Dielectrics and Polarization

In many capacitors there is an insulating material between the plates, called a dielectric. They can be used to maintain separation between the plates. It has been found that the capacitance increases when filled with dielectric:

C = k_e C_0,

where $C_0$ is the capacitance without the dielectric and $k_e$ is the dielectric constant. Each dielectric has a maximum value of electric field before the dielectric breaks down, called the dielectric strength. Below is a table of dielectric materials:

Material	$\boldsymbol{\kappa_e}$	Dielectric strength
Air	1.00059	3
Paper	3.7	16
Glass	4-6	9
Water	80	-

There are two types of dielectrics: polar and non polar. Polar have permanent electric dipole moments, an example is water. Non polar dielectrics don't have permanent electric dipole moments but when an external electric field $\boldsymbol{E_0}$ is present, the electric dipole moments are induced.

	$\boldsymbol{E_0} = 0$	$\boldsymbol{E_0} \neq 0$
Polar
Non polar

In both cases $\boldsymbol{E_0}$ points to the right and the molecules generate an electric field $\boldsymbol{E_P}$ in the direction opposite to $\boldsymbol{E_0}$ The total electric field is:

\boldsymbol{E} = \boldsymbol{E_0} + \boldsymbol{E_P},

implying $|\boldsymbol{E}| < |\boldsymbol{E_0}|$ .

Polarization

To calculate the average electric fields that the dipoles produce, suppose a cylinder of surface area $A$ and height $h$ with $N$ uniformly spread electric dipole moments $\boldsymbol{p}$ :

The polarization vector is the net electric dipole moment per unit volume:

\begin{align*} \boldsymbol{P} &= \frac{1}{Ah} \sum_{i=1}^N \boldsymbol{p_i}, \\ P &= \frac{Np}{Ah}. \end{align*}

The inner dipoles cancel each other out and the only dipoles left are at the edges:

The charge $Q_P$ produces the same electric dipole moment as the total electric dipole moments produced by the small dipoles:

\begin{align*} Q_P h &= Np, \\ Q_P &= \frac{Np}{h}. \end{align*}

The surface charge density is equal to:

\sigma_P = \frac{Q_P}{A} = \frac{Np}{Ah} = P.

In general, the surface charge density is equal to:

\sigma_P = \boldsymbol{P} \cdot \boldsymbol{\hat{n}} = P \cos \theta,

where $\boldsymbol{\hat{n}}$ is normal vector of the surface and $\theta$ the angle between the normal and the polarization vector.

The electric field is constant on the surface of the cylinder. By Gauss's law:

\begin{align*} \oiint_S \boldsymbol{E_P} \cdot d\boldsymbol{A} &= \frac{Q}{\epsilon_0}, \\ E_P A &= \frac{Q}{\epsilon_0}, \\ E_P &= \frac{\sigma_P}{\epsilon_0} \\ &= \frac{P}{\epsilon_0}. \end{align*}

Since the electric field is opposite to $\boldsymbol{P}$ , $\boldsymbol{E_P}$ is equal to:

\boldsymbol{E_P} = -\frac{\boldsymbol{P}}{\epsilon_0}.

The total electric field is equal to:

\begin{align*} \boldsymbol{E} &= \boldsymbol{E_0} + \boldsymbol{E_P} \\ &= \boldsymbol{E_0} - \frac{\boldsymbol{P}}{\epsilon_0}. \end{align*}

The polarization $\boldsymbol{P}$ is linearly proportional to $\boldsymbol{E}$ :

\boldsymbol{P} = \epsilon_0 \chi_0 \boldsymbol{E},

where $\chi_0$ is the electric susceptibility.

Substituting into the previous equation:

\begin{align*} \boldsymbol{E} &= \boldsymbol{E_0} - \chi_0 \boldsymbol{E}, \\ \boldsymbol{E_0} &= \boldsymbol{E} (1 + \chi_0) \\ &= \kappa_e \boldsymbol{E}. \end{align*}

where $\kappa_e = 1 + \chi_0$ .

$\kappa_e$ is always greater than one, implying:

\boldsymbol{E} = \frac{\boldsymbol{E_0}}{\kappa_e} < \boldsymbol{E_0}.

The electric field in the presence of dielectric material is lower than the electric field without it.

Gauss's Law for Dielectrics

In the absence of dielectric, the electric field is equal to:

\begin{align*} \oiint_S \boldsymbol{E_0} \cdot d\boldsymbol{A} &= \frac{Q}{\epsilon_0}, \\ E_0 A &= \frac{Q}{\epsilon_0}, \\ E_0 &= \frac{\sigma}{\epsilon_0}. \end{align*}

In the presence of a dielectric a second charge $Q_P$ gets induced:

where the orange border is the gaussian surface. The charge enclosed is $Q - Q_P$ . Using Gauss's law:

\begin{align*} \oiint_S \boldsymbol{E} \cdot d\boldsymbol{A} &= \frac{Q - Q_P}{\epsilon_0}, \\ E A &= \frac{Q - Q_P}{\epsilon_0}, \\ E &= \frac{Q - Q_P}{\epsilon_0 A}. \end{align*}

As found earlier, the electric field is weakened by the dielectric material:

\begin{align*} E &= \frac{E_0}{\kappa_e} = \frac{\sigma}{\kappa_e \epsilon_0} = \frac{Q - Q_P}{\epsilon_0 A}, \\ \frac{\sigma}{\kappa_e \epsilon_0} &= \frac{Q - Q_P}{\epsilon_0 A}, \\ \frac{Q - Q_P}{\epsilon_0} &= \frac{Q}{\kappa_e \epsilon_0}, \end{align*}

substituting into Gauss's law:

\begin{align*} \oiint_S \boldsymbol{E} \cdot d\boldsymbol{A} &= \frac{Q}{\kappa_e \epsilon_0}, \\ \oiint_S \kappa_e \epsilon_0 \boldsymbol{E} \cdot d\boldsymbol{A} &= Q, \\ \oiint_S \boldsymbol{D} \cdot d\boldsymbol{A} &= Q, \end{align*}

wher\epsilone $\boldsymbol{D} = \kappa_e \epsilon_0 \boldsymbol{E}$ is called the dielectric displacement vector.

Capacitance with Dielectrics

Consider a parallel plate capacitor with the plates having a distance $d$ between them and a dielectric of thickness $s$ and dielectric constant $\kappa_e$ inserted between them:

The voltage is equal to:

\begin{align*} U &= -\int_+^- \boldsymbol{E} \cdot d\boldsymbol{l} \\ &= -\int_0^d E\ dl \\ &= -E_0 (d - s) - E_D s \\ &= -\frac{\sigma}{\epsilon_0} (d - s) - \frac{\sigma}{\epsilon_0 \kappa_e} s \\ &= -\frac{\sigma}{\epsilon_0} \left(d - s + \frac{s}{\kappa_e}\right) \\ &= -\frac{\sigma}{\epsilon_0} \left[d - s\left(1 - \frac{1}{\kappa_e}\right)\right] \\ &= -\frac{Q}{A \epsilon_0} \left[d - s\left(1 - \frac{1}{\kappa_e}\right)\right] < 0, \\ |U| &= \frac{Q}{A \epsilon_0} \left[d - s\left(1 - \frac{1}{\kappa_e}\right)\right]. \end{align*}

The capacitance is equal to:

\begin{align*} C &= \frac{Q}{|U|} \\ &= \frac{A \epsilon_0}{d - s\left(1 - \frac{1}{\kappa_e}\right)}. \end{align*}

It could be useful to check the following limits:

$s$ approaching zero:
$\lim_{s \to 0} C = \frac{A \epsilon_0}{d},$
this is identical to the case without dielectric.
dielectric constant approaching zero:
$\lim_{\kappa_e \to 1} C = \frac{A \epsilon_0}{d},$
this is also identical to the case without dielectric.
$s$ approaching $d$ - filled with dielectric:
$\lim_{s \to d} C = \frac{A \epsilon_0 \kappa_e}{d} = \kappa_e C_0,$
this is identical to the first equation.