Methods for solving electrostatic problems

The multipole expansion

For a charge distribution located near the origin we have

with

.

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.

In spherical coordinates we have

,

with

.

ql-m=(-1)mqlm*

The qlm are called the multipole moments

Relationship between Cartesian and spherical multipoles:

            

The energy of a charge distribution near the origin in an external field is given by

.

This expansion shows how the various multipoles interact with the external field.

The method of images

Consider a charge q at  .
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 f=0.
Assume a grounded conducting sphere of radius R is centered on the origin.  Then placing an image charge on the z-axis makes the sphere an equipotential with f=0.  Add q’’ at the center of the sphere to change the potential or the total charge on the sphere.
Consider a line charge l parallel to the x-axis at .
Assume a conducting cylinder of radius R has as its axis the x-axis.  Then placing an image charge parallel to the x-axis makes the sphere an equipotential with    .   Add V0 everywhere to change the potential of the cylinder.
Assume the z = 0 plane is a plane interface between two dielectrics
Consider a charge q at   in e1. Then placing an image charge  at  d’=-d   on the z-axis in a medium with e1   gives the potential and field in dielectric 1.  And placing an image charge at d’’=d  on the z-axis  in a medium with e2  gives the potential and field in dielectric 2.

Boundary value problems

Let in spherical coordinates. Then

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Let in cylindrical coordinates.  Then

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

Useful mathematical formulas

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