## Electrostatics

The fundamental equations of electrostatics are linear equations,
·E = ρ/ε0×E = 0, (SI units).
The principle of superposition holds.

The electrostatic force on a particle with charge q at position r is F = qE(r).
×E = 0 <==> E = -Φ,  ∇2Φ = -ρ/ε0.
Φ is the electrostatic potential.

### Important formulas:

The field at r due to a point charge at r':  E(r) = [1/(4πε0)] q(r - r')/|r - r'|3.
The field of a charge distribution:
E(r) = [1/(4πε0)][∫v' dV' ρ(r')(r - r')/|r - r'|3 + ∫A' dA' σ(r')(r - r')/|r - r'|3
+ ∫l' dl' λ(r')(r - r')/|r - r'|3 + ∑iqi(r - ri)/|r - ri|3].
(We consider volume, surface, and line charge distributions and point charges.)

The potential at r due to a point charge at r':  Φ(r) = [1/(4πε0)] q/|r - r'|.
The potential of a charge distribution:
Φ(r) = [1/(4πε0)][∫v' dV' ρ(r')/|r - r'| + ∫A' dA' σ(r')/|r - r'|
+ ∫l' dl' λ(r')/|r - r'| + ∑iqi/|r - ri|].

Gauss' law: closed surface dA = Qinside0 .
In situations with enough symmetry, Gauss' law alone can be used to find E.

The electrostatic energy of a charge distribution:
U = ½∫Φ(r)dq(r) = ½[∫vdV ρ(r)Φ(r) + ∫AdA σ(r)Φ(r) + ∫l dl λ(r)Φ(r) + ∑iqiΦ(ri)],
or, for a continuous charge distribution, U = (ε0/2)∫all space E·E dV.

### Dipoles

The field of a dipole at the origin:  E(r) = [1/(4πε0)](1/r3)[3(p·r)r/r2 - p].
The potential of a dipole at the origin:  Φ(r) = [1/(4πε0)](p·r)/r3.
The force on a dipole:  F = (p·E).
The torque on a dipole:  τ = p×E.
The energy of a dipole in an external field:  U = -p·E.

### Properties of conductors in electrostatics

• E = 0 inside.
• ρ = 0 inside.
• Any excess charge resides on the surface.
• E on the surface is perpendicular to the surface.
• E = (σ/ε0)n just outside the surface.

### Boundary conditions in electrostatics

(E2 - E1n2 = (σ/ε0),
(E2 - E1t = 0,
(D2 - D1n2 = σf,
Φ is continuous across the boundary.

### Methods for solving electrostatic problems

The multipole expansion

For a charge distribution located near the origin we have
Φ(r) = [1/(4πε0)][(1/r)[∫v' dV' ρ(r') + (1/r2 )(r/r)·[∫v' dV' ρ(r')r'
+ (1/r3 )[∫v' dV'ρ(r')(3(r·r')2/r2 - r'2)/2 + ... ]
= 1/(4πε0)][Q/r) + p·(r/r)/r2 + (1/(2r5))∑ij3xixjQij + ... ],
with p = ∫v' dV' ρ(r')r' and
Qij = ∫(xi'xj' - (1/3)δijr'2) ρ(r')dV',  Qij = Qji,  ∑iQii = 0.

Here Q = total charge, p = dipole moment, Qij = quadrupole moment tensor of the charge distribution.  If the problem has rotational symmetry about the z-axis, such that Qxx = Qyy = -½Qzz, then Qzz is called the quadrupole moment.

The energy of a charge distribution near the origin in an external field is given by
W = qΦ(0) - p·E(0) - ½∑ijQij ∂Ej/∂xi|xi=0 + ... .
This expansion shows how the various multipoles interact with the external field.

### The method of images

Consider a charge q at z = d on the z - axis.

• Assume the z = 0 plane is a grounded conducting plane.  Then placing an image charge q' = -q at  d' = -d on the z-axis makes the z = 0 plane an equipotential with Φ = 0.
• Assume a grounded conducting sphere of radius R < d is centered on the origin.  Then placing an image charge q' = -qR/d at d' = R2/d on the z-axis makes the sphere an equipotential with Φ = 0.  Add q'' at the center of the sphere to change the potential or the total charge on the sphere.

Consider a line charge λ parallel to the x-axis at d = d k.

• Assume a conducting cylinder of radius R < d has as its symmetry axis the x-axis.  Then placing an image line charge λ' = -λ at d' k with d' = R2/d parallel to the x-axis makes the cylinder an equipotential with Φ = [λ/(2πε0)]ln(R/d).  Add V0 everywhere to change the potential of the cylinder.

### The uniqueness theorem

When solving electrostatic problems, we often rely on the uniqueness theorem.  If the charge density ρ is specified throughout a volume V, and Φ or its normal derivatives are specified at the boundaries of a volume V, then a unique solution exists for Φ inside V.

### Boundary value problems

In regions with ρ = 0 we have ∇2Φ = 0.  We are looking for a solution for ∇2Φ = 0, subject to boundary conditions.
Let Φ =  Φ(r,θ),  0 ≤ θ ≤ π  in spherical coordinates, i.e. let the problem have azimuthal symmetry.
Then the most general solution for ∇2Φ = 0 is
Φ(r,θ) = ∑n=0[Anrn + Bn/rn+1]Pn(cosθ).

Let Φ = Φ(ρ,φ),  0 ≤ θ ≤ 2π  in cylindrical coordinates.
Then the most general solution for ∇2Φ = 0 is
Φ(ρ,φ) = ∑n=1(Ancos(nφ) +  Bnsin(nφ))ρn + ∑n=1(A'ncos(nφ) +  B'nsin(nφ))ρ-n + a0 + b0lnρ.