= Electrostatic acceleration energy gain
{title2=$K_f-K_i=q(\Phi_i-\Phi_f)$}
For a fixed <electric charge> $q$, a time-independent <electric potential> $\Phi$ and negligible collisions or radiation losses, conservation of <energy> gives
$$
K_f-K_i=-\Delta(q\Phi)=q(\Phi_i-\Phi_f)=qV.
$$
Here $V$ is the signed potential drop along the trajectory. A <proton> starting nearly at rest at $+V$ and arriving at ground gains $eV$: a chosen $500\,\mathrm{kV}$ gives $500\,\mathrm{keV}$, by the definition of the <electronvolt>. Several intermediate electrodes can distribute that drop along a tube, but the gains telescope to the same endpoint difference. Passing through more electrodes does not multiply the available <kinetic energy> beyond $qV$.
This is the distinction from a <cyclotron>. In a static <electric field>, a closed trip gives zero net <work>. A <cyclotron> changes the gap <electric field> between successive visits, so correctly timed particles repeatedly gain <kinetic energy>. Its limit involves the <cyclotron resonance condition> and <relativistic detuning of a cyclotron>; a single-pass <electrostatic accelerator> instead requires the full accelerating <electric potential difference> to be sustained at once. Neither machine's beam energy by itself specifies the <energy> released by a <nuclear reaction>.
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