PX284 - Q1b - polarization of a slab

considering a dielectric medium between two parallel plate of opposite charges

applying gauss' law for a 'pill box' containing a small part of one of the sides:

| {\vec{E}}_{P} | \times 2 A = σ_{P} \frac{A}{ε_{0}} = | \vec{P} | \frac{A}{ε_{0}}

total polarization field inside the slab is the sum of those from the top and the bottom sides:

{\vec{E}}_{P} = - \frac{\vec{P}}{2 ε_{0}} - \frac{\vec{P}}{2 ε_{0}} = - \frac{\vec{P}}{ε_{0}}

total field:

\begin{aligned} \vec{E} & = {\vec{E}}_{a p p} + {\vec{E}}_{P} \\ = {\vec{E}}_{a p p} - \frac{\vec{P}}{ε_{0}} \\ = {\vec{E}}_{a p p} - χ \vec{E} \\ ⟹ \vec{E} & = \frac{{\vec{E}}_{a p p}}{1 + χ} = \frac{{\vec{E}}_{a p p}}{ε_{r}} \end{aligned}

reduction of ${\vec{E}}_{a p p}$ to $\vec{E}$ is called 'screening'
${\vec{E}}_{P}$ is also called a 'depolarization field', but it means the same as polarization field
for geometries other than a slab (cuboid), eg: an ellipsoid:

{\vec{E}}_{P} = - α \frac{\vec{P}}{ε_{0}}

where, $α$ is called the depolarization factor

for the slab, $α = 1$

\begin{matrix} \vec{P} = \frac{ε_{0} χ}{1 + α χ} {\vec{E}}_{a p p} \\ \vec{E} = \frac{1}{1 + α χ} {\vec{E}}_{a p p} \end{matrix}