi\epsilon_0} \cdot \frac 1 {\sqrt{ \gamma^2 \left( x-vt \right)^2 + y^2 + z^2 }} \end{align}$$
According to the formula cited above, I arrived at the correct result. But I noticed that I didn't include the transformation of the $\left(\rho;\mathbf j\right)$ four-vector (that is, $\left(\mbox{charge density};\mbox{current density}\right)$). If I included it, I would make this extra substitution in the formula for the potential: $$q' = q/\gamma$$ where $q'$ is the particle's proper charge. Then I would find an extra $\gamma$ in the formula for the field of a moving particle. Since there is no such extra $\gamma$ factor (according to Feynman's formula), there must be something wrong in my $q = \gamma q'$ argument. Where did I misunderstand the Lorentz transformation of four-vectors?