# What is the relation between charge and polarization?

On one hand, in this "Measurement of polarisation" lab manual, polarization is surface charge density, $P=\frac{Q}{A}$, in other words, charge an polarization are essentially the same thing.

On the other hand, in this "Ferroelectric Hysteresis Measurement & Analysis" report, the authors say:

Charge and Polarisation are strictly speaking different but for materials with a high relative permittivity we can assume they are equal.

So if "charge", Q, and "polarization", P, are different (by not just the area involved, A), what is the difference? And how the value of the permittivity is related?

EDIT:

According to Wikipedia, the related equation is for charge displacement field, $\mathbf{D} \equiv \varepsilon_{0} \mathbf{E} + \mathbf{P}$, from which the polarization can be extracted: $\mathbf{P} = \mathbf{D} - \varepsilon_{0} \mathbf{E}$.

However, according to Hall et al., $\mathbf{P}\simeq \mathbf{D} = \varepsilon_0 \varepsilon_r^* \mathbf{E}$, which seems contrary to the previous equation.

Anyway, how the charge Q and the relative permittivity $\varepsilon_r$ of the sample are related and what is the correct equation?

The relation $P=Q/A$ only holds for very special circumstances: when the body is overall charge neutral and the polarization is uniform within the material and points (polarization is a vector, really) perpendicular to the surface. Then the charge in the vicinity of the surface is given by that formula. It's the first condition (charge neutrality) that is relevant to the question about $\epsilon_{r}$ (better known as the dielectric constant). If a material has nonzero net charge, that charge will also need to be accounted for. However, if the dielectric constant is large, the polarization is large, and most of the charge in a given region can be attributed to the effects of polarization. How large, exactly, the dielectric constant must be depends on the actual values of $Q_{TOT}$ and $P$.

Charge and polarisation have different units, they are not the same.

Polarisation is defined as dipole moment per unit volume.

The relationship between polarisation charge on a surface and the polarisation field is

$$\oint \vec{P} \cdot d\vec{A} = -Q_p$$

In linear, isotropic and homogeneous media, we can also say $$\vec{P} = \chi_e \epsilon_0 \vec{E} = (\epsilon_r - 1)\epsilon_0 \vec{E}$$ $$\vec{P} = \vec{D} - \epsilon_0\vec{E},$$ where $\chi_e$ is the electric susceptibility.

If $\epsilon_r \gg 1$, then $$\vec{P} \simeq \epsilon_r \epsilon_0 \vec{E} = \vec{D}$$

• so for a parallel plate capacitor the closed integral reduces to PA=Q? Commented Dec 24, 2015 at 17:18
• @Sparkler yes, you just have to watch that you get the sign of the charge right. Commented Dec 24, 2015 at 17:22