Past exam of the physics 2 paper of the University of Oxford / 2013 / 11

Ignoring edge fields, Gauss's law gives $E=\sigma /{ϵ}_0$ between the parallel plates. Their voltage is $V=Ea=\sigma a/{ϵ}_0$, and their charge magnitude is $Q=\sigma s^2$. Therefore

Take $x\in [-s/2,s/2]$ along the tilt, with local separation $h(x)=a+2\delta x/s$. Since $s\gg a$ and the tilt is small, each narrow strip is a local parallel-plate capacitor at the same voltage $V$. Its surface charge densities are approximatelyParallel strip capacitances add, so for $|\delta |<a$,Using $\log [(1+z)/(1-z)]=2(z+z^3/3+z^5/5+\cdots )$ givesThe denser charge at the smaller gap explains why tilting increases the capacitance. The local-strip approximation neglects fringing and slope corrections of order $(\delta /s{)}^2$, smaller than the retained $(\delta /a{)}^2$ term.

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.