= Cyclotron gap energy gain
{title2=$\Delta K=q\Delta V$}
Define $\Delta V=\Phi_{\mathrm{entry}}-\Phi_{\mathrm{exit}}$ as the <electric potential> drop along the particle's passage. If the <electric field> changes negligibly during that short crossing, the <work> gives
$$
\Delta K=q\Delta V.
$$
The sign matters: a positive <ion> gains <kinetic energy> while going to lower <electric potential>; a negative <ion> needs the opposite drop. A fixed-sign field would accelerate on one crossing and decelerate on the next. The alternating field must reverse before the particle comes back.
For a fixed <electric charge> sign and circulation direction, absorb this alternating crossing direction into the definition of the arrival <phase> $\phi_n$. If $V_0$ is the peak <electric potential difference> across the whole gap, rather than the amplitude of either electrode relative to ground, a thin-gap model is
$$
K_{n+1}=K_n+|q|V_0\cos\phi_n.
$$
Arrival at $\phi_n=0$ gives the largest accelerating kick; $|\phi_n|<\pi/2$ gives a positive kick. Arrival at $\phi_n=\pi$ gives deceleration. For an ideal classical particle initially on crest, the <cyclotron resonance condition> keeps successive arrival <phases> equal. It does not automatically bring an initially mistimed particle onto crest.
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