Capacitance of ferroelectric capacitor in saturated regime Consider a ferroelectric plate capacitor connected to an AC source in the presence of a strong static external electric field which sets the ferroelectric medium in the saturated regime.
The question:
Does the static polarisation $P$ of the ferroelectric have an influence on the capacitance $C$ of the capacitor?
Two contradicting attempts:

*

*Considering the electrical energy of the system, there is an influence:
The electrical energy $\mathcal{E}$ is given by $\delta\mathcal{E}=E\delta D$ with $E$ the electric field and $D$ the electric displacement which results in $\mathcal{E}=\epsilon_0E^2/2 + \epsilon_0EP$ for the saturated ferroelctric.  Electrical energy is also given by $\mathcal{E}=QV$, and thus the capacitance becomes
$$C=\frac{Q}{V} = \frac{\mathcal{E}}{V^2} = \frac{\epsilon_0E^2/2 + \epsilon_0EP}{V^2}\,.$$
Hence, according to this equation, there is an influence of the ferroelectric medium.


*Considering the classical capacitance equation, there is no influence of the ferroelectric medium: In the saturation regime of the ferroelectric, the relative permittivity \epsilon_r is 1. So, according to
$$C=\epsilon_0 \epsilon_r \frac{A}{d} = \epsilon_0 \frac{A}{d}$$
there is no influence of the ferroelectric medium...
Which approach is the correct one?
 A: Your second approach is correct.
There are a few problems with the first one. First of all, $Q=CV$ no longer applies, because when you have no charge on the capacitor, there is still a non-zero E-field within the ferroelectric, and thus a non-zero potential difference across the capacitor plates. The appropriate definition of the capacitance would be
$$C = \frac{dQ}{dV}.$$
Apart from the fact that you seem to be mixing up electrical energy and electrical energy density, the former is not $QV$, this is not even true in an ordinary dielectric capacitor ($\mathcal{E}=\frac{1}{2}QV$). In general, energy density must be found from the relation $$\delta u=E\delta D,$$ or energy from $$\delta\mathcal{E} = V\delta{q}.$$
With energy density $u$ taken to be zero when $E=0$,
$$u=\int\limits_0^EE'\frac{d}{dE'}(\epsilon_0E'+P)\ dE'=\frac{1}{2}\epsilon_0E^2$$
$$\mathcal{E}=Adu=\frac{1}{2}\epsilon_0AdE^2$$
$$V=Ed$$
$$C = \frac{dQ}{d\mathcal{E}}\frac{d\mathcal{E}}{dE}/\frac{dV}{dE}=\frac{1}{V}(\epsilon_0AdE)/(d)=\frac{\epsilon_0 A}{d}. $$
This is obviously not the most direct way to calculate the capacitance but it follows how your first approach attempts to do it.
