Electromagnetic Field Theory and Capacitors

Dielectric Materials and Electric Fields

Polarization

• When an external electric field E_{ext} is applied to a dielectric material, the material becomes polarized.
• Involves the alignment of electric dipoles within the material.
• Atom behavior:
  • Without an external electric field (E_{ext} = 0), the center of the electron cloud coincides with the nucleus.
  • With an external electric field (E_{ext} \neq 0), the electron cloud shifts, creating an electric dipole.
  • This shift results in positive and negative surface charges on the material.
• Polarization Field P: Represents the electric flux density induced by the electric field E. P = \varepsilon0 \varepsilonr E = \varepsilon E

Electric Breakdown

• Dielectric Strength (E_{ds}): The maximum electric field magnitude that a material can withstand without experiencing breakdown.
• Table of Relative Permittivity and Dielectric Strength of Common Materials:
  • Air (at sea level): \varepsilonr = 1.0006, E{ds} = 3 MV/m
  • Petroleum oil: \varepsilonr = 2.1, E{ds} = 12 MV/m
  • Polystyrene: \varepsilonr = 2.6, E{ds} = 20 MV/m
  • Glass: \varepsilonr = 4.5-10, E{ds} = 25-40 MV/m
  • Quartz: \varepsilonr = 3.8-5, E{ds} = 30 MV/m
  • Bakelite: \varepsilonr = 5, E{ds} = 20 MV/m
  • Mica: \varepsilonr = 5.4-6, E{ds} = 200 MV/m
• Permittivity: \varepsilon = \varepsilon0 \varepsilonr and \varepsilon_0 = 8.854 \times 10^{-12} F/m

Joule’s Law

• Power Dissipated: The power dissipated in a volume due to an electric field E and current density J is given by: P = \intV E \cdot J dV = \intS E \cdot I dS
• For a Resistor: Joule’s law simplifies to P = VI
• Homogeneous Conductor:
  • Constant cross-section A.
  • Uniform electric field E.

Boundary Conditions

• Perfect Dielectric: J = 0
• Perfect Conductor: E = 0 (Net electric field inside a conductor is zero).
• Electric field direction is always perpendicular to the conductor surface.
• General Equations
  • D = \varepsilon E
  • D{1n} = \rhos
  • E_{1t} = 0

Boundary Conditions for Different Media

• Normal component of D changes abruptly at a charged boundary between two different media by an amount equal to the surface charge density.
  • {\hat{w}2} \cdot (D1 - D2) = \rhos (C/m²)
  • D{1n} - D{2n} = \rho_s (C/m²)
• Tangential components of the Electric field are the same
  • E{1t} = E{2t}, or \frac{D{1t}}{\varepsilon1} = \frac{D{2t}}{\varepsilon2}

Summary of Boundary Conditions

• \rho_s is the surface charge density at the boundary.
• Normal components of E1, D1, E2, and D2 are along {\hat{n}_2}, the outward normal unit vector of medium 2.
• Remember E = 0 in a good conductor.

General Rule reminder

• \sigma1 J{1n} = \sigma2 J{2n}
• J{1t} = J{2t}

Boundary Conditions: Scenarios

Scenario Setup

• Two media are separated by an interface with an electric field E_1 in medium 1.
• Need to find E_2 given the properties of both media.
• Medium 1: Permittivity \varepsilon1, conductivity \sigma1
• Medium 2: Permittivity \varepsilon2, conductivity \sigma2
• Angle between E1 and the interface normal: \theta1
• Angle between E2 and the interface normal: \theta2
• Electric Field Vector Equations:
  • E1 = E1 \cos(\theta1) \hat{x} + E1 \sin(\theta_1) \hat{z}
  • E2 = E2 \cos(\theta2) \hat{x} - E2 \sin(\theta_2) \hat{z}

Scenario 1: Medium 1 is a Perfect Dielectric, Medium 2 is a Perfect Conductor

• \rhos \neq 0; \sigma1 = 0; \sigma_2 = \infty
• E_2 = 0
• J_1 = 0
• \hat{n}2 \cdot (D1 - D2) = \rhos: -\varepsilon1 E1 \sin(\theta1) + \varepsilon2 E2 \sin(\theta2) = \rho_s
• \rhos = -\varepsilon1 E1 \sin(\theta1)

Scenario 2: Medium 1 is a Perfect Dielectric, Medium 2 is a Perfect Dielectric

• \rhos = 0; \sigma1 = 0; \sigma_2 = 0
• Tangential components are equal: E{1t} = E{2t} \Rightarrow E1 \cos(\theta1) = E2 \cos(\theta2)
• Normal components of displacement vector: \hat{n}2 \cdot (D1 - D2) = 0 \Rightarrow -\varepsilon1 E1 \sin(\theta1) + \varepsilon2 E2 \sin(\theta_2) = 0
• \frac{E1}{E2} = \frac{\cos(\theta2)}{\cos(\theta1)} = \frac{\varepsilon2 \sin(\theta2)}{\varepsilon1 \sin(\theta1)}
• \frac{\tan(\theta2)}{\tan(\theta1)} = \frac{\varepsilon1}{\varepsilon2}

Scenario 3: Medium 1 is a Lossy Dielectric, Medium 2 is a Lossy Dielectric

• \rhos \neq 0; \sigma1 \neq 0; \sigma_2 \neq 0
• Tangential E field is continuous: E1 \cos(\theta1) = E2 \cos(\theta2)
• Normal J is continuous: \sigma1 E1 \sin(\theta1) = \sigma2 E2 \sin(\theta2)
• \frac{E1}{E2} = \frac{\cos(\theta2)}{\cos(\theta1)} = \frac{\sigma2 \sin(\theta2)}{\sigma1 \sin(\theta1)}
• \frac{\tan(\theta2)}{\tan(\theta1)} = \frac{\sigma1}{\sigma2}
• s field behavior: \hat{n}2 \cdot (D1 - D2) = \rhos: -\varepsilon1 E1 \sin(\theta1) + \varepsilon2 E2 \sin(\theta2) = \rho_s
• \rhos = E2 \sin(\theta2) \left( \frac{\sigma1 \varepsilon2}{\sigma2} - \varepsilon_1 \right)

Capacitance

• Definition: The capacitance C of a two-conductor configuration is defined as: C = \frac{Q}{V} (C/V or F)
• Where:
  • Q is the charge
  • V is the voltage

Calculating Capacitance

• General Formula:
  • Q{incl} = \ointS \varepsilon E \cdot dS
  • V = - \int_{ab} E \cdot dl
• Capacitance Formula:
  • C = \frac{\oint_S \varepsilon E \cdot dS}{-\int E \cdot dl} (F)
• Resistance Formula:
  • R = \frac{\int E \cdot dl}{\int \sigma E \cdot dS} (\Omega)
• RC Product: For a medium with uniform \sigma and \varepsilon, RC = \frac{\varepsilon}{\sigma}

Example Parellel Plate Capacitor

• Assumptions:
  • Neglect fringing effects (A >> d).
• Relationship
  • Q = \varepsilon E A
  • V = - \int (-E) \cdot dz = Ed
  • C = \frac{Q}{V} = \frac{\varepsilon A}{d}
  • V = E d , At breakdown: V{br} = E{ds} d

Example: Coaxial Line

• Electric Field: The electric field between the conductors is: E = -\hat{r} \frac{Q}{2 \pi \varepsilon \rho l}
• Potential Difference: The potential difference V between the outer and inner conductors is:
  • V = - \inta^b E \cdot dl = - \inta^b -\hat{r} \frac{Q}{2 \pi \varepsilon \rho l} \cdot (\hat{r} d\rho) = \frac{Q}{2 \pi \varepsilon l} \ln(\frac{b}{a})
• Capacitance: The capacitance C is then given by:
  • C = \frac{Q}{V} = \frac{2 \pi \varepsilon l}{\ln(b/a)}
• Capacitance per Unit Length: C' = \frac{C}{l} = \frac{2 \pi \varepsilon}{\ln(b/a)} (F/m)