Concepts and formulas

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.


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


imageBoundary 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.

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


The method of images with dielectrics

Assume the z = 0 plane is a plane interface between two dielectrics
Consider a charge q at z = d on the z-axis in ε1.  Then placing an image charge q' = -q(ε2 - ε1)/(ε2 + ε1) at z' = -d on the z-axis in a medium with ε1 gives the potential and field in dielectric 1.  And replacing q with an image q'' = q(2ε2)/(ε2 + ε1) charge at z''= d  on the z-axis in a medium with ε2 gives the potential and field in dielectric 2.


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.