= Solution
An accelerating <point charge> has a transverse <radiation field> in addition to its near field. At nonrelativistic <speed> and large distance $R$,
$$
\mathbf E_{\rm rad}(\mathbf r,t)=\frac{q}{4\pi\epsilon_0c^2R}
\widehat{\mathbf n}\times[\widehat{\mathbf n}\times\mathbf a(t-R/c)],\qquad
\mathbf B_{\rm rad}=\frac1c\widehat{\mathbf n}\times\mathbf E_{\rm rad}.
$$
The <acceleration> is evaluated at <retarded time>. <Radiation fields> decay as $1/R$, so their outward <energy> <flux> integrated over a <sphere> is independent of <radius>; the near <electric field> instead decays as $1/R^2$. The radiation vanishes along the <acceleration> and is strongest perpendicular to it.<The Feynman Lectures on Physics, Volume I by Richard Feynman and others>[§28–2, Eqs. (28.5)–(28.6): transverse acceleration field and inverse-distance radiation]{ref}
Using the <Poynting vector> gives
$$
\frac{dP}{d\Omega}=\frac{q^2a^2\sin^2\vartheta}{16\pi^2\epsilon_0c^3},\qquad
\boxed{P=\frac{q^2a^2}{6\pi\epsilon_0c^3}.}
$$
The integral uses $\int\sin^2\vartheta\,d\Omega=8\pi/3$. This is the <Larmor formula>. A harmonically accelerated charge with peak <acceleration> $a_0$ has mean <power> $q^2a_0^2/(12\pi\epsilon_0c^3)$; external work supplies this radiated <energy> in steady operation. Uniform inertial motion does not radiate.
\Image[/past-exam-of-the-optics-and-electrodynamics-course-of-the-university-of-cambridge/2012/paper-1-oe2012-charge-radiation.png]
{title=Original nonrelativistic radiation pattern, with zero power along the acceleration and maximum power transverse to it}
At relativistic <speed> the angular pattern becomes directionally concentrated and retarded-source factors matter. The total instantaneous radiated <power> generalizes to $q^2\gamma^6[a^2-|\boldsymbol\beta\times\mathbf a|^2]/(6\pi\epsilon_0c^3)$, with $\boldsymbol\beta=\mathbf v/c$. The displayed simple $\sin^2\vartheta$ pattern is restricted to the <nonrelativistic limit>.
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