Dielectric Materials

Local field

The local field $E_{loc}$ is the actual electric field acting on an individual atom or molecule inside a dielectric. It is not equal to the applied (macroscopic) field $E$, because the induced dipoles of all the surrounding atoms also produce a field at that point. In a polarised solid, $E_{loc} > E$.

Lorentz method for the local field

Imagine a small spherical cavity around the chosen atom, inside a dielectric between capacitor plates. The field at the centre is

$$E_{loc} = E_1 + E_2 + E_3$$

• $E_1$: field due to charges on the plates and the polarization charges on the outer dielectric surfaces; together this is the applied macroscopic field $E$.
• $E_2$: field due to polarization charges on the surface of the spherical cavity.
• $E_3$: field of the dipoles inside the sphere; it is zero for a cubic (symmetric) arrangement.

  +  +  +  +  +  +  +   plate
  ------------------
      .--------.
     /  - - -  \     surface charge
    |  (atom)   |    on cavity: E2
     \  + + +  /
      '--------'
  ------------------
  -  -  -  -  -  -  -   plate

Field of cavity surface charge ($E_2$): on a ring of the sphere at angle $\theta$ to the field, the surface charge density is $P\cos\theta$. The ring of radius $r\sin\theta$, width $r\,d\theta$ has area $2\pi r^2\sin\theta\, d\theta$, and its field component along $E$ at the centre is

$$dE_2 = \frac{(P\cos\theta)(2\pi r^2 \sin\theta\, d\theta)}{4\pi\varepsilon_0 r^2}\cos\theta$$ $$E_2 = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\cdot\frac{2}{3} = \frac{P}{3\varepsilon_0}$$

So the Lorentz local field is

$$E_{loc} = E + \frac{P}{3\varepsilon_0}$$

Derivation of Clausius-Mossotti equation

For electronic polarization, each atom acquires an induced dipole $p = \alpha_e E_{loc}$. With $N$ atoms per unit volume:

$$P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right)$$

Also, from the macroscopic definition, $P = \varepsilon_0(\varepsilon_r - 1)E$, so $E = \dfrac{P}{\varepsilon_0(\varepsilon_r - 1)}$. Then

$$\begin{aligned} P &= N\alpha_e\left[\frac{P}{\varepsilon_0(\varepsilon_r - 1)} + \frac{P}{3\varepsilon_0}\right] \\ 1 &= \frac{N\alpha_e}{\varepsilon_0}\left[\frac{3 + \varepsilon_r - 1}{3(\varepsilon_r - 1)}\right] = \frac{N\alpha_e}{3\varepsilon_0}\cdot\frac{\varepsilon_r + 2}{\varepsilon_r - 1} \end{aligned}$$ $$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}}$$

This is the Clausius-Mossotti equation. It links a microscopic property (polarizability $\alpha_e$ of one atom) to a macroscopic property (relative permittivity $\varepsilon_r$).

Notes:

• It is valid for non-polar solids with cubic symmetry and for gases and liquids where $E_3 = 0$.
• From it, $E_{loc} = \dfrac{\varepsilon_r + 2}{3}E$. Example: Si ($\varepsilon_r = 11.9$) has $E_{loc} = 4.63E$.
• For a dilute gas, $\varepsilon_r \approx 1$, and it reduces to $\varepsilon_r = 1 + N\alpha_e/\varepsilon_0$.

Loss tangent

In an ideal dielectric the current leads the voltage by exactly 90°, so no power is lost. In a real dielectric, polarization cannot follow an AC field instantly, so part of the energy is turned into heat. This is described by a complex relative permittivity

$$\varepsilon_r = \varepsilon_r' - j\varepsilon_r''$$

where $\varepsilon_r'$ is the ordinary (storage) part and $\varepsilon_r''$ is the loss part. The loss tangent is

$$\tan\delta = \frac{\varepsilon_r''}{\varepsilon_r'}$$

$\delta$ is the angle by which the current falls short of leading the voltage by 90°. Good insulators have $\tan\delta \sim 10^{-4}$ (polyethylene); lossy ones are 0.01–0.1.

      I_total
        ^   /
  I_cap |  /
  (wC'V)| /  delta
        |/________> V
          I_loss (wC''V)

Power loss per unit volume

Consider a parallel-plate capacitor with area $A$, spacing $d$, filled with the dielectric, and an applied voltage $V = V_0 e^{j\omega t}$.

Capacitance with complex permittivity:

$$C = \frac{\varepsilon_0(\varepsilon_r' - j\varepsilon_r'')A}{d}$$

Current:

$$I = j\omega C V = \frac{\omega\varepsilon_0 A}{d}\left(j\varepsilon_r' + \varepsilon_r''\right)V$$

The term $\frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V$ is in phase with $V$ and causes power loss; the $j\varepsilon_r'$ term is the lossless capacitive current. So the dielectric acts like a capacitor with a parallel conductance

$$G_p = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}$$

Power dissipated (using rms voltage $V_{rms}$):

$$W = G_p V_{rms}^2 = \frac{\omega\varepsilon_0\varepsilon_r'' A}{d}V_{rms}^2$$

Divide by the volume $Ad$ and use $E_{rms} = V_{rms}/d$:

$$W_{vol} = \frac{W}{Ad} = \omega\varepsilon_0\varepsilon_r'' E_{rms}^2$$

Since $\varepsilon_r'' = \varepsilon_r'\tan\delta$:

$$\boxed{W_{vol} = \omega\, \varepsilon_0\, \varepsilon_r'\, E_{rms}^2 \tan\delta}$$

Conclusion

The dielectric loss per unit volume is directly proportional to:

• the frequency $\omega = 2\pi f$ of the applied field,
• the loss tangent $\tan\delta$ (or loss factor $\varepsilon_r'\tan\delta$),
• the square of the field strength.

So insulation for high-frequency or high-voltage use (cables, RF capacitors) must have a very low $\tan\delta$. The same heating is used usefully in microwave ovens and dielectric heating of plastics and wood.

Local field and polarization

Polarization $P$ is the net induced dipole moment per unit volume of a dielectric: $P = N p$, where $N$ is the number of dipoles per unit volume and $p$ the average dipole moment.

The local field $E_{loc}$ is the field that actually acts on an ion or atom. It equals the applied field $E$ plus the field produced by all the surrounding polarized ions. Since the dipoles of the neighbours point the same way, $E_{loc}$ is greater than $E$, and the dipole moment of each unit is $p = \alpha E_{loc}$, not $\alpha E$.

Ionic polarization and ionic polarizability

In an ionic crystal such as NaCl, an applied field pushes the cations along the field and anions opposite to it. Each pair is displaced from its equilibrium spacing by a small amount $x$, which induces a dipole.

  No field:   Na+  ---a---  Cl-
  With field E  -->
              Na+ -->   <-- Cl-
              (spacing changes by x)

The restoring force behaves like a spring of constant $\beta$: $F = \beta x = eE_{loc}$. The induced dipole is

$$p_i = e x = \frac{e^2}{\beta}E_{loc} = \alpha_i E_{loc}, \qquad \alpha_i = \frac{e^2}{\beta}$$

Each ion is also electronically polarized, so the total polarizability per ion pair is

$$\alpha = \alpha_i + \alpha_{e+} + \alpha_{e-}$$

Lorentz local field

Take a spherical cavity around the reference ion. The field at its centre is

$$E_{loc} = E + E_{sphere} + E_{inside}$$

The polarization charge density on the cavity wall at angle $\theta$ is $P\cos\theta$. Summing the axial field of all rings:

$$E_{sphere} = \int_0^{\pi}\frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{3\varepsilon_0}$$

For a cubic crystal the dipoles inside the sphere give $E_{inside} = 0$. Hence

$$E_{loc} = E + \frac{P}{3\varepsilon_0}$$

Clausius-Mossotti equation

With $N_i$ ion pairs per unit volume:

$$P = N_i\alpha E_{loc} = N_i\alpha\left(E + \frac{P}{3\varepsilon_0}\right) \;\Rightarrow\; P = \frac{N_i\alpha E}{1 - \dfrac{N_i\alpha}{3\varepsilon_0}}$$

Macroscopically $P = \varepsilon_0(\varepsilon_r - 1)E$. Equating:

$$\begin{aligned} \varepsilon_0(\varepsilon_r - 1) &= \frac{N_i\alpha}{1 - N_i\alpha/3\varepsilon_0} \\ (\varepsilon_r - 1)\left(1 - \frac{N_i\alpha}{3\varepsilon_0}\right) &= \frac{N_i\alpha}{\varepsilon_0} \\ \varepsilon_r - 1 &= \frac{N_i\alpha}{3\varepsilon_0}\left(3 + \varepsilon_r - 1\right) = \frac{N_i\alpha}{3\varepsilon_0}(\varepsilon_r + 2) \end{aligned}$$ $$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N_i(\alpha_i + \alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}}$$

Low and high frequency forms

• At low frequency (DC up to infrared), both ionic and electronic polarizations follow the field: $\varepsilon_r = \varepsilon_r(0)$ and the full $\alpha$ is used.
• At optical frequencies the heavy ions cannot follow, so only electronic terms remain; $\varepsilon_r = \varepsilon_{op} = n^2$ (refractive index squared):

$$\frac{\varepsilon_{op} - 1}{\varepsilon_{op} + 2} = \frac{N_i(\alpha_{e+} + \alpha_{e-})}{3\varepsilon_0}$$

Subtracting the two gives $\alpha_i$ alone. Example: NaCl has $\varepsilon_r(0) \approx 5.9$ and $\varepsilon_{op} \approx 2.25$, showing that ionic polarization contributes more than half of its static permittivity.

Significance

The equation relates the microscopic polarizability of each ion pair to the measurable permittivity, explains why $E_{loc} = \frac{\varepsilon_r + 2}{3}E$, and shows why ionic crystals have higher static permittivity than optical permittivity. It holds for cubic ionic crystals; it fails for polar liquids with permanent dipoles, where dipole interactions are strong.

Dielectric breakdown is the sudden large increase of current through an insulator when the applied field exceeds a critical value (the dielectric strength), so that the material loses its insulating property, usually permanently in solids.

Thermal breakdown

1. When AC or DC voltage is applied, a small leakage current and dielectric loss ($W = \omega\varepsilon_0\varepsilon_r'E^2\tan\delta$ per unit volume) heat the insulator.
2. The conductivity of an insulator rises steeply (exponentially) with temperature, and $\tan\delta$ also rises.
3. More conductivity gives more current and more heat. If heat is generated faster than it can be conducted away to the surroundings, temperature rises without limit (thermal runaway).
4. The material melts, chars or burns, forming a conducting path, usually at the hottest point in the middle.

Features: it takes seconds to minutes; breakdown voltage falls with higher frequency, higher ambient temperature, longer application time and thicker insulation (poorer cooling). It is the most common mode in power equipment insulation (e.g. cable insulation).

 heat
  ^       generated
  |         /
  |        /   ...lost (cooling)
  |  .....'
  |  stable / unstable -> runaway
  |____________________> T

Electromechanical breakdown

1. A voltage $V$ across a soft solid dielectric of thickness $d$ produces an electrostatic attraction between the electrodes (Maxwell stress), $\frac{1}{2}\varepsilon_0\varepsilon_r\frac{V^2}{d^2}$ per unit area.
2. This compressive force squeezes the material, so $d$ decreases.
3. The field $V/d$ then increases, the force grows further, and the material is compressed more.
4. If the elastic restoring stress (Young's modulus $Y$) cannot balance it, the insulation collapses mechanically and then breaks down.

Balancing the stresses, the material becomes unstable when $d$ falls to about $0.6\,d_0$, giving the highest apparent field

$$E_{max} \approx 0.6\sqrt{\frac{Y}{\varepsilon_0\varepsilon_r}}$$

It occurs mainly in soft materials such as polymers (polyethylene, rubber) and at high temperature, where $Y$ is low.

Silicon is a covalent, non-polar crystal, so its permittivity comes only from electronic polarization of the valence electrons. Use the Clausius-Mossotti equation and the Lorentz local field.

Given: $\varepsilon_r = 11.9$, $N = 5 \times 10^{28}\ \text{m}^{-3}$, $\varepsilon_0 = 8.854 \times 10^{-12}\ \text{F/m}$. Each Si atom has $Z = 4$ valence electrons.

Electronic polarizability per Si atom

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0} \;\Rightarrow\; \alpha_e = \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2}$$ $$\alpha_e = \frac{3 \times 8.854 \times 10^{-12}}{5 \times 10^{28}} \times \frac{10.9}{13.9} = 4.17 \times 10^{-40}\ \text{F m}^2$$

Local field versus applied field

Lorentz field: $E_{loc} = E + \dfrac{P}{3\varepsilon_0}$ with $P = \varepsilon_0(\varepsilon_r - 1)E$:

$$E_{loc} = E + \frac{(\varepsilon_r - 1)E}{3} = \frac{\varepsilon_r + 2}{3}E = \frac{13.9}{3}E = 4.63E$$

So the local field is about 4.63 times the applied field (greater by $3.63E$).

Resonant frequency

In the simple model the electron cloud ($Ze$, mass $Zm_e$) is bound to the nucleus by a spring, and

$$\alpha_e = \frac{Ze^2}{m_e\omega_0^2} \;\Rightarrow\; \omega_0 = \sqrt{\frac{Ze^2}{m_e\alpha_e}}$$ $$\omega_0 = \sqrt{\frac{4 \times (1.602 \times 10^{-19})^2}{9.109 \times 10^{-31} \times 4.17 \times 10^{-40}}} = 1.64 \times 10^{16}\ \text{rad/s}$$ $$f_0 = \frac{\omega_0}{2\pi} = 2.62 \times 10^{15}\ \text{Hz}$$

This lies in the ultraviolet region, so electronic polarization in Si follows the field up to optical frequencies.

Answer: $\alpha_e = 4.17 \times 10^{-40}\ \text{F m}^2$; $E_{loc} = 4.63E$; $\omega_0 = 1.64 \times 10^{16}\ \text{rad/s}$ ($f_0 \approx 2.6 \times 10^{15}$ Hz, taking $Z = 4$).

Pure Si has only electronic polarization (covalent, no permanent dipoles, cubic structure), so the Clausius-Mossotti equation with the Lorentz local field applies.

Given: $\varepsilon_r = 11.9$, $N = 5 \times 10^{28}\ \text{m}^{-3}$, $\varepsilon_0 = 8.85 \times 10^{-12}\ \text{F/m}$.

(i) Electronic polarizability per Si atom

Clausius-Mossotti equation:

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}$$ $$\begin{aligned} \alpha_e &= \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2} \\ &= \frac{3 \times 8.85 \times 10^{-12}}{5 \times 10^{28}} \times \frac{11.9 - 1}{11.9 + 2} \\ &= 5.31 \times 10^{-40} \times 0.7842 = 4.16 \times 10^{-40}\ \text{F m}^2 \end{aligned}$$

(If the local field were ignored, $\alpha_e = \varepsilon_0(\varepsilon_r - 1)/N = 1.93 \times 10^{-39}$ F m², about 4.6 times too large; this shows why the local field matters.)

(ii) Local field compared with applied field

The Lorentz local field is

$$E_{loc} = E + \frac{P}{3\varepsilon_0}, \qquad P = \varepsilon_0(\varepsilon_r - 1)E$$ $$E_{loc} = E\left(1 + \frac{\varepsilon_r - 1}{3}\right) = \frac{\varepsilon_r + 2}{3}E = \frac{13.9}{3}E = 4.63E$$

So the field acting on each Si atom is about 4.63 times the applied field, i.e. greater than it by $3.63E$ (363%).

Answer: (i) $\alpha_e \approx 4.16 \times 10^{-40}\ \text{F m}^2$; (ii) $E_{loc} = 4.63E$.

Electric dipole moment

Two equal and opposite charges $+Q$ and $-Q$ separated by a small distance $a$ form an electric dipole. Its dipole moment is a vector from $-Q$ to $+Q$:

$$\mathbf{p} = Q\mathbf{a} \qquad (\text{unit: C m})$$

When an atom is placed in a field, its electron cloud shifts relative to the nucleus, creating an induced dipole moment proportional to the field acting on it:

$$p = \alpha_e E_{loc}$$

where $\alpha_e$ is the electronic polarizability. The polarization $P$ is the dipole moment per unit volume, $P = Np$.

 no field:   ( +)  cloud centred on nucleus, p = 0
 field E --> 
             (-- +)  cloud shifted left
              <-a->   p = Q a  (points along E)

Local electric field

The local field $E_{loc}$ is the actual field experienced by one atom inside a dielectric. It is the applied field plus the fields of all the other induced dipoles around it. In a solid it is larger than the applied macroscopic field $E$.

Lorentz expression for the local field

Cut an imaginary sphere around the reference atom, large compared with atomic size. The field at its centre is

$$E_{loc} = E + E_{sph} + E_{in}$$

• $E$: macroscopic field (from plate charges and outer surface polarization charges).
• $E_{sph}$: field from bound charges on the inner surface of the spherical cavity.
• $E_{in}$: field from dipoles inside the sphere; zero for cubic symmetry or random arrangement.

            E -->
         .------.
       -'   ^    '+       surface charge
      -  theta    +       = P cos(theta)
      -    atom   +
       -.        .+
         '-....-'

The bound surface charge density at angle $\theta$ is $\sigma = P\cos\theta$. A ring between $\theta$ and $\theta + d\theta$ has area $dA = 2\pi r^2\sin\theta\,d\theta$. The field component along $E$ at the centre is

$$dE_{sph} = \frac{P\cos\theta \cdot 2\pi r^2\sin\theta\, d\theta}{4\pi\varepsilon_0 r^2}\cos\theta = \frac{P}{2\varepsilon_0}\cos^2\theta\sin\theta\, d\theta$$ $$E_{sph} = \frac{P}{2\varepsilon_0}\int_0^{\pi}\cos^2\theta\sin\theta\, d\theta = \frac{P}{2\varepsilon_0}\left[-\frac{\cos^3\theta}{3}\right]_0^{\pi} = \frac{P}{3\varepsilon_0}$$

Hence

$$E_{loc} = E + \frac{P}{3\varepsilon_0}$$

Clausius-Mossotti equation

With $N$ atoms per unit volume:

$$P = N\alpha_e E_{loc} = N\alpha_e\left(E + \frac{P}{3\varepsilon_0}\right) \quad\cdots (1)$$

From electrostatics, $D = \varepsilon_0E + P = \varepsilon_0\varepsilon_rE$, so

$$P = \varepsilon_0(\varepsilon_r - 1)E \quad\Rightarrow\quad E = \frac{P}{\varepsilon_0(\varepsilon_r - 1)} \quad\cdots (2)$$

Put (2) in (1) and divide by $P$:

$$\begin{aligned} 1 &= N\alpha_e\left[\frac{1}{\varepsilon_0(\varepsilon_r - 1)} + \frac{1}{3\varepsilon_0}\right] = \frac{N\alpha_e}{3\varepsilon_0}\cdot\frac{3 + (\varepsilon_r - 1)}{\varepsilon_r - 1} \end{aligned}$$ $$\boxed{\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0}}$$

Meaning and use

• It relates the microscopic polarizability $\alpha_e$ to the macroscopic, measurable $\varepsilon_r$.
• It gives $E_{loc} = \frac{\varepsilon_r + 2}{3}E$; for Si ($\varepsilon_r = 11.9$, $N = 5 \times 10^{28}$ m⁻³) it gives $\alpha_e = 4.17 \times 10^{-40}$ F m² and $E_{loc} = 4.63E$.
• At optical frequency $\varepsilon_r = n^2$, giving the Lorentz-Lorenz equation for refractive index.
• Valid for non-polar dielectrics with cubic symmetry, gases and non-polar liquids; not for polar materials where permanent dipoles interact strongly.

Polarization is the alignment or creation of electric dipoles in a dielectric by an applied field; $P$ is the net dipole moment per unit volume, $P = \varepsilon_0(\varepsilon_r - 1)E$. There are four main mechanisms, and the total polarization is their sum: $P = P_e + P_i + P_d + P_{if}$.

1. Electronic polarization

• The field shifts the electron cloud of each atom relative to its nucleus, creating an induced dipole.
• Present in all materials (atoms, molecules, ions).
• $p_e = \alpha_e E_{loc}$, with $\alpha_e = Ze^2/(m_e\omega_0^2)$; for a hydrogen-like atom $\alpha_e = 4\pi\varepsilon_0 R^3$.
• Very fast (follows up to about $10^{15}$ Hz, UV); independent of temperature.
• Example: Si, Ge, diamond, inert gases.

 E=0:  ( (+) )        E -->:  ( (+))
       cloud centred          cloud shifted left

2. Ionic (atomic) polarization

• In ionic crystals (NaCl, KCl), the field displaces positive ions along $E$ and negative ions opposite, changing the inter-ionic spacing and creating a net dipole per ion pair.
• $p_i = \alpha_i E_{loc}$, $\alpha_i = e^2/\beta$ where $\beta$ is the bond "spring" constant.
• Ions are heavy, so it follows the field only up to the infrared (about $10^{12}$–$10^{13}$ Hz); nearly independent of temperature.

 E -->    Na+ ->  <- Cl-  ->Na+  <- Cl-

3. Orientational (dipolar) polarization

• Occurs in polar molecules with a permanent dipole $p_0$ (H₂O, HCl, nitrobenzene).
• Without a field dipoles point randomly ($P = 0$); a field tends to rotate them into line, while thermal motion opposes this.
• Average dipole along field $= \dfrac{p_0^2}{3kT}E$, so $\alpha_d = \dfrac{p_0^2}{3kT}$, which decreases with temperature.
• Slow (rotation of molecules): follows up to about $10^{6}$–$10^{10}$ Hz; large losses near the relaxation frequency.

4. Interfacial (space-charge) polarization

• Charges (ions, electrons) drift and pile up at interfaces, grain boundaries, defects or electrodes in non-homogeneous dielectrics.
• The accumulated charge acts like a large dipole.
• Slowest: important only at low frequency (below about $10^{3}$ Hz); it gives very large apparent $\varepsilon_r$ at DC and low frequencies.

Comparison

As frequency rises, mechanisms drop out one after another, so $\varepsilon_r'$ falls in steps (from interfacial at low frequency down to electronic only at optical frequency).

Dielectric breakdown occurs when the electric field in the insulator exceeds its dielectric strength $E_{br}$ (the maximum field it can withstand, e.g. about 3 kV/mm for air, 20–40 kV/mm for oil and polymers).

For a parallel-plate capacitor with voltage $V$ and plate spacing $d$, the field in the dielectric is

$$E = \frac{V}{d}$$

So for a given voltage, the field is inversely proportional to the spacing. The breakdown voltage is

$$V_{br} = E_{br}\,d$$

Why small spacing breaks down sooner

1. When $d$ is reduced, the same voltage produces a larger field, so $E$ reaches $E_{br}$ at a lower voltage.
2. A higher field gives the few free electrons more energy between collisions ($e E \lambda$), so they can ionize atoms and start an avalanche at a lower voltage.
3. A higher field also means more leakage current and dielectric heating per unit volume ($\propto E^2$), leading to thermal breakdown sooner.
4. In thin soft films, the electrostatic pressure $\frac{1}{2}\varepsilon E^2$ is larger and can cause electromechanical collapse.

Example

For air ($E_{br} \approx 3\ \text{kV/mm}$):

So a capacitor with closer plates fails at a much lower voltage. This is why high-voltage capacitors use either larger spacing or a dielectric with very high strength (mica, polypropylene), and each capacitor has a rated working voltage. (Very thin films often show a slightly higher $E_{br}$ because they contain fewer defects, but their breakdown voltage $E_{br}d$ is still lower.)

Dipolar (orientational) polarization occurs in materials with permanent molecular dipoles $p_0$ (e.g. H₂O, HCl). Without a field the dipoles point randomly, so $P = 0$. An applied field tries to align them, while thermal agitation tries to keep them random. The result depends on temperature.

 E = 0                E -->
  \  /  |  -           ->  /  ->  \
  -  \  /  |           ->  ->  /  ->
  random, P = 0        partly aligned, P > 0

Energy of a dipole in a field

A dipole at angle $\theta$ to the field $E$ has potential energy

$$U = -p_0E\cos\theta$$

and its component along the field is $p_0\cos\theta$.

Average dipole moment (Boltzmann statistics)

The probability of a dipole having orientation $\theta$ is proportional to $e^{-U/kT}$. The number of dipoles in the solid angle $d\Omega = 2\pi\sin\theta\,d\theta$ is proportional to $e^{p_0E\cos\theta/kT}\sin\theta\,d\theta$. So

$$\langle p_0\cos\theta \rangle = p_0\frac{\int_0^{\pi}\cos\theta\, e^{a\cos\theta}\sin\theta\, d\theta}{\int_0^{\pi}e^{a\cos\theta}\sin\theta\, d\theta}, \qquad a = \frac{p_0E}{kT}$$

Put $x = \cos\theta$, $dx = -\sin\theta\,d\theta$:

$$\langle p_0\cos\theta \rangle = p_0\frac{\int_{-1}^{1}x e^{ax}dx}{\int_{-1}^{1}e^{ax}dx} = p_0\left[\coth a - \frac{1}{a}\right] = p_0 L(a)$$

$L(a)$ is the Langevin function.

Usual condition (weak field, normal temperature)

Normally $p_0E \ll kT$ (e.g. $p_0 \sim 10^{-30}$ C m, $E = 10^{6}$ V/m gives $p_0E \approx 10^{-24}$ J, while $kT \approx 4 \times 10^{-21}$ J at 300 K), so $a \ll 1$ and $L(a) \approx a/3$. Then

$$\langle p \rangle = \frac{p_0^2 E}{3kT}$$

so the dipolar polarizability and polarization are

$$\alpha_d = \frac{p_0^2}{3kT}, \qquad P_d = N\frac{p_0^2}{3kT}E$$

Conclusion

$\alpha_d \propto 1/T$: as temperature rises, thermal agitation disorders the dipoles more and dipolar polarization falls. This is the Debye relation. Including electronic polarization, the total polarizability is

$$\alpha = \alpha_e + \frac{p_0^2}{3kT}$$

A plot of $\alpha$ (or $\varepsilon_r - 1$) against $1/T$ is a straight line: its slope gives $p_0$ and its intercept gives $\alpha_e$. For non-polar materials the slope is zero, which shows that only dipolar polarization depends on temperature.

Ferroelectricity is the property of some crystals to have a spontaneous polarization (permanent net dipole moment) even without an applied field, whose direction can be reversed by an external field. They show a P-E hysteresis loop and lose this property above the Curie temperature $T_C$. Examples: barium titanate (BaTiO₃, $T_C \approx 120°\text{C}$), Rochelle salt, PZT.

Piezoelectricity is the property of some crystals to develop electric polarization (and a voltage) when mechanically stressed (direct effect), and to change shape when an electric field is applied (converse effect). It needs a crystal with no centre of symmetry. Examples: quartz, PZT, Rochelle salt, ZnO.

Similarities

1. Both occur only in crystals without a centre of symmetry (non-centrosymmetric).
2. Both are linked to displacement of ions creating dipoles.
3. Every ferroelectric is also piezoelectric (and pyroelectric).
4. Both are used in transducers, sensors and capacitors; PZT is both.

Differences

  P                    P
  ^   ____             ^      /
  |  /   /             |     /  slope = d
 -+-/---/--> E        -+----/----> stress
  |/___/               |   /
 Ferroelectric loop    Piezoelectric line

Piezoelectric materials are crystals that become electrically polarized when mechanically stressed, and change dimensions when an electric field is applied. The crystal must lack a centre of symmetry. Examples: quartz (SiO₂), Rochelle salt, barium titanate, PZT (lead zirconate titanate), ZnO.

Piezoelectric effect in terms of polarization

In an unstressed piezoelectric crystal, the positive and negative ions are arranged so that the centres of positive and negative charge in each unit cell coincide. The dipole moments cancel and the net polarization is zero.

When a stress $T$ is applied, the ions are displaced unequally. Because the cell has no centre of symmetry, the centre of positive charge moves away from the centre of negative charge. Each cell then has a net dipole moment, and the crystal acquires a polarization $P$. Bound charges appear on opposite faces and a voltage can be measured between electrodes.

 Unstressed         Compressed (force down)
     +                  +
   -   -              -   -
     +    P=0         -+-     P != 0
   +   +             + . +   (+ centre moves)
     -                  -

For small stresses the induced polarization is proportional to stress:

$$P = d\,T$$

where $d$ is the piezoelectric coefficient (C/N). Reversing the stress (tension instead of compression) reverses $P$.

Converse effect: an applied field $E$ shifts the ions and produces a strain $S = dE$; the crystal expands or contracts.

Applications

• Direct effect: gas lighters and spark igniters, microphones, pressure and vibration sensors, accelerometers.
• Converse effect: quartz crystal oscillators (in watches and computers), ultrasonic transducers, buzzers, precision actuators.

Dielectric breakdown is the sudden loss of insulating property of a dielectric when the applied field exceeds a critical value called the dielectric strength $E_{br}$; a large current flows and solids are usually damaged permanently.

Types of breakdown in solids

1. Intrinsic (electronic) breakdown
2. Thermal breakdown
3. Electromechanical breakdown
4. Discharge (partial-discharge) breakdown in internal voids
5. Electrochemical breakdown (slow chemical ageing of the insulation)

Intrinsic breakdown

• Depends only on the material itself, under pure, defect-free conditions; it occurs in about $10^{-8}$ s.
• The few free electrons in the conduction band gain energy from the field. When the field is high enough, they gain more energy than they lose to the lattice and can ionize atoms by impact, freeing more electrons. This chain reaction is an electron avalanche, and the current rises sharply.
• It sets the highest possible strength of a material (e.g. about $10^{8}$–$10^{9}$ V/m for good insulators). It is almost independent of temperature and thickness.

 e- -> * -> 2e- -> * -> 4e- -> 8e- ...
       ionising collisions (avalanche)

Thermal breakdown

• Leakage current and dielectric loss ($\propto \omega E^2\tan\delta$) heat the dielectric.
• Conductivity of insulators rises exponentially with temperature, so the current and heat increase further.
• If heat generated exceeds heat conducted away, temperature runs away until the material melts or chars.
• Slow (seconds to minutes); breakdown voltage falls with higher frequency, ambient temperature and longer stress. Common in power cables and capacitors.

Dielectric breakdown is the sudden, large rise in current through an insulator when the applied field exceeds a critical value (the dielectric strength), after which the material stops insulating.

Thermal breakdown

• Every real dielectric has a small conduction current and dielectric (AC) loss, which produce heat: $P = \omega \varepsilon_0 \varepsilon_r \tan\delta\, E^2$ per unit volume.
• The conductivity of insulators rises exponentially with temperature, so more heat gives more current, which gives still more heat.
• If heat produced is more than heat that can be conducted away to the surroundings, temperature runs away, the material melts, chars or burns, and a conducting path forms.
• Features: takes seconds to minutes, depends on frequency (worse at high frequency), ambient temperature, thickness and cooling; breakdown voltage falls as temperature rises.

Electromechanical breakdown

• The applied voltage puts opposite charges on the two electrodes, which attract each other with an electrostatic pressure $p = \tfrac{1}{2}\varepsilon_0\varepsilon_r E^2$.
• Soft solids (polymers, rubber, especially when warm) are compressed by this force, so the thickness $d$ decreases.
• For the same voltage, smaller $d$ means larger field $E = V/d$, more pressure and more compression.
• When the mechanical restoring stress (Young's modulus) can no longer balance the electric pressure (at about $d \approx 0.6 d_0$), the material collapses mechanically and breaks down.

Electronic vs orientational polarization

The total polarizability of a dielectric is $\alpha = \alpha_e + \alpha_i + \alpha_d$ (+ $\alpha_{sc}$ interfacial). Each mechanism needs a certain time to respond. When the applied AC field changes faster than a mechanism can follow, that mechanism drops out, so $\alpha$ (and $\varepsilon_r$) falls in steps as frequency rises.

 alpha (or er)
   ^
   |____
   |    \  interfacial (space charge)
   |     \_______
   |             \   orientational
   |              \________
   |                       \/\  ionic
   |                          \_______
   |                                  \/\ electronic
   |                                     \______
   +----+--------+--------+--------+-------> f (Hz)
       1e2     1e6-1e9   1e13    1e15
       power   radio/MW  infra-   UV/
       freq.             red      optical

Explanation of each region

1. Interfacial (space-charge) polarization: charges drift to grain boundaries and electrodes. Very slow; it follows the field only up to about $10^2$–$10^3$ Hz.
2. Orientational (dipolar) polarization: permanent dipoles rotate. Molecules have inertia and suffer collisions, so they follow the field up to about $10^{6}$–$10^{10}$ Hz (radio/microwave). Above this, $\alpha_d \to 0$.
3. Ionic polarization: positive and negative ions are displaced. Ions are heavy, so a resonance occurs at infrared frequencies (~$10^{13}$ Hz).
4. Electronic polarization: the light electron cloud shifts. It follows the field up to UV/optical frequencies (~$10^{15}$ Hz). Above that, nothing can respond and $\varepsilon_r \to 1$.

Key points

• Near each relaxation frequency (orientational) the fall is smooth and the dielectric loss ($\varepsilon_r''$) shows a peak.
• Near each resonance (ionic, electronic) there is a small rise then a sharp drop, also with a loss peak.
• At optical frequencies only electronic polarization remains, so $\varepsilon_r = n^2$ (n = refractive index). Example: water has $\varepsilon_r \approx 80$ at low frequency but $n^2 \approx 1.77$ at optical frequency, because dipolar polarization has dropped out.

Polarization is the displacement of bound charges in a dielectric under an electric field, giving a net dipole moment per unit volume, $P = N\alpha E_{loc}$.

Types of polarization

1. Electronic polarization: the electron cloud of each atom shifts opposite to the field, while the nucleus shifts slightly with it. It occurs in all materials. $\alpha_e = 4\pi\varepsilon_0 R^3$, independent of temperature.
2. Ionic polarization: in ionic solids (NaCl, KCl), positive ions move along the field and negative ions against it, changing the bond lengths and creating a net dipole moment.
3. Orientational (dipolar) polarization: polar molecules (H₂O, HCl, nitrobenzene) have permanent dipoles pointing randomly. The field tends to align them, and thermal motion opposes this, so $\alpha_d = p_0^2/(3kT)$.
4. Interfacial (space-charge) polarization: mobile charges pile up at grain boundaries, impurities or electrodes, as in polycrystalline and multiphase materials.

 No field:   (+)    E ->   ( +)->   electron cloud
  atom       (o)           (o )     shifts left,
                                    nucleus right
 Polar molecules:  /  \  |  -  ->  -> -> ->  aligned

Electronic vs orientational polarization

Polarization $\vec P$ is the net induced dipole moment per unit volume of a dielectric: $P = Np$, where $N$ is the number of molecules per m³ and $p$ the average dipole moment of each. It is related to the field by $P = \varepsilon_0(\varepsilon_r - 1)E$.

Step 1: Polarization in terms of $\varepsilon_r$

From $D = \varepsilon_0 E + P = \varepsilon_0\varepsilon_r E$:

$$P = \varepsilon_0(\varepsilon_r - 1)E \quad (1)$$

Step 2: Polarization in terms of polarizability

A molecule feels the local field $E_{loc}$, not the applied field, so

$$P = N\alpha_e E_{loc} \quad (2)$$

Step 3: Local (Lorentz) field

Imagine a small spherical cavity around a molecule. The field at its centre is

$$E_{loc} = E + E_1 + E_2 + E_3$$

• $E_1$: field of charges on the plates (included in $E$)
• $E_2$: field of polarization charges on the cavity surface $= \dfrac{P}{3\varepsilon_0}$ (Lorentz field)
• $E_3$: field of dipoles inside the cavity $= 0$ for cubic symmetry

$$E_{loc} = E + \frac{P}{3\varepsilon_0} \quad (3)$$

   +  +  +  +  +  +   plate
       - - - -
     -  ( mol )  +    spherical cavity,
       + + + +        surface charges give
   -  -  -  -  -  -   P/(3 e0)

Step 4: Combine

Put (1) in (3):

$$\begin{aligned} E_{loc} &= E + \frac{\varepsilon_0(\varepsilon_r - 1)E}{3\varepsilon_0} = E\,\frac{\varepsilon_r + 2}{3} \end{aligned}$$

Equate (1) and (2):

$$\begin{aligned} \varepsilon_0(\varepsilon_r - 1)E &= N\alpha_e\, E\,\frac{\varepsilon_r + 2}{3} \\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{N\alpha_e}{3\varepsilon_0} \end{aligned}$$

This is the Clausius–Mossotti equation. It links the macroscopic quantity $\varepsilon_r$ with the microscopic quantity $\alpha_e$.

Notes

• Valid for non-polar materials with cubic symmetry (gases, elemental solids like Si, Ge, diamond).
• If $N\alpha_e$ is known, $\varepsilon_r = \dfrac{1 + 2N\alpha_e/3\varepsilon_0}{1 - N\alpha_e/3\varepsilon_0}$.
• At optical frequency $\varepsilon_r = n^2$, giving the Lorentz–Lorenz equation.

Use the Clausius–Mossotti relation for a covalent (non-polar) solid, where only electronic polarization exists.

Data and assumptions

• $N = 6\times10^{22}\ \text{cm}^{-3} = 6\times10^{28}\ \text{m}^{-3}$, taken as the number of polarizable Si units per m³ (as given).
• Relative permittivity of Si, $\varepsilon_r = 11.9$ (standard value, not given in the question).
• $\varepsilon_0 = 8.854\times10^{-12}$ F/m.

i) Electronic polarizability

$$\frac{\varepsilon_r - 1}{\varepsilon_r + 2} = \frac{N\alpha_e}{3\varepsilon_0} \;\Rightarrow\; \alpha_e = \frac{3\varepsilon_0}{N}\cdot\frac{\varepsilon_r - 1}{\varepsilon_r + 2}$$ $$\begin{aligned} \frac{3\varepsilon_0}{N} &= \frac{3\times8.854\times10^{-12}}{6\times10^{28}} = 4.427\times10^{-40} \\ \frac{\varepsilon_r - 1}{\varepsilon_r + 2} &= \frac{10.9}{13.9} = 0.7842 \\ \alpha_e &= 4.427\times10^{-40}\times0.7842 = 3.47\times10^{-40}\ \text{F m}^2 \end{aligned}$$

Answer (i): $\alpha_e \approx 3.47\times10^{-40}\ \text{F m}^2$ per Si atom.

(If the given figure is read strictly as valence electrons, the atom density is $6\times10^{28}/4 = 1.5\times10^{28}$ m⁻³ and the polarizability per atom becomes $1.39\times10^{-39}$ F m².)

ii) Local field compared with applied field

For a sample with electrodes on opposite faces, the Lorentz local field is

$$E_{loc} = E + \frac{P}{3\varepsilon_0} = E + \frac{(\varepsilon_r - 1)E}{3} \;\Rightarrow\; \frac{E_{loc}}{E} = \frac{\varepsilon_r + 2}{3}$$ $$\frac{E_{loc}}{E} = \frac{11.9 + 2}{3} = \frac{13.9}{3} = 4.63$$

Answer (ii): the local field is about 4.63 times the applied field.

Dielectrics are chosen by their permittivity, loss, strength and temperature range. The main fields of application are:

Summary by function

• Insulation: gases, oils, porcelain, polymers (prevent current flow, high dielectric strength).
• Energy storage: capacitors use mica, paper, ceramics and plastic films (high $\varepsilon_r$, low loss).
• Transducers: piezo/pyroelectric materials convert mechanical or thermal energy to electrical.

HCl is a polar molecule with a permanent dipole moment $p_0$. In a field $E$, a dipole at angle $\theta$ to the field has potential energy

$$U = -p_0E\cos\theta$$

Thermal motion randomizes the dipoles, so the number of dipoles in an energy state follows the Boltzmann distribution, $\propto e^{-U/kT} = e^{p_0E\cos\theta/kT}$.

Average of $\cos\theta$

The number of dipoles in the solid angle between $\theta$ and $\theta + d\theta$ is proportional to $2\pi\sin\theta\,d\theta\, e^{x\cos\theta}$, where $x = p_0E/kT$.

$$\langle\cos\theta\rangle = \frac{\int_0^{\pi}\cos\theta\, e^{x\cos\theta}\sin\theta\,d\theta}{\int_0^{\pi} e^{x\cos\theta}\sin\theta\,d\theta}$$

Put $u = \cos\theta$:

$$\langle\cos\theta\rangle = \frac{\int_{-1}^{1}u\,e^{xu}du}{\int_{-1}^{1}e^{xu}du} = \coth x - \frac{1}{x} = L(x)$$

$L(x)$ is the Langevin function.

Small-field approximation

At normal fields $p_0E \ll kT$ ($x \ll 1$), so $L(x) \approx x/3$:

$$\langle\cos\theta\rangle = \frac{p_0E}{3kT}$$

Average dipole energy

$$\begin{aligned} \langle U\rangle &= -p_0E\langle\cos\theta\rangle \\ &= -p_0E\cdot\frac{p_0E}{3kT} \\ &= -\frac{p_0^2E^2}{3kT} \end{aligned}$$

Related results

• Average dipole moment along the field: $p_{av} = p_0\langle\cos\theta\rangle = \dfrac{p_0^2E}{3kT}$
• Orientational polarizability: $\alpha_d = \dfrac{p_0^2}{3kT}$

The average energy is negative (alignment lowers energy), grows as $E^2$, and falls as $1/T$ because heat opposes alignment.