Because of their spontaneous polarization, ferroelectrics exhibit hysteresis properties: their response to an external field (the dependence of the polarization $P$ vs. the external electric field $E$) depends on their previous state.

Consider a ferroelectric material that had polarization $P_1$ in an external electric field $E_1$ at the initial time. The field was then increased to $E_2$, and the polarization increased to $P_2$, following the law $P_-(E)$. Then the field was again reduced to $E_1$, and the polarization of the ferroelectric returned to its initial state following the law $P_+(E)$.

A1 Derive an expression for the elementary work $\delta A$ that the external field has to do to change the polarization of a unit volume of ferroelectric by $\mathrm dP$.

A2 Derive an equation relating the heat $\delta Q$ released inside the ferroelectric, the work $\delta A$ done by external field and the change $\mathrm dU$ in ferroelectric internal energy.

A3 Express as some proper integral the total work $A$ that the external field has to do to a unit volume of the material during the whole cycle of field's increase and decrease. By how much $\Delta U$ does the internal energy density changes during the cycle? How much heat $Q$ is released in a unit volume?

Now consider a hysteresis loop of one of the most widely used ferroelectric materials, barium titanate (BTO). The loop is plotted on a figure below.

On the horizontal axis is the external electric field strength in $kV/cm$, on the vertical axis is the polarization of the segnetoelectric in $μC/cm^2$.

Suppose a large $(>300~V)$ negative voltage is applied to a thin BTO film of thickness $d=10~µm$ and area $S=1~cm^2$. Then it's changed to a large positive voltage and finally returned back.

B1 Calculate what amount of heat $Q$ was released in the film.

Now suppose an alternating voltage with effective voltage $V_\mathrm{eff}=220~V$ and frequency $\nu=50~Hz$ is applied to the film.

B2 Calculate the average thermal power $\dot Q$ released in the film.