Electrostatics

The fundamental equations of electrostatics are linear equations,

Ñ×E = r/e0,   Ñ´E = 0, (SI units),   Ñ×E = 4pr,   Ñ´E = 0, (Gaussian units).

The principle of superposition holds.

The electrostatic force on a particle with charge q at position r is F = qE(r).

Ñ´E = 0 <==> E = -Ñf.

f is the electrostatic potential.

Important formulas:

The field at r due to a point charge at r’:   

(Omit the factor to obtain the corresponding expressions in Gaussian units.)

The field of a charge distribution:

(We consider volume, surface, and line charge distributions and point charges.)

The potential at r due to a point charge at r’:     

The potential of a charge distribution:

Gauss’ law:

(SI units),    (Gaussian units).

In situations with enough symmetry, Gauss’ law alone can be used to find E.

The electrostatic energy of a charge distribution:

or, for a continuous charge distribution,

(SI units),      (Gaussian units).

Dipoles

The field of a dipole at the origin:

The potential of a dipole at the origin:

The force on a dipole:

F = Ñ(p×E)

The torque on a dipole:

t = p´E

The energy of a dipole in an external field:

U = -(p×E)

Properties of conductors in electrostatics

E = 0 inside.
r = 0 inside.
Any excess charge resides on the surface.
E on the surface is perpendicular to the surface.
(SI units),    (Gaussian units),  just outside the surface.

Boundary conditions in electrostatics

  (SI units),        

,                                

,                         

f is continuous across the boundary.   

 (Gaussian units),

,                                

,

f is continuous across the boundary.

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.

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  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 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.

Boundary value problems

In regions with r = 0 we have Ñ2f = 0.  We then are looking for a solution for Ñ2f = 0, subject to the boundary conditions.
Let in spherical coordinates, i.e. let the problem have azimuthal symmetry. Then the most general solution for Ñ2f = 0 is

.

Let in cylindrical coordinates.
Then the most general solution for Ñ2f = 0 is

.

The uniqueness theorem:

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