= Solution
Isolation fixes charge, not voltage: $Q=C_0V_0$. The <electrostatic energy> is therefore $U(t)=Q^2/[2C(\delta(t))]$. Inverting the small-tilt capacitance gives
$$
\boxed{U(t)=\frac{\epsilon_0s^2V_0^2}{2a}\left[1-\frac{\delta_0^2}{3a^2}\sin^2\omega_0t+O\left(\frac{\delta_0^4}{a^4}\right)\right].}
$$
The leading variation has frequency $2\omega_0$, since tilts of opposite sign give the same capacitance. With $U_0=C_0V_0^2/2$, the result is $U_0[1-\delta_0^2/(6a^2)+(\delta_0^2/(6a^2))\cos2\omega_0t]$ to this order. Energy decreases as the gap becomes less uniform at fixed charge; the difference is exchanged mechanically with the oscillator.
\Image[/past-exam-of-the-physics-2-paper-of-the-university-of-oxford/2013-tilted-capacitor.png]
{title=Capacitance increase and fixed-charge energy decrease under a small tilt}
{description=The exact local-strip capacitance is compared with its quadratic expansion. The isolated capacitor has fixed total charge and an energy modulation at twice the mechanical frequency.}
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