Conductors and Dielectrics

Why it matters

Conductors and dielectrics are fundamental in designing and analyzing electronic circuits and devices. Understanding their properties helps in predicting how materials will behave in electric fields, which is crucial for applications ranging from capacitors to transmission lines.

Key ideas

• Conductors: Materials that allow the flow of electric charge with minimal resistance. In conductors, free electrons move easily under the influence of an electric field.

  • Properties: High electrical conductivity, low resistivity.
  • Examples: Metals like copper, aluminum.

• Dielectrics: Insulating materials that do not conduct electricity but can be polarized by an electric field.

  • Properties: High resistivity, low conductivity, characterized by dielectric constant (permittivity).
  • Examples: Glass, plastic, ceramics.

• Polarization: In dielectrics, the electric field causes a shift in the position of bound charges, leading to induced dipoles.

• Permittivity (ε): The constitutive coefficient relating electric displacement D to E in a linear isotropic medium. It is the product of the permittivity of free space (ε₀) and the relative permittivity (εᵣ).

• Conductivity (σ): A measure of a material's ability to conduct electric current.

Formulas

In electrostatic equilibrium an ideal conductor has zero internal electric field and is equipotential. Real dielectrics can have small leakage and finite breakdown strength. The scalar constitutive relations below assume linear isotropic media.

• J = σ·E

  • J: Current density (A/m²)
  • σ: Conductivity (S/m)
  • E: Electric field (V/m)

• D = ε·E

  • D: Electric displacement field (C/m²)
  • ε: Permittivity (F/m)
  • E: Electric field (V/m)

• ε = ε₀·εᵣ

  • ε: Permittivity of the material (F/m)
  • ε₀: Permittivity of free space (8.854 x 10⁻¹² F/m)
  • εᵣ: Relative permittivity (dimensionless)

Worked example

Given: A dielectric material with a relative permittivity of 4 is subject to an internal electric field of 100 V/m. Calculate the electric displacement field.

1. Identify the given values:

   • εᵣ = 4
   • E = 100 V/m
   • ε₀ = 8.854 x 10⁻¹² F/m

2. Calculate the permittivity (ε):

   • Formula: ε = ε₀·εᵣ
   • Calculation: ε = 8.854 x 10⁻¹² F/m · 4 = 3.5416 x 10⁻¹¹ F/m

3. Calculate the electric displacement field (D):

   • Formula: D = ε·E
   • Calculation: D = 3.5416 x 10⁻¹¹ F/m · 100 V/m = 3.5416 x 10⁻⁹ C/m²

Final Answer: 3.5416 x 10⁻⁹ C/m²

Common mistakes

• Confusing conductivity with resistivity.
• Forgetting to convert units, especially when dealing with permittivity and electric fields.
• Misapplying the concept of polarization in dielectrics.

For GATE EC

Questions often involve calculating electric fields, displacement fields, and understanding the behavior of materials in electric fields. Practice problems on permittivity, conductivity, and polarization effects.

Quick check

1. What is the primary difference between conductors and dielectrics?
2. How does the relative permittivity affect the electric field in a dielectric?
3. What is the unit of electric displacement field?

Answers: 1. Conductors allow electric charge flow; dielectrics do not but can be polarized. 2. For fixed free charge in an ideal filled parallel-plate capacitor, larger permittivity reduces E; for fixed voltage and geometry E = V/d remains unchanged. 3. C/m².