 Solution

Isolation fixes charge, not voltage: $Q=C_0{V}_0$. The electrostatic energy is therefore $U(t)=Q^2/[2C(\delta (t))]$. Inverting the small-tilt capacitance givesThe leading variation has frequency $2{\omega }_0$, since tilts of opposite sign give the same capacitance. With $U_0=C_0{V}_0^2/2$, the result is $U_0[1-{\delta }_0^2/(6a^2)+({\delta }_0^2/(6a^2))\cos 2{\omega }_{0}t]$ to this order. Energy decreases as the gap becomes less uniform at fixed charge; the difference is exchanged mechanically with the oscillator.