Electromagnetic Fields and Waves: Notes, Solved Problems and Interactive Lab

3.4.11 Recap

Recap

Matter answers the field. Bound charges $ρ_{b} = - \nabla \cdot P$ and $ρ_{S b} = P \cdot a_{n}$ , with $P$ the dipole moment per unit volume (3.4.7).
Flux density. $D = ε_{0} E + P$ , and in a linear isotropic medium $P = ε_{0} χ E$ and $D = ε E$ , $ε = ε_{0} ε_{r}$ , $ε_{r} = 1 + χ > 1$ (3.4.10). $D$ counts free charge, $ε_{0} E$ all charge.
Field equations $\nabla \times E = 0$ , $\nabla \cdot D = ρ_{v}$ are universal, while $D = ε E$ is a constitutive relation. $E = - \nabla V$ , $\nabla \cdot (ε \nabla V) = - ρ_{v}$ , and $\nabla^{2} V = - ρ_{v} / ε$ only in a simple medium.
Dielectric strength $E_{bd}$ : the largest field a material withstands (air $\approx 3$ MV/m).
Boundary conditions: $E_{1 τ} = E_{2 τ}$ and $D_{2 n} - D_{1 n} = ρ_{S}$ (free charge, $a_{n}$ from 1 to 2). Refraction $tan θ_{2} / tan θ_{1} = ε_{2} / ε_{1}$ for $ρ_{S} = 0$ .
Conductors: $E = 0$ and $V =$ const inside, charge on the surface, $E_{τ} = 0$ and $D_{n} = ρ_{S}$ just outside. A closed conductor shields an empty cavity.
Uniqueness and images: a potential that satisfies Poisson's equation and the boundary conditions is the solution. $q$ above a grounded plane $\equiv$ $q$ and $- q$ (valid above the plane only), with $Q_{ind} = - q$ and $F = q^{2} / (16 π ε_{0} h^{2})$ .
Energy: $W = \frac{1}{2} \int ρ_{v} V d v = \frac{1}{2} \int D \cdot E d v$ , $w = \frac{1}{2} ε E^{2}$ , and for a capacitor $W = \frac{1}{2} Q V = \frac{1}{2} C V^{2} = Q^{2} /2 C$ .
Capacitance $C = Q / V$ depends only on geometry and $ε$ : $εS / d$ , $2 π ε L / ln (b / a)$ , $4 π ε ab / (b - a)$ , $4 π ε a$ . Layers $∥$ plates: series, side by side: parallel.