= Solution
For <electrode> area $A$, the change in spontaneous <polarization density> supplies a charge scale $\Delta Q=pA\Delta T$, while the film's <capacitance> is $C=\varepsilon A/d$. The voltage magnitude is therefore $|\Delta V|=|p|d|\Delta T|/\varepsilon$. More explicitly, for an initially compensated film with no change in free <electrode> charge, $\Delta D=\varepsilon\Delta E+p\Delta T=0$. Taking $z$ from the lower <electrode> to the upper one gives $\Delta V_{\rm upper-lower}=-d\Delta E=pd\Delta T/\varepsilon$.
$$
\Delta V_{\rm upper-lower}
=\frac{(-30\times10^{-6})(10^{-6})(1)}{10\varepsilon_0}
=\boxed{-0.339\ \mathrm V}.
$$
The measurable magnitude is \b[$0.339\ \mathrm V$], with polarity reversed by reversing the poling direction or the <temperature> change. This is an ideal open-circuit estimate: leakage, the measuring instrument's input <capacitance> and finite charge-relaxation time can reduce or reshape the observed voltage.
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