= Solution
After <field electron emission>, let $h$ denote the plate separation of the <parallel-plate capacitor>, which the question does not specify. The uniform capacitor <electric field> has magnitude $E_c=|V|/h$, and a nonrelativistic <Electron> of <mass> $m_e$ has constant <acceleration> magnitude $a=eE_c/m_e$. The sign of its charge affects the direction of the radiation field but not its radiated <power>.
In the supplied nonrelativistic <Liénard-Wiechert field>, $\dot{\boldsymbol\beta}=\mathbf a/c$. At polar angle $\theta$ to the <acceleration>, $|\widehat{\mathbf n}\times(\widehat{\mathbf n}\times\mathbf a)|=a\sin\theta$, hence
$$
E_{\rm rad}=\frac{ea\sin\theta}{4\pi\epsilon_0c^2R}.
$$
The outgoing <Poynting vector> has magnitude $S=\epsilon_0cE_{\rm rad}^2$. Its radiated <power> per <solid angle> is therefore
$$
\frac{dP}{d\Omega}=R^2S=\frac{e^2a^2}{16\pi^2\epsilon_0c^3}\sin^2\theta.
$$
Using $\int\sin^2\theta\,d\Omega=8\pi/3$ gives the <Larmor formula> and the requested constant <power> during the uniform-field part of the flight:
$$
\boxed{P=\frac{e^2a^2}{6\pi\epsilon_0c^3}
=\frac{e^4E_c^2}{6\pi\epsilon_0m_e^2c^3}
=\frac{e^4V^2}{6\pi\epsilon_0m_e^2c^3h^2}.}
$$
The brackets in the given field require the <acceleration> at <retarded time>; the radiation reaches the far observer after the propagation delay. This derivation assumes that <radiation reaction> is a small correction to the <acceleration> and that the flight is nonrelativistic, for example $e|V|\ll m_ec^2$ for release from rest. Entry, emission and impact transients are not described by the uniform-field model. A result solely in terms of $V$ cannot be obtained without a plate spacing or a specified internal field.
Back to article page
