= Solution
Let $\mathbf R$ run from the charge's present position to the observation point, and let $\boldsymbol\beta=\mathbf v/c$ be constant. Lorentz-transforming the Coulomb field gives
$$
\boxed{\mathbf E(\mathbf R,t)=\frac{q}{4\pi\epsilon_0}
\frac{(1-\beta^2)\mathbf R}{[R^2-|\boldsymbol\beta\times\mathbf R|^2]^{3/2}},\qquad
\mathbf B=\frac{\mathbf v\times\mathbf E}{c^2}.}
$$
At fixed R, the radial magnitude is proportional to $(1-\beta^2)/(1-\beta^2\sin^2\theta)^{3/2}$. It is reduced by $\gamma^{-2}$ along the motion and enhanced by $\gamma$ perpendicular to it. The field is therefore concentrated around the transverse plane at high <speed>. For negative charge all electric-field arrows reverse.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2013/paper-1-oe2013-moving-charge.png]
{title=Original angular field-strength comparison for a uniformly moving charge, measured from its instantaneous position, with the direction of motion marked}
Although the fields depend causally on retarded motion, for constant <velocity> the electric-field direction points along the separation from the present position. This does not imply instantaneous action: the velocity-dependent part of the retarded field removes a naive directional lag. Both fields fall as $R^{-2}$; \b[uniform straight-line motion produces no acceleration-radiation term proportional to $R^{-1}$].
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