= Solution
For <quartz>, $A=10^{-4}\,\mathrm{m^2}$ and $t=5.0\times10^{-4}\,\mathrm m$. With $g=9.81\,\mathrm{m\,s^{-2}}$, the load and <stress> are $F=4.905\,\mathrm N$ and $\Delta\sigma=4.905\times10^4\,\mathrm{Pa}$. Therefore
$$
|\Delta V|=\frac{(2.3\times10^{-12})(5.0\times10^{-4})(4.905\times10^4)}{(8.854\times10^{-12})(3.8)}=\boxed{1.68\,\mathrm V.}
$$
As 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 $Q/(C+C_{\rm in})$, while finite leakage makes it decay: it is not a guaranteed permanent DC output.
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