Past exam of the materials science course of the University of Cambridge / 2017 / Paper 1 / 2 / c

For quartz, $A=10^{-4}{\mathrm{m}}^2$ and $t=5.0\times 10^{-4}\mathrm{m}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the load and stress are $F=4.905\mathrm{N}$ and $\Delta \sigma =4.905\times 10^4\mathrm{Pa}$. ThereforeAs a check, $Q=dF=11.28\mathrm{pC}$ and $C=6.73\mathrm{pF}$ give the same result. This is the ideal maximum open-circuit change on applying the load. Input capacitance reduces it to $Q/(C+C_{\mathrm{in}})$, while finite leakage makes it decay: it is not a guaranteed permanent DC output.