= Solution
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 approximately
$$
\boxed{\sigma_\pm(x)=\pm\frac{\epsilon_0V}{a+2\delta x/s}.}
$$
Parallel strip capacitances add, so for $|\delta|<a$,
$$
C(\delta)=\epsilon_0s\int_{-s/2}^{s/2}\frac{dx}{a+2\delta x/s}
=\frac{\epsilon_0s^2}{2\delta}\log\frac{a+\delta}{a-\delta}.
$$
Using $\log[(1+z)/(1-z)]=2(z+z^3/3+z^5/5+\cdots)$ gives
$$
\boxed{C(\delta)=\frac{\epsilon_0s^2}{a}\left[1+\frac{\delta^2}{3a^2}+O\left(\frac{\delta^4}{a^4}\right)\right].}
$$
The 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.
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