PX284 - Q1a - polarization

considering an atom: a positively charged nucleus ( $+ q$ ) surrounded by a negatively charged cloud $(- q)$
if an electric field is applied, $+ q$ and $- q$ shift in opposite directions, by a distance, $s$ , ie. a dipole: $\vec{P} = q \vec{s}$
now, considering in a larger scale, with many atoms
defining polarization as the net dipole moment per unit volume:

\vec{P} = \frac{\sum_{i} {\vec{P}}_{i}}{V}

considering a polarizable material
upon exposure to an electric field, it will have an excess of positive charges in one end, and an excess of negative in the other
the surface charge density:

σ_{P} = \frac{Q}{A} = \frac{(q n) \times (A s)}{A}

where, $q$ is the charge of an atom, $n$ is the number density, $A$ is the area, and $s$ is the displacement, so $A s$ is the volume

⟹ σ_{P} = q n s = | \vec{P} |

in general:

σ_{P} = \vec{P} \cdot \hat{n}

where, $\hat{n}$ is the outward facing unit normal to the surface

now considering a non-uniform polarization
two small cuboids centred at $x_{0}$ , separated by $δ x$

Q (x_{0}) = A (P_{x} (x_{0} - \frac{δ x}{2}) - P_{x} (x_{0} + \frac{δ x}{2}))

using taylor's theorem:

f (x) ≃ f (x_{0}) + δ x \frac{d f}{d x} |_{x = 0} + \dots

Q (x_{0}) ≃ - A δ_{x} \frac{\partial P_{x}}{\partial x} |_{x_{0}}

volume charge density:

ρ_{P} = \frac{Q}{A δ x} = - \frac{\partial P_{x}}{\partial x} |_{x_{0}}

including contributions from $P_{y}$ and $P_{z} :$

ρ_{P} = - \vec{\nabla} \cdot \vec{P}

differentiating it, the polarization current is obtained (and the continuity equation):

\frac{\partial ρ_{P}}{\partial t} + \vec{\nabla} \cdot {\vec{J}}_{P} = 0

where, ${\vec{J}}_{P} = \frac{\partial \vec{P}}{\partial t}$ is the polarization current density

in many materials:

| \vec{P} | \propto | \vec{E} |

$\vec{P} \propto \vec{E}$ - 'linear', usually true at low $| \vec{E} |$
'isotropic':

\vec{P} ∥ \vec{E} ⟹ \vec{P} = ε_{0} χ \vec{E}

where, $χ$ is the susceptibility

\vec{E} = {\vec{E}}_{a p p l i e d} + {\vec{E}}_{P} = {\vec{E}}_{a p p l i e d} - χ \vec{E}

\vec{E} = \frac{{\vec{E}}_{a p p}}{1 + χ} = \frac{{\vec{E}}_{a p p}}{ε_{r}}

where, $ε_{r}$ is called the relative permittivity