Rajesh Kumar

Dielectric Properties of Materials

by Rajesh Kumar 13 min read

Macroscopic Dielectric Constant

Learning Objectives:

  • Understand the concept of macroscopic dielectric constant in materials.
  • Explore the relationship between electric field, polarization, and dielectric behavior.
  • Apply the concept to calculate fields, capacitance, and material properties.

Key Concepts / Definitions:

  • Dielectric Constant ($\varepsilon_r$): The ratio of the permittivity of a material to the permittivity of free space. It measures the material’s ability to reduce the electric field: \(\varepsilon_r = \frac{\varepsilon}{\varepsilon_0}\)
  • Polarization ($\vec{P}$): The dipole moment per unit volume of a dielectric, induced by an external electric field.
  • Electric Susceptibility ($\chi_e$): A dimensionless quantity that describes how easily a material can be polarized: \(\vec{P} = \varepsilon_0 \chi_e \vec{E}\)
  • $E$ (Macroscopic Electric Field): The average electric field in the dielectric material due to both external sources and the material’s polarization.
  • $E_{\text{ext}}$ (External Field): The electric field applied from outside the dielectric, typically produced by free charges (e.g., on capacitor plates), before considering the dielectric response.
  • $E_{\text{pol}}$ (Polarization Field): The electric field arising from the polarized bound charges within the dielectric material; it generally opposes the external field.
  • $E_{\text{local}}$ (Local Field): The actual microscopic electric field experienced by an individual molecule or atom, including contributions from both the external field and nearby polarized molecules.

Theory and Explanation:

When an external electric field is applied to a dielectric material, the bound charges within atoms or molecules slightly displace, creating induced dipoles. The collective effect of these dipoles leads to a net polarization $\vec{P}$ in the material.

This polarization generates an internal electric field that partially cancels the applied field, resulting in a reduced macroscopic electric field $\vec{E}$ inside the material.

To describe the behavior of a dielectric, we define the electric displacement field $\vec{D}$: \(\vec{D} = \varepsilon_0 \vec{E} + \vec{P}\)

In linear, isotropic, and homogeneous dielectrics, polarization is directly proportional to the electric field: \(\vec{P} = \varepsilon_0 \chi_e \vec{E}\)

Substituting this into the equation for $\vec{D}$ gives: \(\vec{D} = \varepsilon_0 (1 + \chi_e) \vec{E} = \varepsilon \vec{E}\)

Thus, the permittivity of the dielectric is: \(\varepsilon = \varepsilon_0 (1 + \chi_e)\)

And the dielectric constant (relative permittivity) is: \(\varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = 1 + \chi_e\)

This quantity indicates how much the presence of the dielectric reduces the internal electric field compared to vacuum.

Mathematical Formulation:

  • Electric displacement field: \(\vec{D} = \varepsilon_0 \vec{E} + \vec{P}\)

  • Polarization in linear media: \(\vec{P} = \varepsilon_0 \chi_e \vec{E}\)

  • Displacement field in terms of dielectric constant: \(\vec{D} = \varepsilon \vec{E}\)

  • Dielectric constant: \(\varepsilon_r = \frac{\varepsilon}{\varepsilon_0} = 1 + \chi_e\)

  • Capacitance with dielectric: \(C = \varepsilon_r \cdot C_0 = \varepsilon_r \cdot \frac{\varepsilon_0 A}{d}\)

Where:

  • $C_0$ = Capacitance in vacuum
  • $A$ = Area of capacitor plates
  • $d$ = Distance between plates

Solved Examples:

  • Example 1:
    Problem: A parallel-plate capacitor has vacuum between the plates and a capacitance of 10 pF. What will be its capacitance if a dielectric with dielectric constant $\varepsilon_r = 4$ is inserted?
    Solution:
    \(C = \varepsilon_r \cdot C_0 = 4 \cdot 10 = 40 \, \text{pF}\)
    Answer: 40 pF
  • Example 2:
    Problem: A dielectric has a polarization $\vec{P} = 2 \times 10^{-7} \, \text{C/m}^2$ and is subjected to an electric field $E = 500 \, \text{V/m}$. Find its dielectric constant.
    Solution:
    \(\chi_e = \frac{P}{\varepsilon_0 E} = \frac{2 \times 10^{-7}}{8.85 \times 10^{-12} \times 500} \approx 45.2\) \(\varepsilon_r = 1 + \chi_e = 1 + 45.2 = 46.2\)
    Answer: $\varepsilon_r \approx 46.2$

Important Points / Summary:

  • The dielectric constant describes the material’s ability to reduce an internal electric field.
  • It is dimensionless and always greater than or equal to 1.
  • A higher $\varepsilon_r$ implies better electric insulation and greater capacitance.

Practice Questions:

  • Short Answer:
    1. Define the macroscopic dielectric constant and explain how it differs from permittivity.
    2. What is the physical meaning of electric susceptibility?
  • Numerical:
    1. A dielectric has $\chi_e = 2.5$. Calculate its dielectric constant.
    2. A capacitor has plate area $0.01 \, \text{m}^2$ and plate separation $2 \, \text{mm}$. Calculate its capacitance with $\varepsilon_r = 3$.
  • MCQs:
    1. The dielectric constant $\varepsilon_r$ is:
      • (a) less than 1
      • (b) equal to 1
      • (c) always greater than 1
      • (d) can be negative
        Answer: (c)
    2. The correct relation between $\vec{D}$, $\vec{E}$, and $\vec{P}$ is:
      • (a) $\vec{D} = \vec{E} + \vec{P}$
      • (b) $\vec{D} = \varepsilon_0 \vec{E} + \vec{P}$
      • (c) $\vec{D} = \varepsilon_r \vec{P}$
      • (d) $\vec{D} = \vec{P} - \varepsilon \vec{E}$
        Answer: (b)

The mechanisms of polarization include:

  1. Electronic Polarization:
    • Occurs in all dielectric materials.
    • Caused by the displacement of the electron cloud relative to the nucleus in atoms.
    • Dominant at high frequencies (e.g., optical range).
  2. Ionic Polarization:
    • Present in ionic crystals (e.g., NaCl).
    • Results from the relative displacement of positive and negative ions.
  3. Orientation Polarization:
    • Occurs in materials with permanent dipole moments.
    • Dipoles align with the external electric field.
    • Temperature-dependent and significant at lower frequencies.
  4. Space Charge Polarization:
    • Arises from charge accumulation at interfaces or grain boundaries.
    • Prominent at low frequencies.

Solved Examples:

  • Example 1:
    Problem: A dielectric with electric susceptibility $\chi_e = 2.5$ is placed in an electric field of magnitude $E = 10^5$ V/m. Find the polarization vector $\vec{P}$.
    Solution:
    Using $\vec{P} = \epsilon_0 \chi_e \vec{E}$:
    \(\vec{P} = (8.85 \times 10^{-12} \, \text{F/m})(2.5)(10^5 \, \text{V/m}) = 2.21 \times 10^{-6} \, \text{C/m}^2\)

  • Example 2:
    Problem: Determine the electric displacement $\vec{D}$ for a dielectric with $\epsilon_0 = 8.85 \times 10^{-12}$ F/m, $\chi_e = 3$, and $E = 2 \times 10^4$ V/m.
    Solution:
    First calculate $\vec{P}$:
    \(\vec{P} = \epsilon_0 \chi_e \vec{E} = (8.85 \times 10^{-12})(3)(2 \times 10^4) = 5.31 \times 10^{-7} \, \text{C/m}^2\)
    Then,
    \(\vec{D} = \epsilon_0 \vec{E} + \vec{P} = (8.85 \times 10^{-12})(2 \times 10^4) + 5.31 \times 10^{-7} = 7.08 \times 10^{-7} \, \text{C/m}^2\)

Practice Questions:

  • Short Answer:
    1. What is electronic polarization?
    2. How does temperature affect orientation polarization?
  • Numerical:
    1. A dielectric with $\chi_e = 4$ is placed in a field of $E = 3 \times 10^5$ V/m. Calculate $\vec{P}$.
    2. Given $\vec{E} = 10^4$ V/m and $\vec{P} = 1.77 \times 10^{-6}$ C/m², find $\vec{D}$.
  • MCQs:

    1. Which type of polarization is dominant in the optical frequency range?
      a) Ionic
      b) Electronic
      c) Orientation
      d) Space charge

    2. The unit of polarization vector $\vec{P}$ is:
      a) V/m
      b) C/m²
      c) F/m
      d) N/C

Clausius-Mossotti Equation

Learning Objectives:

  • Understand the relationship between microscopic polarizability and macroscopic dielectric constant.
  • Derive the Clausius-Mossotti equation.
  • Apply the equation to determine the polarizability of molecules in a dielectric.

Key Concepts / Definitions:

  • Polarizability ($\alpha$): The measure of how easily an electron cloud of a molecule is distorted by an external electric field.
  • Number Density ($N$): Number of molecules per unit volume in the dielectric material.

Theory and Explanation: The Clausius-Mossotti equation connects the microscopic property of molecules (polarizability $\alpha$) with the macroscopic property of the dielectric (relative permittivity $\epsilon_r$).

When an external electric field is applied, the molecules of the dielectric become polarized. The total polarization $\vec{P}$ depends on the polarizability $\alpha$ and the number of molecules per unit volume $N$.

\[\vec{P} = N \vec{p} = N \alpha \vec{E}_{\text{local}}\]

However, the field experienced by a molecule is not simply the applied field. It is modified due to the field created by other polarized molecules. This local field is often approximated using the Lorentz model, assuming the molecule is inside a spherical cavity within a uniformly polarized medium.

Step-by-Step Derivation:

  1. Macroscopic Polarization:

    From basic electromagnetism, polarization in a linear dielectric is given by:

    \[\vec{P} = \epsilon_0 \chi_e \vec{E}\]

    where $\vec{E}$ is the macroscopic field.

  2. Microscopic Polarization:

    For an individual molecule with polarizability $\alpha$, the induced dipole moment is:

    \[\vec{p} = \alpha \vec{E}_{\text{local}}\]

    Therefore, polarization becomes:

    \[\vec{P} = N \vec{p} = N \alpha \vec{E}_{\text{local}}\]

  3. Local Field Estimation:

    The local electric field is the sum of the macroscopic electric field \(\vec{E}\) and the field due to surrounding polarized molecules $\vec{E}_{\text{pol}}$.

    Using Lorentz’s method, the field at the center of a uniformly polarized spherical cavity is:

    \[\vec{E}_{\text{pol}} = \frac{\vec{P}}{3 \epsilon_0}\]

    Hence, the local field becomes:

    \[\vec{E}_{\text{local}} = \vec{E}+ \frac{\vec{P}}{3 \epsilon_0}\]

  4. Substituting into Microscopic Expression:

    Substituting $\vec{E}_{\text{local}}$ into the microscopic equation:

    \[\vec{P} = N \alpha \left( \vec{E} + \frac{\vec{P}}{3 \epsilon_0} \right)\]

  5. Solving for $\vec{P}$:

    Expand and isolate $\vec{P}$:

    \[\vec{P} = N \alpha \vec{E} + \frac{N \alpha}{3 \epsilon_0} \vec{P}\]

    Rearranged:

    \[\vec{P} \left(1 - \frac{N \alpha}{3 \epsilon_0} \right) = N \alpha \vec{E}\]

  6. Expressing $\vec{P}$:

    \[\vec{P} = \frac{N \alpha}{1 - \frac{N \alpha}{3 \epsilon_0}} \vec{E}\]

  7. Relating to Macroscopic Susceptibility:

    Since $\vec{P} = \epsilon_0 \chi_e \vec{E}$,

    \[\epsilon_0 \chi_e = \frac{N \alpha}{1 - \frac{N \alpha}{3 \epsilon_0}}\]

    Multiply both sides by the denominator:

    \[\epsilon_0 \chi_e \left(1 - \frac{N \alpha}{3 \epsilon_0}\right) = N \alpha\]

    Expand:

    \[\epsilon_0 \chi_e - \frac{N \alpha \chi_e}{3} = N \alpha\]

    Bring all terms to one side:

    \[\epsilon_0 \chi_e = N \alpha \left(1 + \frac{\chi_e}{3} \right)\]

  8. Solving for $\alpha$:

    \[\alpha = \frac{3 \epsilon_0}{N} \cdot \frac{\chi_e}{3 + \chi_e}\]

  9. In Terms of Relative Permittivity $\epsilon_r = 1 + \chi_e$:

    Replace $\chi_e = \epsilon_r - 1$:

    \[\alpha = \frac{3 \epsilon_0}{N} \cdot \frac{\epsilon_r - 1}{\epsilon_r + 2}\]

    This is the Clausius-Mossotti Equation.

Mathematical Formulation: The Clausius-Mossotti equation is:

\[\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{N \alpha}{3 \epsilon_0}\] \[\alpha = \frac{3 \epsilon_0}{N} \cdot \frac{\epsilon_r - 1}{\epsilon_r + 2}\]

Solved Examples:

  • Example 1:
    Problem: A gas has a relative permittivity $\epsilon_r = 1.0006$ and molecular density $N = 2.5 \times 10^{25}$ molecules/m³. Find the polarizability $\alpha$.
    Solution:
    Using: \(\alpha = \frac{3 \epsilon_0}{N} \cdot \frac{\epsilon_r - 1}{\epsilon_r + 2}\)
    \(\alpha = \frac{3 \times 8.85 \times 10^{-12}}{2.5 \times 10^{25}} \cdot \frac{0.0006}{1.0006 + 2}\)
    \(\alpha \approx \frac{2.655 \times 10^{-11}}{2.5 \times 10^{25}} \cdot \frac{0.0006}{3.0006}\)
    \(\alpha \approx 2.12 \times 10^{-40} \, \text{F·m}^2\)

  • Example 2:
    Problem: Determine the relative permittivity $\epsilon_r$ of a dielectric with polarizability $\alpha = 1 \times 10^{-39}$ F·m² and number density $N = 5 \times 10^{28}$ m⁻³.
    Solution:
    Using: \(\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{N \alpha}{3 \epsilon_0}\)
    Calculate RHS: \(\frac{(5 \times 10^{28}) (1 \times 10^{-39})}{3 \times 8.85 \times 10^{-12}} = \frac{5 \times 10^{-11}}{2.655 \times 10^{-11}} \approx 1.884\)
    So, \(\frac{\epsilon_r - 1}{\epsilon_r + 2} = 1.884\)
    Solve for $\epsilon_r$: \((\epsilon_r - 1) = 1.884 (\epsilon_r + 2)\)
    \(\epsilon_r - 1 = 1.884 \epsilon_r + 3.768\)
    \(-0.884 \epsilon_r = 4.768 \Rightarrow \epsilon_r \approx -5.39\)
    (Negative value indicates inconsistency—check units or values used; realistic $\epsilon_r$ should be > 1)

Important Points / Summary:

  • The Clausius-Mossotti equation links microscopic and macroscopic dielectric behavior.
  • Valid primarily for dilute gases and non-polar materials.
  • Fails when interactions between molecules are strong or in polar materials at high density.

Practice Questions:

  • Short Answer:
    1. What does the Clausius-Mossotti equation represent physically?
    2. List the assumptions made in deriving the Clausius-Mossotti relation.
  • Numerical:
    1. Calculate $\alpha$ for a dielectric with $\epsilon_r = 1.0008$ and $N = 1.5 \times 10^{25}$ m⁻³.
    2. Given $\alpha = 2 \times 10^{-40}$ F·m² and $N = 2 \times 10^{26}$ m⁻³, find $\epsilon_r$.
  • MCQs:

    1. The Clausius-Mossotti equation is applicable when:
      a) Material is metallic
      b) Intermolecular interactions are strong
      c) The medium is dilute and non-polar
      d) The dielectric is ferroelectric

    2. In the Clausius-Mossotti equation, the term $\alpha$ represents:
      a) Dielectric constant
      b) Electric susceptibility
      c) Molecular polarizability
      d) Local field intensity

Frequency Dependence of Polarizabilities, Dielectric Constant in Alternating Fields, and Clausius-Mossotti Catastrophe

Learning Objectives:

  • Understand how various types of polarizabilities respond to different frequency ranges and timescales.
  • Analyze the behavior of dielectric constant as a function of frequency, and interpret complex dielectric response.
  • Explore the Clausius-Mossotti relation and the physical significance of the conditions leading to the catastrophe.

Key Concepts / Definitions:

  • Clausius-Mossotti Catastrophe: A theoretical prediction from the Clausius-Mossotti equation where the dielectric constant becomes infinite if $N\alpha \to 3\epsilon_0$, signaling a breakdown of linear dielectric behavior or phase transition.

Frequency Dependence:

As the frequency of the applied alternating electric field increases, different polarization mechanisms in a dielectric respond differently depending on their intrinsic time scales. These mechanisms include:

(a) Electronic Polarizability

  • Origin: Displacement of the electron cloud relative to the nucleus in an atom or molecule.
  • Response Time: Very fast (~$10^{-15}$ s).
  • Active Range: Remains active up to optical and ultraviolet frequencies (~$10^{15}$ Hz).
  • Remarks: Since electrons are light and bound by strong restoring forces, they can respond to very high-frequency fields.

(b) Ionic Polarizability

  • Origin: Displacement of positive and negative ions relative to each other in ionic crystals.
  • Response Time: Moderate (~$10^{-13}$ to $10^{-14}$ s).
  • Active Range: Prominent in the infrared frequency range (~$10^{13}$ Hz).
  • Remarks: Ionic motion becomes too slow to respond at higher frequencies due to inertia.

(c) Dipolar (Orientation) Polarizability

  • Origin: Alignment of permanent dipole moments in molecules (e.g., H$_2$O, HCl) with the external field.
  • Response Time: Slow (~$10^{-9}$ to $10^{-12}$ s).
  • Active Range: Effective at microwave and radio frequencies (~$10^9$ Hz).
  • Remarks: Thermal agitation and molecular rotation limit the ability of dipoles to reorient at high frequency.

(d) Space Charge (Interfacial) Polarizability

  • Origin: Accumulation of charges at interfaces or grain boundaries in heterogeneous materials.
  • Response Time: Very slow (milliseconds or longer).
  • Active Range: Significant only at very low frequencies (below ~$10^3$ Hz).
  • Remarks: These charges cannot follow rapid field reversals due to low mobility.

As frequency increases:

  • At low frequencies: all polarization mechanisms contribute, so $\epsilon_r$ is large.
  • At intermediate frequencies: dipolar and space charge mechanisms cannot respond quickly, their contribution vanishes.
  • At high frequencies: only electronic (and sometimes ionic) polarizability remains.
  • At optical frequencies: dielectric constant reduces to $\epsilon_\infty$, corresponding to pure electronic polarization.

This results in a stepwise decrease in dielectric constant with increasing frequency, known as dielectric dispersion.

Dielectric Loss and Complex Permittivity:

In alternating fields, some energy is dissipated due to lag in polarization response. The dielectric constant becomes a complex quantity:

\[\epsilon(\omega) = \epsilon'(\omega) - i \epsilon''(\omega)\]
  • $\epsilon’(\omega)$: stores energy (capacitive behavior).
  • $\epsilon’’(\omega)$: represents energy loss (resistive behavior), also known as dielectric loss.

The loss tangent or dissipation factor is defined as:

\[\tan \delta = \frac{\epsilon''}{\epsilon'}\]

Clausius-Mossotti Relation and Catastrophe:

The Clausius-Mossotti equation connects microscopic polarizability $\alpha$ with macroscopic dielectric constant $\epsilon_r$:

\[\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{N \alpha}{3 \epsilon_0}\]

Rewriting:

\[\epsilon_r = \frac{1 + 2\left(\frac{N \alpha}{3 \epsilon_0}\right)}{1 - \left(\frac{N \alpha}{3 \epsilon_0}\right)}\]

This shows that:

  • As $N \alpha \to 3 \epsilon_0$, $\epsilon_r \to \infty$
  • This is the Clausius-Mossotti Catastrophe

Interpretation:

  • This condition implies that polarization grows uncontrollably.
  • Indicates a phase transition, such as the onset of ferroelectricity or dielectric breakdown.
  • Physically, the system can no longer support a linear dielectric response.

Mathematical Formulation:

  • Complex dielectric function: \(\epsilon(\omega) = \epsilon'(\omega) - i \epsilon''(\omega)\)
  • Clausius-Mossotti relation: \(\frac{\epsilon_r - 1}{\epsilon_r + 2} = \frac{N \alpha}{3 \epsilon_0}\)

Solved Examples:

  • Example 1:
    Problem: A dielectric has $\epsilon_s = 10$, $\epsilon_\infty = 2$, and $\tau = 10^{-6}$ s. Find $\epsilon(\omega)$ at $f = 10^6$ Hz.
    Solution:
    \(\omega = 2 \pi f = 2 \pi \times 10^6 \, \text{rad/s}\)
    \(\epsilon(\omega) = 2 + \frac{8}{1 + i (2\pi)}\)
    Rationalize the denominator:
    \(\epsilon(\omega) = 2 + \frac{8(1 - i 2\pi)}{1 + (2\pi)^2}\)
    Compute numeric real and imaginary parts for final values.

  • Example 2:
    Problem: Determine if Clausius-Mossotti catastrophe occurs for $N = 5 \times 10^{28}$ m⁻³, $\alpha = 1.6 \times 10^{-40}$ F·m².
    Solution:
    \(\frac{N \alpha}{3 \epsilon_0} = \frac{(5 \times 10^{28})(1.6 \times 10^{-40})}{3 \times 8.85 \times 10^{-12}} \approx 0.30\)
    Since the value < 1, no catastrophe. Catastrophe occurs when the ratio → 1.

Important Points / Summary:

  • Polarizability mechanisms respond over different frequency ranges; the total dielectric constant depends on which are active.
  • In AC fields, dielectric constant becomes complex and shows dispersion and loss.
  • The Clausius-Mossotti catastrophe signals the breakdown of linear dielectric behavior and may indicate phase transitions in materials.

Practice Questions:

  • Short Answer:
    1. Explain why $\epsilon’’(\omega)$ becomes zero at high frequencies.
    2. Why does $\epsilon_r$ decrease with increasing frequency?
  • Numerical:
    1. Calculate $\epsilon(\omega)$ for a material with $\epsilon_s = 12$, $\epsilon_\infty = 4$, and $\tau = 5 \times 10^{-7}$ s at $f = 1$ MHz.
    2. For a dielectric with $N = 4 \times 10^{28}$ m⁻³ and $\alpha = 2 \times 10^{-40}$ F·m², compute $\epsilon_r$ using the Clausius-Mossotti equation.
  • MCQs:

    1. At high frequencies (e.g., optical range), which type of polarization dominates?
      a) Dipolar
      b) Ionic
      c) Space charge
      d) Electronic

    2. The Clausius-Mossotti relation predicts divergence in $\epsilon_r$ when:
      a) $N \alpha = \epsilon_0$
      b) $N \alpha = 2 \epsilon_0$
      c) $N \alpha = 3 \epsilon_0$
      d) $N \alpha = 0$