= Solution
Treat the film as a <parallel-plate capacitor>, initially compensated and then isolated electrically. Its <capacitance> and temperature-induced <electric polarization> charge are
$$
C=\frac{\epsilon_0\epsilon A}{d},\qquad\Delta Q=A p\Delta T.
$$
The corresponding ideal open-circuit signal, with the <electrode> polarity defined consistently with this charge convention, is
$$
\boxed{V_{\rm oc}=\frac{\Delta Q}{C}=\frac{p\,d\,\Delta T}{\epsilon_0\epsilon}}.
$$
Area cancels because both charge and <capacitance> are proportional to it. Substitution gives
$$
V_{\rm oc}=\frac{(-35\times10^{-6})(2\times10^{-6})(1)}{(8.854\times10^{-12})(8)}
=\boxed{-0.988\,\mathrm V}.
$$
The <voltage> magnitude is approximately \b[$0.99\,\mathrm V$]; interchanging the voltmeter leads reverses its sign. This is an ideal capacitive result, before leakage, mobile-charge screening or loading by the measurement circuit reduces the signal.
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