Electromagnetic Field Theory and Capacitors

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. $P = \varepsilon0 \varepsilonr E = \varepsilon E$

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

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.

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 $\frac{D{1t}}{\varepsilon1} = \frac{D{2t}}{\varepsilon2}$

Field Component	Any Two Media	Conductor-Dielectric Boundary
	Medium 1: Dielectric $\varepsilon_1$
Tangential E	$E<em>{1t} = E</em>{2t}$	$E<em>{1t} = E</em>{2t} = 0$
Tangential D	$\frac{D<em>{1t}}{\varepsilon</em>1} = \frac{D<em>{2t}}{\varepsilon</em>2}$	$D<em>{1t} = D</em>{2t} = 0$
Normal E	$\varepsilon<em>1 E</em>{1n} - \varepsilon<em>2 E</em>{2n} = \rho_s$	$E<em>{1n} = \frac{\rho</em>s}{\varepsilon_1}$
$E_{2n} = 0$
Normal D	$D<em>{1n} - D</em>{2n} = \rho_s$	$D<em>{1n} = \rho</em>s$
$D_{2n} = 0$

$\rho_s$ 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.

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:
- $E1 = E1 \cos(\theta1) \hat{x} + E1 \sin(\theta_1) \hat{z}$
- $E2 = E2 \cos(\theta2) \hat{x} - E2 \sin(\theta_2) \hat{z}$

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

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

$\rhos \neq 0; \sigma1 \neq 0; \sigma_2 \neq 0$
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: $\hat{n}2 \cdot (D1 - D2) = \rhos$ : $-\varepsilon1 E1 \sin(\theta1) + \varepsilon2 E2 \sin(\theta2) = \rho_s$
$\rhos = E2 \sin(\theta2) \left( \frac{\sigma1 \varepsilon2}{\sigma2} - \varepsilon_1 \right)$

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

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 $\varepsilon$ , $RC = \frac{\varepsilon}{\sigma}$

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: $V{br} = E{ds} d$

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)