Page 331 - Intro to Tensor Calculus

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§2.6 ELECTRIC AND MAGNETIC FIELDS
Introduction
In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most
often encountered. In this section the equations will be given in the mks system of units. If you want the
equations in the Gaussian system of units make the replacements given in the column 3 of Table 1.

Table 1. MKS AND GAUSSIAN UNITS

MKS MKS Replacement GAUSSIAN
symbol units symbol units
~
~
E (Electric field) volt/m E statvolt/cm
~
B (Magnetic field) weber/m 2 ~ B gauss
c
~
D (Displacement field) coulomb/m 2 ~ D statcoulomb/cm 2
4π
~
H (Auxiliary Magnetic field) ampere/m c ~ H oersted
4π
~
J (Current density) ampere/m 2 J ~ statampere/cm 2
~
A (Vector potential) weber/m ~ A gauss-cm
c
V (Electric potential) volt V statvolt
(Dielectric constant)
4π
4πµ
µ (Magnetic permeability) 2
c
Electrostatics
~
A basic problem in electrostatic theory is to determine the force F on a charge Q placed a distance r
from another charge q. The solution to this problem is Coulomb’s law
1 qQ
~
F = b e r (2.6.1)
4π 0 r 2
2
2
where q, Q are measured in coulombs, 0 =8.85 × 10 −12 coulomb /N · m is called the permittivity in a
~
vacuum, r is in meters, [F] has units of Newtons and b e r is a unit vector pointing from q to Q if q, Q have
~
~
the same sign or pointing from Q to q if q, Q are of opposite sign. The quantity E = F/Q is called the
~
~
electric field produced by the charges. In the special case Q =1, we have E = F and so Q = 1 is called
a test charge. This tells us that the electric field at a point P can be viewed as the force per unit charge
exerted on a test charge Q placed at the point P. The test charge Q is always positive and so is repulsed if
q is positive and attracted if q is negative.
The electric field associated with many charges is obtained by the principal of superposition. For
example, let q 1 ,q 2 ,...,q n denote n-charges having respectively the distances r 1 ,r 2 ,...,r n from a test charge
Q placed at a point P. The force exerted on Q is
~ ~ ~ ~
F =F 1 + F 2 + ··· + F n

1 q 1 Q q 2 Q q n Q
~
F = 2 b e r 1 + 2 b e r 2 + ··· + 2 b e r n
4π 0 r 1 r 2 r n (2.6.2)
n
~
F 1 X q i
~
~
or E = E(P)= = 2 b e r i
Q 4π 0 r
i=1 i