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Electrostatics

Gauss's Law¶

\begin{matrix} Φ = Q_{e n c} \\ Φ = \oint_{S} \vec{D} \cdot d \vec{S} = \int_{v} \nabla \cdot \vec{D} d v = \int_{v} ρ_{v} d v \\ \nabla \cdot \vec{D} = ρ_{v} \end{matrix}

Electric Field¶

Electric dipole¶

Dipole moment: $\vec{p}$

\begin{matrix} \vec{p} = Q \vec{d} \\ V = \frac{Q}{4 π ε} \cdot \frac{d \cos θ}{r^{2}} = \frac{\vec{p} \cdot \hat{a_{r}}}{4 π ε r^{2}} = \frac{\vec{p} \cdot \vec{r}}{4 π ε r^{3}} \end{matrix}

note that $\hat{a_{r}}$ is unit vector.

\begin{aligned} \vec{E} & = - \nabla V \\ = - [\frac{\partial V}{\partial r} \hat{r} + \frac{1}{r} \frac{\partial V}{\partial θ} \hat{θ}] \\ = \frac{p}{4 π ε r^{3}} (2 \cos θ \hat{r} + \sin θ \hat{θ}) \end{aligned}

Polarization¶

Polarization vector: $\vec{P}$

\begin{matrix} \vec{P} \equiv χ_{e} ε_{0} \vec{E} \end{matrix}

surface bound charge density: $ρ_{p s}$
- the direction $\hat{a_{n}}$ is outgoing the volume which contain charge

\begin{matrix} ρ_{p s} = \vec{P} \cdot \vec{a_{n}} \end{matrix}

volume bound charge density: $ρ_{p v}$

\begin{matrix} ρ_{p v} = - \nabla \cdot \vec{P} \end{matrix}

Electric flux density field $D$

\begin{aligned} ρ_{v} & = ρ_{total} - ρ_{p v} \\ ⟹ \nabla \cdot \vec{D} & = \nabla \cdot (ε_{0} \vec{E}) - (- \nabla \cdot \vec{P}) \\ ⟹ \vec{D} & = ε_{0} \vec{E} + \vec{P} \\ = ε_{0} (1 + χ_{e}) \vec{E} \\ = ε_{r} ε_{0} \vec{E} \\ = ε \vec{E} \end{aligned}

Electric Potential¶

\begin{aligned} V_{A B} & = - \int_{A}^{B} \vec{E} \cdot d \vec{l} = V_{B} - V_{A} \\ = \int_{A}^{B} d V = \int_{A}^{B} \vec{\nabla} V \cdot d \vec{l} \end{aligned}

implies,

\begin{matrix} \vec{E} = - \nabla V \end{matrix}

by Stokes' Theorem

\begin{matrix} \oint_{L} \vec{E} \cdot d \vec{l} = \int_{S} (\nabla \times \vec{E}) \cdot d \vec{S} = 0 \\ ⟹ \nabla \times \vec{E} = 0 \end{matrix}

Electric Energy¶

$W_{E}$ : Energy stored in the capacitor
discrete charges

\begin{matrix} W_{E} = \frac{1}{2} \sum Q_{k} V_{k} \end{matrix}

continuous charges

\begin{aligned} W_{E} & = \int_{v} ρ_{v} V (\vec{r}) d v \\ = \frac{1}{2} \int_{v} (\nabla \cdot \vec{D}) V d v = \frac{1}{2} (\int_{v} \nabla \cdot (\vec{D} V) d v - \int_{v} \vec{D} \cdot (\nabla V) d v) \\ = \frac{1}{2} (\oint_{S} V \vec{D} \cdot d \vec{S} + \int_{v} \vec{D} \cdot \vec{E} d v) \end{aligned}

consider $S = U$ (Universe), then

\begin{matrix} \oint_{S} V \vec{D} \cdot d \vec{S} \to 0 \\ ⟹ W_{E} = \frac{1}{2} \int_{U} \vec{D} \cdot \vec{E} d v \end{matrix}

Conduction Current¶

Definition: Charge flows due to $\vec{E}$
force on electron $F_{e}$

\begin{matrix} {\vec{F}}_{e} = - e \vec{E} \end{matrix}

friction $F_{r}$

\begin{matrix} {\vec{F}}_{r} = \frac{Δ p}{Δ t} = - \frac{m \vec{u}}{τ} \end{matrix}

\begin{matrix} {\vec{F}}_{e} + {\vec{F}}_{r} = 0 \\ ⟹ \frac{m \vec{u}}{τ} = - e \vec{E} \end{matrix}

mean free time $τ$
average drift velocity $\vec{u}$

\begin{matrix} \vec{u} = - \frac{e τ}{m} \vec{E} = - μ_{e} \vec{E} \end{matrix}

Current Density¶

current view:

\begin{matrix} \vec{J} = \frac{\partial I}{\partial \vec{S}} \end{matrix}

convection current view:

\begin{matrix} \vec{J} = ρ_{v} \vec{u} = (- n e) (- \frac{e τ}{m}) \vec{E} \end{matrix}

Ohm's law view:

\begin{matrix} ∵ I = Y V \\ \vec{J} = σ \vec{E} \end{matrix}

conductivity $σ$

Continuity Equation¶

\begin{matrix} I_{n e t} = - \frac{d}{d t} Q = - \frac{d}{d t} \int_{v} ρ_{v} d v \\ I_{n e t} = \oint_{S} \vec{J} \cdot \vec{S} = \int_{v} \nabla \cdot \vec{J} d v \\ ⟹ \nabla \cdot \vec{J} = - \frac{d}{d t} ρ_{v} \end{matrix}

note

collapse: true

Recall that the physical meaning of divergence,

\begin{matrix} \nabla \cdot \vec{A} 0 & (source) \\ \nabla \cdot \vec{A} < 0 & (sink) \end{matrix}

It is reasonable to write $- ρ_{v}^{'}$ for a charge source.

Also intuitive for steady current have

\begin{matrix} \nabla \cdot \vec{J} = 0 \end{matrix}

Relaxation Time¶

relaxation time $τ$
view from solving differential equation

\begin{matrix} \nabla \cdot \vec{J} = - \frac{d}{d t} ρ_{v} = \nabla \cdot (σ \vec{E}) = \frac{σ}{ε} \nabla \cdot \vec{D} = \frac{σ}{ε} ρ_{v} \\ - \frac{σ}{ε} d t = \frac{d ρ_{v}}{ρ_{v}} \\ ⟹ ρ_{v} = ρ_{o} \exp (- \frac{σ}{ε} t) = ρ_{o} e^{- t / τ} \\ ⟹ τ = \frac{ε}{σ} \end{matrix}

view from having $τ = R C$

\begin{matrix} τ = R C = \frac{V}{I} \frac{Q}{V} = \frac{\int ε \vec{E} \cdot d \vec{S}}{\int σ \vec{E} \cdot d \vec{S}} = \frac{ε}{σ} \end{matrix}

Boundary Condition¶

for there is no surface charge density

\begin{matrix} \nabla \cdot (ρ_{S} \vec{a_{n}}) = 0 \\ ⟹ \nabla \cdot \vec{D} = 0 \\ \vec{J} = σ \vec{E} \\ ⟹ \vec{\nabla} \cdot \vec{J} = 0 \end{matrix}

implies

\begin{matrix} D_{1 n} = D_{2 n} \\ J_{1 n} = J_{2 n} \end{matrix}

tangent direction

for the closed-loop in conservative field, we have

\begin{matrix} ∵ \oint_{c} \vec{E} \cdot d \vec{l} = 0 \end{matrix}

then

\begin{matrix} {\vec{E}}_{1} = {\vec{E}}_{2} \\ E_{1 n} = E_{2 n} = 0 \\ ⟹ E_{1 t} = E_{2 t} \end{matrix}

for $\vec{J} = σ \vec{E}$

\begin{matrix} \frac{J_{1 t}}{σ_{1}} = \frac{J_{2 t}}{σ_{2}} \end{matrix}