Griffiths Ch 2: Electrostatics

We get the first two of Maxwell's Equations in this chapter!
Electric fieldWe begin with Coulomb's law:
\vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{s^2} \hat{s} dq
E⃗(r⃗)=14πϵ0∫1s2s^dq\vec{E}(\vec{r}) = \frac{1}{4\pi \epsilon_0} \int \frac{1}{s^2} \hat{s} dqE(r)=4πϵ0​1​∫s21​s^dq
where the separation vector 
 s = \vec{r} - \vec{r'} s=r⃗−r′⃗ s = \vec{r} - \vec{r'} s=r−r′
. This can be turned into a volume, surface or line integral with  dq = \rho \: d\tau = \sigma \: d\vec{a} = \lambda \: d\vec{l} dq=ρ dτ=σ da⃗=λ dl⃗ dq = \rho \: d\tau = \sigma \: d\vec{a} = \lambda \: d\vec{l} dq=ρdτ=σda=λdl
, where  \rho, \sigma, \lambda ρ,σ,λ \rho, \sigma, \lambda ρ,σ,λ
 are the corresponding charge densities.
This form isn't exactly in the chapter, but we also have the force law 
 \vec{F} = Q \vec{E} F⃗=QE⃗ \vec{F} = Q \vec{E} F=QE
 for electric force, and more generally  \vec{F} = Q(\vec{E} + \vec{v} \times \vec{B}) F⃗=Q(E⃗+v⃗×B⃗) \vec{F} = Q(\vec{E} + \vec{v} \times \vec{B}) F=Q(E+v×B)
 to include magnetic force, which will come later.
From this we get Maxwell's equations for the divergence and curl of electric fields.
Gauss' law:
\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0} \implies \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}
∮E⃗⋅da⃗=Qencϵ0  ⟹  ∇⃗⋅E⃗=ρϵ0\oint \vec{E} \cdot d\vec{a} = \frac{Q_{enc}}{\epsilon_0} \implies \vec{\nabla} \cdot \vec{E} = \frac{\rho}{\epsilon_0}∮E⋅da=ϵ0​Qenc​​⟹∇⋅E=ϵ0​ρ​
and the curl of an electric field is always 0:
\oint \vec{E} \: d \vec{l} = 0 \implies \vec{\nabla} \times \vec{E} = \vec{0}
∮E⃗ dl⃗=0  ⟹  ∇⃗×E⃗=0⃗\oint \vec{E} \: d \vec{l} = 0 \implies \vec{\nabla} \times \vec{E} = \vec{0}∮Edl=0⟹∇×E=0
Electric potentialThe scalar potential 
 V(\vec{r}) = -\int_O^{\vec{r}} \vec{E} \cdot d \vec{l} V(r⃗)=−∫Or⃗E⃗⋅dl⃗ V(\vec{r}) = -\int_O^{\vec{r}} \vec{E} \cdot d \vec{l} V(r)=−∫Or​E⋅dl
 is path independent. Note that  \vec{E} = -\vec{\nabla} V \implies \vec{\nabla} ^2 V = -\frac{\rho}{\epsilon_0} E⃗=−∇⃗V  ⟹  ∇⃗2V=−ρϵ0 \vec{E} = -\vec{\nabla} V \implies \vec{\nabla} ^2 V = -\frac{\rho}{\epsilon_0} E=−∇V⟹∇2V=−ϵ0​ρ​
, and for a region of no charge  \vec{\nabla}^2 V = 0 ∇⃗2V=0 \vec{\nabla}^2 V = 0 ∇2V=0
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Generally for a charge distribution,
V(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{1}{s} dq
V(r⃗)=14πϵ0∫1sdqV(\vec{r}) = \frac{1}{4 \pi \epsilon_0} \int \frac{1}{s} dqV(r)=4πϵ0​1​∫s1​dq
with 
 s, dq s,dq s, dq s,dq
  same as for Coulomb's law.
Work and energyThe energy of a charge distribution can be calculated by the work it takes to bring in each charge. Individually 
 W = \int_a^b \vec{F} \cdot d\vec{l} = Q[V(b) - V(a)] W=∫abF⃗⋅dl⃗=Q[V(b)−V(a)] W = \int_a^b \vec{F} \cdot d\vec{l} = Q[V(b) - V(a)] W=∫ab​F⋅dl=Q[V(b)−V(a)]
; with a reference point at infinity,  W = QV(\vec{r}) W=QV(r⃗) W = QV(\vec{r}) W=QV(r)
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Generally 
 W = \frac{1}{2} \int \rho V d\tau W=12∫ρVdτ W = \frac{1}{2} \int \rho V d\tau W=21​∫ρVdτ
, or over all space,  W = \frac{\epsilon_0}{2} \int E^2 d\tau W=ϵ02∫E2dτ W = \frac{\epsilon_0}{2} \int E^2 d\tau W=2ϵ0​​∫E2dτ
.
ConductorIn a conductor, electrons are free to roam. They have the following properties:
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 \vec{E} = 0 E⃗=0 \vec{E} = 0 E=0
 inside because charges move to induce a field in opposition to whatever field is applied
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 \rho = 0 ρ=0 \rho = 0 ρ=0
 inside due to Gauss' law. Then all net charge must be on the surface of the conductor.
Also, 
 \vec{E} E⃗ \vec{E} E
 must be perpendicular to the surface of the conductor, so from Gauss' law
\vec{E}  = \frac{\sigma}{\epsilon_0} \hat{n}
E⃗=σϵ0n^\vec{E}  = \frac{\sigma}{\epsilon_0} \hat{n}E=ϵ0​σ​n^
on the surface. Also, 
 \sigma = -\epsilon_0 \frac{\partial V}{\partial n} σ=−ϵ0∂V∂n \sigma = -\epsilon_0 \frac{\partial V}{\partial n} σ=−ϵ0​∂n∂V​
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CapacitorsCapacitance is defined as 
 C = \frac{Q}{V} C=QV C = \frac{Q}{V} C=VQ​
. For parallel plates,  C = \frac{A \epsilon_0}{d} C=Aϵ0d C = \frac{A \epsilon_0}{d} C=dAϵ0​​
 for plates of area A with separation from each other d.
The work it takes to charge a capacitor up to a potential V is 
 W = \frac{1}{2} CV^2 W=12CV2 W = \frac{1}{2} CV^2 W=21​CV2
.
Griffiths Ch 2: Electrostatics

Electric field

Electric potential

Work and energy

Conductor

Capacitors

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