= Electric field of a uniformly moving point charge
Let $\mathbf R$ join the instantaneous position of charge q to the field point, with constant $\boldsymbol\beta=\mathbf v/c$. Transforming its rest-frame Coulomb field gives
$$
\mathbf E=\frac{q(1-\beta^2)\mathbf R}{4\pi\epsilon_0[R^2-|\boldsymbol\beta\times\mathbf R|^2]^{3/2}},\qquad
\mathbf B=\mathbf v\times\mathbf E/c^2.
$$
The <electric field> is radial from the current position, reduced along the <velocity> and enhanced transversely. This present-position form is compatible with the causal retarded solution specifically because motion is uniform. There is no acceleration-dependent radiation field, although the moving charge has a <magnetic field> and transports electromagnetic <energy>.
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