2 Electrostatics

In the equations below, ζ=1 in SI units and ζ=4π in Gaussian units.

2.1 Fundamental laws

e211=Grid[{{"SI","Gaussian"},
{"curlE = 0 ","curl E = 0"},
{"div D = ρ ","div D = 4πρ"}},
Frame->All]//TraditionalForm

eq214=Grid[{{"SI","Gaussian"},
{"(∇^2)Φ = - ρ/ϵ","(∇^2)Φ = - 4πρ/ϵ"}},Frame->All]//TraditionalForm

eq215="Φ=- 1/4π ∫V (∇^2)Φ 1/r dV + 1/4π ∫S[ 1/r ∇Φ=Φ∇(1/r)]▪dS"//TraditionalForm

eq216=Grid[{{"SI","Gaussian"},
{"Φ = 1/4πϵ  ∫  =\!\(\*FractionBox[\(\(ρ\)\(\\ \)\(dV\)\(\\ \)\), \(r\)]\)","Φ = 1/ϵ  ∫  =\!\(\*FractionBox[\(\(ρ\)\(\\ \)\(dV\)\(\\ \)\), \(r\)]\)"}},
Frame->All]//TraditionalForm

eq217=Grid[{{"SI","Gaussian"},
{"Φ =\!\(\*FractionBox[\(\(q\)\(\\ \)\), \(4  πϵr\)]\)","Φ =\!\(\*FractionBox[\(\(q\)\(\\ \)\), \(ϵr\)]\)"}},
Frame->All]//TraditionalForm

eq57=Grid[{{"","SI","Gaussian"},
{"polarization","P=D-ϵ0E","P=1/4π ( D=ϵ0 E}"},
{"magnetization","M=B/μ0-H","P=1/4π ((B/μ0-H)}"},
{"magn susceptivility","χm = M/H","χm=M/H"},
{"elec susceptbility","χe=P/ϵ0E","χe=P/ϵ0 E"}},Frame->All]

eq121="J=σE"
eq122="R=L/σA = lρ/A"

eq141=Grid[{{"field equations","SI","Gaussian"},
{"conservation of charge","divJ=-∂ρ/∂t","divJ=-∂ρ/∂t"},
{"Faraday's law","curlE = - ∂B/∂t","curlB = - 1/c∂B/∂t"},
{"Ampere's law","curlH = J + ∂D/∂t","curlH = 4π/c J +1/c ∂D/∂t"},
{"Gauss's law","divD = ρ","divD=4πρ"},
{"Gauss's law","divB=0","divB=0"}},Frame->All]

eq151=Grid[{{"SI","Gaussian"},
{"(OverHat[n])•(B2-B1) = 0","(OverHat[n])•(B2-B1) = 0"},
{"(OverHat[n])×(H2-H1) = K","(OverHat[n])×(H2-H1) = 4π/c K"},
{"(OverHat[n])×(E2-E1) = 0","(OverHat[n])•(E2-E1) = 0"},
{"(OverHat[n])▪(D2-D1) = σ","(OverHat[n])▪(D2-D1) = 4π σ"}},Frame->All]//TraditionalForm

eq153=Grid[{{"SI"},
{"(∇^2)Φ = ϵμ (∂^2)(∂^2) - ∂∂(divA+ϵμ∂∂)-ρ/ϵ"},
{"Gaussian"},
{"(∇^2)Φ = ϵμ/(c^2) (∂^2)(∂^2) - 1/c ∂∂(divA+ϵμ/c ∂∂)-4πρ/ϵ"}},
Frame->All]//TraditionalForm

eq154=Grid[{{"SI"},
{"(∇^2)A = ϵμ (∂^2)(∂^2) + ∇(divA+ϵμ∂∂)-μJ"},
{"Gaussian"},
{"(∇^2)A = ϵμ/(c^2) (∂^2)(∂^2) + ∇(divA+ϵμ/c ∂∂)-4πμ/c J"}},
Frame->All]//TraditionalForm

eq171=Grid[
{{"SI","Gaussian"},
{"div E=1/ϵ0 ( ρ - div P)","div E=4π/ϵ0 ( ρ - div P)"}},
Frame->All]//TraditionalForm

eq181=Grid[
{{"SI","Gaussian"},
{"curl B=μ0 ( J + curl M)","curl E=4πμ0 ( J + curl M)"}},
Frame->All]//TraditionalForm

2.2 Fields of some simple charge distributions

2.2a Line of charge

eq221="Φ=–\!\(\*FractionBox[\(\(ζτ\)\(\\ \)\), \(2  πϵ\)]\) ln R"//TraditionalForm

2.2b Sphere of charge

eq222=StringJoin["Φ=\!\(\*FractionBox[\(\(ζq\)\(\\ \)\), \(4  πϵr\)]\) for r ≥ a ;",
"     Φ=\!\(\*FractionBox[\(\(ζq\)\(\\ \)\), \(4  πϵa\)]\) for r ≤ a ."]//TraditionalForm

2.2c Ring of charge

eq223a=StringJoin["Φ=\!\(\*FractionBox[\(\(ζq\)\(\\ \)\), \(4  πϵd\)]\) Σ Pn Pn for r > d ;"]//TraditionalForm
eq223b=StringJoin["Φ=\!\(\*FractionBox[\(\(ζq\)\(\\ \)\), \(4  πϵr\)]\) Σ Pn Pn for r < d ."]//TraditionalForm

Figure 1

2.3 Electric multipoles; the double layer

eq233="Φ0 = ζ/4πϵ ▫ p0/r (monopole)"//TraditionalForm

eq234="Φ1 = - ζ/4πϵ ▫ p1 ∇(1/r) p0/r (dipole)"//TraditionalForm

eq235="Φ2 = - ζ/4πϵ ▫ p2/2 u1▪∇[ u0▪∇(1/r) p0/r (quadrupole)"//TraditionalForm

Electrostatic boundary values problems

2.5a Point charge and a conducting sphere

StringJoin["Φ= ζ/(4πϵ) [ q/r2 + r1φ2/r - q Σ Pn(cosθ)/(r^(n + 1)) (r1^(2  n + 1))/(Z^(n + 1))",
" for r>r1"]//TraditionalForm
StringJoin["Φ= (Subscript[φ, s]) =  ζ/(4πϵ) (q1/r1+q/Z)"," for r<r1"]//TraditionalForm

2.5b Point charge and a dielectric sphere

StringJoin["Φ= ζ/(4πϵ) [ q/r2 + q Σ [..] Pn(cosθ)/(r^(n + 1)) (r1^(2  n + 1))/(Z^(n + 1))",
" for r>r1"]//TraditionalForm
StringJoin["Φ=  ζ/(4πϵ) q Σ [..] (r^n)/(Z^(n + 1)) Pn(cosθ](q1/r1+q/Z)"," for r<r1"]//TraditionalForm

2.5c Conducting spehre in a uniform electric field

eq5c=StringJoin["Φ= -(Subscript[E, 0]) r cosθ + (Subscript[E, 0])\!\(\*FractionBox[\(\*SubsuperscriptBox[\(r\), \(1\), \(3\)]\(\\ \)\(cosθ\)\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\) +\!\(\*FractionBox[\(\*SubscriptBox[\(ζq\), \(1\)]\(\\ \)\), \(4  πϵr\)]\)",
" ⟵ for r>(Subscript[r, 1])"]//TraditionalForm
eq5cb=StringJoin["Φ=  (Subscript[ζq, 1])/(4πϵ (Subscript[r, 1])) "," ⟵ for r<r1"]//TraditionalForm

Electrostatic boundary values problems

2.5a Point charge and a conducting sphere

StringJoin["Φ= ζ/(4πϵ) [ q/r2 + r1φ2/r - q Σ Pn(cosθ)/(r^(n + 1)) (r1^(2  n + 1))/(Z^(n + 1))",
" for r>r1"]//TraditionalForm
StringJoin["Φ= (Subscript[φ, s]) =  ζ/(4πϵ) (q1/r1+q/Z)"," for r<r1"]//TraditionalForm

2.5b Point charge and a dielectric sphere

StringJoin["Φ= ζ/(4πϵ) [ q/r2 + q Σ [..] Pn(cosθ)/(r^(n + 1)) (r1^(2  n + 1))/(Z^(n + 1))",
" for r>r1"]//TraditionalForm
StringJoin["Φ=  ζ/(4πϵ) q Σ [..] (r^n)/(Z^(n + 1)) Pn(cosθ](q1/r1+q/Z)"," for r<r1"]//TraditionalForm

2.5c Conducting sphere in a uniform electric field

eq5c=StringJoin["Φ= -(Subscript[E, 0]) r cosθ + (Subscript[E, 0])\!\(\*FractionBox[\(\*SubsuperscriptBox[\(r\), \(1\), \(3\)]\(\\ \)\(cosθ\)\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\) +\!\(\*FractionBox[\(\*SubscriptBox[\(ζq\), \(1\)]\(\\ \)\), \(4  πϵr\)]\)",
" ⟵ for r>(Subscript[r, 1])"]//TraditionalForm
eq5cb=StringJoin["Φ=  (Subscript[ζq, 1])/(4πϵ (Subscript[r, 1])) "," ⟵ for r<r1"]//TraditionalForm

2.5d Dielectric sphere in a uniform electric field

eq5d=StringJoin["(Subscript[Φ, 0]) = -(Subscript[E, 0]) r cosθ"]//TraditionalForm
eq5db=StringJoin["Φ = -(Subscript[E, 0]) r cos θ + ",
"((Subscript[ϵ, 1])-ϵ)/((Subscript[ϵ, 1])+2ϵ)((Subscript[r, 1])^3) (Subscript[E, 0])\!\(\*FractionBox[\(\(cosθ\)\(\\ \)\), SuperscriptBox[\(r\), \(2\)]]\)","for r > (Subscript[r, 1]) "]//TraditionalForm 
eq5dc=StringJoin["Φ = - (3ϵ)/(Subscript[ϵ, 1])+2ϵ) (Subscript[E, 0]) r cos θ ",
"for r < (Subscript[r, 1]) "]//TraditionalForm

2.5 The method of images

For an electrostatic system with a regularly shaped conductor or dielectric in hte field of a simple distrbution of chrge, it is often possible to construct a potential distribnu which satisfies tje boundary conditiopnms outsidfe th conducti0n or dielectric by replacing the dielectric or conduction by a distribution of charges located inside its boundaries. Since the field so constructed satisfies 
Poisson;s equation as well as rth boundary conditions, it must represent the actural field oiutside the conductor. Such oimage chartge distributions are called image charges.

Examples

(1) Point charge outside an infinite grounded plane conductor
fig 2 image charge = q’ = -q.

(2) Point charge outside a pair of intersecting grounded plane conductors
Fig 3

(3) Point charge outside a conducting sphere
Fig 4
q’ = image charge = - aq/[OQ]
[OQ]=a^2/[OQ]

(4) Line of charge outside a conducting cylinder
Fig 5
τ’ = - τ  and [OR]=(a^2)/[OQ]

(5) Point charge near a plane boundary between two semi-infinite dielectrics
Fig 6
q’’ = ((2k)/(k+1)) q
q’ = ((k-1)/(k+1)) q

2.7 Capacitors

A capacitro is a system consisting with two insulated conductiors bearing equal and opposite charge. Let ΔΦ be the potential difference between the two conductors, and let q be the charge magnitude on each conductor. Then the capacitance is defined tyo be C = q/ΔV.

The capacitance of a single conductor is defione as the limit of the capacitance of a system, of two such conductors as their separation becomes infinite.

The work required to charge a capacitor is
W = 1/2 C (((ΔV))^2) = 1/2 (q^(2/C))

On the equation follolwing, ζ = 1 for SI units, or ζ = 4 π  for Gaussian units.

2.7a Parallel plates

StirngJoin["C =\!\(\*FractionBox[\(\(ϵA\)\(\\ \)\), \(ζA\)]\)"," assuming A ≫ 
(d^2).

2.7a Parallel plates

StringJoin["C =\!\(\*FractionBox[\(\(ϵA\)\(\\ \)\), \(ζd\)]\)"," assuming A ≫ (d^2)."]//TraditionalForm

2.7 b Concentric speheres

StringJoin["C =\!\(\*FractionBox[\(\(4\)  \(π\)\(\\ \)\(ϵ\)\(\\ \)\), \(ζ \((\\ 1/\*SubscriptBox[\(r\), \(1\)] - \\ 1/\*SubscriptBox[\(r\), \(2\)])\)\)]\)"]//TraditionalForm

2.7c Concentric cylinders

StringJoin["C =\!\(\*FractionBox[\(\(4\)  \(π\)\(\\ \)\(ϵ\)\(\\ \)\(L\)\(\\ \)\), \(ζ\\ 2\\ ln \((\*SubscriptBox[\(r\), \(2\)]/\*SubscriptBox[\(r\), \(1\)])\)\)]\)"," assuming L ≫ (Subscript[r, 2])-(Subscript[r, 1])"]//TraditionalForm

2.7d Parallel curcular cylinedrs

D = separation of the centers; positive when the cylinders are external, negative when one cylined is inside the other.

StringJoin["C =(1/2)\!\(\*FractionBox[\(\(4\)  \(π\)\(\\ \)\(ϵ\)\(\\ \)\(L\)\(\\ \)\), \(ζ\)]\)","{arccosh[]}^(-1)"," assuming L ≫ D and L ≫ |(Subscript[r, 1])-(Subscript[r, 2])|"]//TraditionalForm

2.7e circular cylinedrs of equal radii

StringJoin["C =X"]//TraditionalForm

2.7f cylinedrs and an infinite plane

StringJoin["C =X"]//TraditionalForm

2.7g Curcular disk

StringJoin["C =\!\(\*FractionBox[\(\(4\)  \(πϵ\)\(\\ \)\), \(ζ\)]\)\!\(\*FractionBox[\(\(2\)  \(r\)\(\\ \)\), \(π\)]\)"]//TraditionalForm

2.8 Normal stress on a conductor

Consider a element of surface dA of a conductor. Lert the charge density on dA be σ. There will be an outward for dF normal to dA:

eq28=Grid[
{{"SI units","Gaussian units"},
{"dF =(1/2 σ/ϵ)(σ dA)","dF = (4πσ/2ϵ) (σ dA)"}},Frame->All]//TraditionalForm

Therefore the normal stress on the surface is

eq281=Grid[
{{"SI units","Gaussian units"},
{"dF/dA = (σ^2)/2ϵ","dF/dA = 4((πσ)^2)/2ϵ"}},Frame->All]//TraditionalForm

2.9 Energy density of the electrostatic field

eq291=Grid[{
{"SI","Gaussian"},
{"W = ∫ ((E▪D)/2) dV","W = ∫ ((E▪D)/(8  π)) dV"}},Frame->All]//TraditionalForm

That is, the energy density of the field is

eq292=Grid[{
{"SI","Gaussian"},
{"u = ((E▪D)/2)","u = ((E▪D)/(8  π))"}},Frame->All]//TraditionalForm