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

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.