In electromagnetic theory the mks system of units and the Gaussian system of units are the ones most

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Oct 18, 2013 (3 years and 9 months ago)

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x2.6 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
Replacement
MKS
MKS GAUSSIAN
units units
symbol
symbol
~ ~
E (Electric eld) volt=m E statvolt=cm
~
B
2
~
B (Magnetic eld) weber=m gauss
c
~
2 D 2
~
D (Displacement eld) coulomb=m statcoulomb=cm
4
~
cH
~
H (Auxiliary Magnetic eld) ampere=m oersted
4
2 2
~ ~
J (Current density) ampere=m J statampere=cm
~
A
~
A (Vector potential) weber/m gauss-cm
c
V (Electric potential) volt V statvolt

(Dielectric constant)
4
4
(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 chargeq. The solution to this problem is Coulomb’s law
1 qQ
~
F = b e (2:6:1)
r
2
4 r
0
−12 2 2
where q, Q are measured in coulombs, =8:85 10 coulomb =N m is called the permittivity in a
0
~
b
vacuum, r is in meters, [F] has units of Newtons and e is a unit vector pointing from q to Q if q;Q have
r
~ ~
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 el d 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 eld at a point P can be viewed as the force per unit charge
exerted on a test chargeQ placed at the point P. The test chargeQ is always positive and so is repulsed if
q is positive and attracted if q is negative.
The electric el d associated with many charges is obtained by the principal of superposition. For
example, letq ;q ;:::;q denote n-charges having respectively the distancesr ;r ;:::;r from a test charge
1 2 n 1 2 n
Q placed at a pointP: The force exerted onQ is
~ ~ ~ ~
F =F +F + +F
1 2 n

1 q Q q Q q Q
1 2 n
~
F = b e + b e + + b e
r r r
1 2 n
2 2
2
4 r r r
0 (2:6:2)
1 2 n
n
~ X
F 1 q
i
~ ~
b
or E =E(P)= = e
r
i
2
Q 4 r
0
i
i=1326
~ ~
whereE =E(P) is the electric el d associated with the system of charges. The equation (2.6.2) can be gen-
eralized to other situations by de ning other types of charge distributions. We introduce a line charge density

2 3
, (coulomb=m), a surface charge density , (coulomb=m ), a volume charge density , (coulomb=m ),
then we can calculate the electric eld associated with these other types of charge distributions. For example,

if there is a charge distribution = (s) along a curve C,where s is an arc length parameter, then we
would have
Z
1 b e
r

~
E(P)= ds (2:6:3)
2
4 r
0
C
as the electric eld at a point P due to this charge distribution. The integral in equation (2.6.3) being a
line integral along the curve C and whereds is an element of arc length. Here equation (2.6.3) represents a
continuous summation of the charges along the curve C. For a continuous charge distribution over a surface
S, the electric eld at a point P is
ZZ
1 b e
r

~
E(P)= d (2:6:4)
2
4 r
0
S

where d represents an element of surface area on S. Similarly, if represents a continuous charge distri-
bution throughout a volumeV , then the electric eld is represented
ZZZ
b
1 e
r

~
E(P)= d (2:6:5)
2
4 r
0 V
whered is an element of volume. In the equations (2.6.3), (2.6.4), (2.6.5) we let (x;y;z) denote the position
0 0 0
of the test charge and let (x;y;z ) denote a point on the line, on the surface or within the volume, then
0 0 0
~r=(x−x )b e +(y−y )b e +(z−z )b e (2:6:6)
1 2 3
~r

represents the distance from the point P to an element of charge ds, d or d withr =j~rj and b e = :
r
r
~
If the electric eld is conservative, then r E = 0, and so it is derivable from a potential functionV
by taking the negative of the gradient ofV and
~
E =−rV: (2:6:7)
~
For these conditions note thatrV d~r =−E d~r is an exact dieren tial so that the potential function can
be represented by the line integral
Z
P
~
V =V(P)=− E d~r (2:6:8)

where is some reference point (usually in nity, where V(1) = 0). For a conservative electric eld the line
integral will be independent of the path connecting any two pointsa andb so that

Z Z Z Z
b a b b
~ ~ ~
V(b)−V(a)=− E d~r− − E d~r =− E d~r = rV d~r: (2:6:9)
a a
Let = 1 in equation (2.6.8), then the potential function associated with a point charge moving in
the radial direction b e is
r
Z Z
r r
−q 1 q 1 q
r
~
V(r)=− E d~r = dr = j = :
1
2
4 r 4 r 4 r
0 0 0
1 1327
By superposition, the potential at a pointP for a continuous volume distribution of charges is given by
ZZZ ZZ

1 1
V(P)= d and for a surface distribution of chargesV(P)= d and for a line
4 r 4 r
0 0
V S
Z

1
distribution of chargesV(P)= ds; and for a discrete distribution of point charges
4 r
0 C
N
X
1 q
i
V(P)= . When the potential functions are de ned from a common reference point, then the
4 r
0 i
i=1
principal of superposition applies.
The potential function V is related to the work done W in moving a charge within the electric el d.
The work done in moving a test chargeQ from point a to point b is an integral of the force times distance
~ ~ ~ ~
moved. The electric force on a test chargeQ is F =QE and so the forceF =−QE is in opposition to this
force as you move the test charge. The work done is
Z Z Z
b b b
~ ~
W = F d~r = −QE d~r =Q rV d~r =Q[V(b)−V(a)]: (2:6:10)
a a a
The work done is independent of the path joining the two points and depends only on the end points and
the change in the potential. If one moves Q from in nity to point b, then the above becomes W =QV (b):
~ ~
An electric eld E =E(P) is a vector eld which can be represented graphically by constructing vectors
at various selected points in the space. Such a plot is called a vector eld plot. A eld line associated with
a vector eld is a curve such that the tangent vector to a point on the curve has the same direction as the
vector eld at that point. Field lines are used as an aid for visualization of an electric eld and vector elds
~
in general. The tangent to a eld line at a point has the same direction as the vector eld E at that point.
For example, in two dimensions let~r =xb e +yb e denote the position vector to a point on a eld line. The
1 2
~ ~
tangent vector to this point has the directiond~r =dxb e +dyb e .IfE =E(x;y)=−N(x;y)b e +M(x;y)b e
1 2 1 2
~
is the vector eld constructed at the same point, then E andd~r must be colinear. Thus, for each point (x;y)
~
on a eld line we require that d~r = KE for some constant K. Equating like components we nd that the
eld lines must satisfy the di erential relation.
dx dy
= =K
−N(x;y) M(x;y)
(2:6:11)
or M(x;y)dx +N(x;y)dy =0:
In two dimensions, the family of equipotential curvesV(x;y)=C =constant, are orthogonal to the family
1
of eld lines and are described by solutions of the di erential equation
N(x;y)dx−M(x;y)dy =0
obtained from equation (2.6.11) by taking the negative reciprocal of the slope. The eld lines are perpendic-
ular to the equipotential curves because at each point on the curveV =C we haverV being perpendicular
1
~
to the curve V = C and so it is colinear with E at this same point. Field lines associated with electric
1
elds are called electric lines of force. The density of the eld lines drawn per unit cross sectional area are
proportional to the magnitude of the vector e ld through that area.328
Figure 2.6-1. Electric forces due to a positive charge at (−a;0) and negative charge at (a;0):
EXAMPLE 2.6-1.
Find the eld lines and equipotential curves associated with a positive charge q located at the point
(−a;0) and a negative charge−q located at the point (a;0):
~
Solution: With reference to the gure 2.6-1, the total electric force E on a test charge Q = 1 place
at a general point (x;y) is, by superposition, the sum of the forces from each of the isolated charges and is
~ ~ ~
E =E +E : The electric force vectors due to each individual charge are
1 2
kq(x +a)b e +kqyb e
1 2
2 2 2
~
E = with r =(x +a) +y
1
1
3
r
1
(2:6:12)
b b
−kq(x−a)e −kqy e
1 2
2 2 2
~
E = with r =(x−a) +y
2
2
3
r
2
1
where k = is a constant. This gives
4
0

kq(x +a) kq(x−a) kqy kqy
~ ~ ~
E =E +E = − b e + − b e :
1 2 1 2
3 3 3 3
r r r r
1 2 1 2
This determines the di erential equation of the eld lines
dx dy
= : (2:6:13)
kq(x+a) kq(x−a) kqy kqy

− 3 3
3 3
r r
r r
1 2
1 2
To solve this di erential equation we make the substitutions
x +a x−a
cos = and cos = (2:6:14)
1 2
r r
1 2329
Figure 2.6-2. Lines of electric force between two opposite sign charges.
as suggested by the geometry from g ure 2.6-1. From the equations (2.6.12) and (2.6.14) we obtain the
relations
r dx− (x +a)dr
1 1
− sin d =
1 1
2
r
1
2r dr =2(x +a)dx+2ydy
1 1
r dx− (x−a)dr
2 2
− sin d =
2 2
2
r
2
2r dr =2(x−a)dx+2ydy
2 2
which implies that
2
(x +a)ydy y dx
− sin d =− +
1 1
3 3
r r
1 1
(2:6:15)
2
(x−a)ydy y dx
− sin d =− +
2 2
3 3
r r
2 2
Now compare the results from equation (2.6.15) with the dieren tial equation (2.6.13) and determine that
y is an integrating factor of equation (2.6.13) . This shows that the dieren tial equation (2.6.13) can be
written in the much simpler form of the exact di erential equation
− sin d +sin d =0 (2:6:16)
1 1 2 2
in terms of the variables and : The equation (2.6.16) is easily integrated to obtain
1 2
cos − cos =C (2:6:17)
1 2
where C is a constant of integration. In terms ofx;y the solution can be written
x +a x−a
p p
− =C: (2:6:18)
2 2 2 2
(x +a) +y (x−a) +y
These eld lines are illustrated in the g ure 2.6-2.330
The di erential equation for the equipotential curves is obtained by taking the negative reciprocal of
the slope of the eld lines. This gives
kq(x−a) kq(x+a)

3 3
dy r r
2 1
= :
kqy kqy
dx

3 3
r r
1 2
This result can be written in the form

(x +a)dx +ydy (x−a)dx +ydy
− + =0
3 3
r r
1 2
which simpli es to the easily integrable form
dr dr
1 2
− + =0
2 2
r r
1 2
in terms of the new variables r andr . An integration produces the equipotential curves
1 2
1 1
− =C
2
r r
1 2
1 1
p p
or − =C :
2
2 2 2 2
(x +a) +y (x−a) +y
The potential function for this problem can be interpreted as a superposition of the potential functions
kq kq
V =− andV = associated with the isolated point charges at the points (−a;0) and (a;0):
1 2
r r
1 2
Observe that the electric lines of force move from positive charges to negative charges and they do not
cross one another. Where eld lines are close together the eld is strong and where the lines are far apart
the eld is weak. If the eld lines are almost parallel and equidistant from one another the eld is said to be
~
uniform. The arrows on the eld lines show the direction of the electric eld E. If one moves along a eld
line in the direction of the arrows the electric potential is decreasing and they cross the equipotential curves
~
at right angles. Also, when the electric eld is conservative we will have r E =0:
In three dimensions the situation is analogous to what has been done in two dimensions. If the electric
~ ~
eld is E =E(x;y;z)=P(x;y;z)b e +Q(x;y;z)b e +R(x;y;z)b e and~r =xb e +yb e +zb e is the position
1 2 3 1 2 3
~
vector to a variable point (x;y;z) on a eld line, then at this point d~r and E must be colinear so that
~
d~r =KE for some constantK. Equating like coe cients gives the system of equations
dx dy dz
= = =K: (2:6:19)
P(x;y;z) Q(x;y;z) R(x;y;z)
From this system of equations one must try to obtain two independent integrals, call them u (x;y;z)=c
1 1
and u (x;y;z)= c . These integrals represent one-parameter families of surfaces. When any two of these
2 2
~
surfaces intersect, the result is a curve which represents a eld line associated with the vector eld E: These
type of eld lines in three dimensions are more di cult to illustrate.
~
The electric flux of an electric eld E over a surface S is de ned as the summation of the normal
E
~
component of E over the surface and is represented
ZZ
2
Nm
~
= E n ^d with units of (2:6:20)
E
C
S331
^
where n is a unit normal to the surface. The flux can be thought of as being proportional to the number
E
of electric eld lines passing through an element of surface area. If the surface is a closed surface we have
by the divergence theorem of Gauss
ZZZ ZZ
~ ~
= r Ed = E n ^d
E
V S
where V is the volume enclosed by S:
Gauss Law
Letd denote an element of surface area on a surfaceS: A cone is formed if all points on the boundary
ofd are connected by straight lines to the origin. The cone need not be a right circular cone. The situation
is illustrated in the g ure 2.6-3.
Figure 2.6-3. Solid angle subtended by element of area.
We let~r denote a position vector from the origin to a point on the boundary ofd and let n ^ denote a
unit outward normal to the surface at this point. We then have n ^ ~r = r cos where r = j~rj and is the
angle between the vectors n ^ and~r. Construct a sphere, centered at the origin, having radiusr. This sphere

intersects the cone in an element of area dΩ. The solid angle subtended by d is de ned as d! = : Note
2
r
that this is equivalent to constructing a unit sphere at the origin which intersect the cone in an element of
aread!. Solid angles are measured in steradians. The total solid angle about a point equals the area of the
sphere divided by its radius squared or 4 steradians. The element of areadΩ is the projection ofd on the
^ ^
n ~r n ~r dΩ
constructed sphere and dΩ= d cos = d so that d! = d = : Observe that sometimes the
3 2
r r r
dot product n ^ ~r is negative, the sign depending upon which of the normals to the surface is constructed.
(i.e. the inner or outer normal.)
The Gauss law for electrostatics in a vacuum states that the flux through any surface enclosing many
charges is the total charge enclosed by the surface divided by : The Gauss law is written
0

ZZ
Q
e
for charges inside S

0
~
E n ^ d = (2:6:21)
0 for charges outsideS
S332
^
whereQ represents the total charge enclosed by the surfaceS with n the unit outward normal to the surface.
e
The proof of Gauss’s theorem follows. Consider a single charge q within the closed surface S: The electric
1 q
~
e ld at a point on the surface S due to the chargeq within S is representedE = b e and so the flux
r
2
4 r
0
integral is
ZZ ZZ ZZ
b ^
q e n q dΩ q
r
~
= E n ^d = d = = (2:6:22)
E
2 2
4 r 4 r
S S 0 0 S 0
ZZ
b ^
e n cos d dΩ
r
since = = =d! and d! =4 : By superposition of the charges, we obtain a similar
2 2 2
r r r
S
n
X
result for each of the charges within the surface. Adding these results gives Q = q: For a continuous
e i
i=1
ZZZ

distribution of charge inside the volume we can write Q = d ,where is the charge distribution
e
V
per unit volume. Note that charges outside of the closed surface do not contribute to the total flux across
the surface. This is because the eld lines go in one side of the surface and go out the other side. In this
ZZ
~
case E n ^d = 0 for charges outside the surface. Also the position of the charge or charges within the
S
volume does not e ect the Gauss law.
The equation (2.6.21) is the Gauss law in integral form. We can put this law in dieren tial form as
follows. Using the Gauss divergence theorem we can write for an arbitrary volume that
ZZ ZZZ ZZZ ZZZ

Q 1
e

~ ~
^
E nd = r Ed = d = = d

0 0 0
S V V V
which for an arbitrary volume implies


~
r E = : (2:6:23)

0
The equations (2.6.23) and (2.6.7) can be combined so that the Gauss law can also be written in the form


2
r V =− which is called Poisson’s equation.

0
EXAMPLE 2.6-2
Find the electric eld associated with an in nite plane sheet of positive charge.

Solution: Assume there exists a uniform surface charge and draw a circle at some point on the plane
surface. Now move the circle perpendicular to the surface to form a small cylinder which extends equal
distances above and below the plane surface. We calculate the electric flux over this small cylinder in the

limit as the height of the cylinder goes to zero. The charge inside the cylinder is A whereA is the area of
the circle. We nd that the Gauss law requires that
ZZ

Q A
e
~
E n ^d = = (2:6:24)

0 0
S
where n ^ is the outward normal to the cylinder as we move over the surface S: By the symmetry of the
situation the electric force vector is uniform and must point away from both sides to the plane surface in the
direction of the normals to both sides of the surface. Denote the plane surface normals by b e and−b e and
n n
~ ~
assume that E = b e on one side of the surface and E =− b e on the other side of the surface for some
n n
constant . Substituting this result into the equation (2.6.24) produces
ZZ
~
E n ^d =2 A (2:6:25)
S333
since only the ends of the cylinder contribute to the above surface integral. On the sides of the cylinder we
will have n ^ b e = 0 and so the surface integral over the sides of the cylinder is zero. By equating the
n


results from equations (2.6.24) and (2.6.25) we obtain the result that = and consequently we can write
2
0


~
E = b e where b e represents one of the normals to the surface.
n n
2
0

Note an electric eld will always undergo a jump discontinuity when crossing a surface charge : As in


~ ~
the above example we haveE = b e and E =− b e so that the di erence is
up n down n
2 2
0


(1) (2)
i i
~ ~
E −E = b e or E n +E n + =0: (2:6:26)
up down n
i i

0 0
It is this di erence which causes the jump discontinuity.
EXAMPLE 2.6-3.
Calculate the electric eld associated with a uniformly charged sphere of radius a.

Solution: We proceed as in the previous example. Let denote the uniform charge distribution over the
surface of the sphere and let b e denote the unit normal to the sphere. The total charge then is written as
r
ZZ
2
q = d =4 a : If we construct a sphere of radiusr>a around the charged sphere, then we have
S
a
by the Gauss theorem
ZZZ
Q q
e
~
b
E e d = = : (2:6:27)
r

0 0
S
r
~
Again, we can assume symmetry for E and assume that it points radially outward in the direction of the
~ ~
surface normal b e and has the form E = b e for some constant : Substituting this value for E into the
r r
equation (2.6.27) we nd that
ZZ ZZ
q
2
~
E b e d = d =4 r = : (2:6:28)
r

0
S S
r r
1 q
~
This gives E = b e where b e is the outward normal to the sphere. This shows that the electric eld
r r
2
4 r
0
outside the sphere is the same as if all the charge were situated at the origin.
i i 1 2 3
ForS a piecewise closed surface enclosing a volume V andF =F (x ;x ;x ) i=1; 2;3, a continuous
vector e ld with continuous derivatives the Gauss divergence theorem enables us to replace a flux integral
i i
of F overS by a volume integral of the divergence of F over the volumeV such that
ZZ ZZZ ZZ ZZZ
i i
~ ~
F n d = F d or F n ^d = divFd : (2:6:29)
i
;i
S V S V
i
IfV contains a simple closed surface where F is discontinuous we must modify the above Gauss divergence
theorem.
EXAMPLE 2.6-4.
We examine the modi cation of the Gauss divergence theorem for spheres in order to illustrate the
concepts. LetV have surface areaS which encloses a surface . Consider the gure 2.6-4 where the volume
V enclosed by S and containing has been cut in half.334
Figure 2.6-4. SphereS containing sphere .
Applying the Gauss divergence theorem to the top half of g ure 2.6-4 gives
ZZ ZZ ZZ ZZZ
i T i b i i
T T
F n d + F n d + F n d = F d (2:6:30)
i i i ;i
S S V
T b1 T T
where the n are the unit outward normals to the respective surfaces S , S and : Applying the Gauss
i T b1 T
divergence theorem to the bottom half of the sphere in gure 2.6-4 gives
ZZ ZZ ZZ ZZZ
i B i b i i
B B
F n d + F n d + F n d = F d (2:6:31)
i i i ;i
S S V
B b2 B B
Observe that the unit normals to the surfacesS andS are equal and opposite in sign so that adding the
b1 b2
equations (2.6.30) and (2.6.31) we obtain
ZZ ZZ ZZZ
(1)
i i i
F n d + F n d = F d (2:6:32)
i
;i
i
S V +V
T B335
whereS =S +S is the total surface area of the outside sphere and = + is the total surface area
T B T B
(1)
of the inside sphere, and n is the inward normal to the sphere when the top and bottom volumes are
i
combined. Applying the Gauss divergence theorem to just the isolated small sphere we nd
ZZ ZZZ
(2)
i i
F n d = F d (2:6:33)
i ;i
V

(2)
where n is the outward normal to : By adding the equations (2.6.33) and (2.6.32) we nd that
i
ZZ ZZ ZZZ

(1) (2)
i i i i
F n d + F n +F n d = F d (2:6:34)
i
;i
i i
S V
where V =V +V +V : The equation (2.6.34) can also be written as
T B
ZZ ZZZ ZZ

(1) (2)
i i i i
F n d = F d − F n +F n d : (2:6:35)
i
;i i i
S V
In the case thatV contains a surface the total electric charge inside S is
ZZZ ZZ

Q = d + d (2:6:36)
e
V

where is the surface charge density on and is the volume charge density throughoutV: The Gauss
theorem requires that
ZZ ZZZ ZZ
Q 1 1
e
i
E n d = = d + d : (2:6:37)
i

0 0 0
S V
In the case of a jump discontinuity across the surface we use the results of equation (2.6.34) and write
ZZ ZZZ ZZ

(1) (2)
i i i i
E n d = E d − E n +E n d : (2:6:38)
i
;i i i
S V
Subtracting the equation (2.6.37) from the equation (2.6.38) gives

ZZZ ZZ


(1) (2)
i i i
E − d − E n +E n + d =0: (2:6:39)
;i i i

0 0
V
For arbitrary surfacesS and , this equation implies the di erential form of the Gauss law


i
E = : (2:6:40)
;i

0
Further, on the surface , where there is a surface charge distribution we have


(1) (2)
i i
E n +E n + =0 (2:6:41)
i i

0

which shows the electric eld undergoes a discontinuity when you cross a surface charge :336
Electrostatic Fields in Materials
When charges are introduced into materials it spreads itself throughout the material. Materials in
which the spreading occurs quickly are called conductors, while materials in which the spreading takes a
long time are called nonconductors or dielectrics. Another electrical property of materials is the ability to
hold local charges which do not come into contact with other charges. This property is called induction.
For example, consider a single atom within the material. It has a positively charged nucleus and negatively
~
charged electron cloud surrounding it. When this atom experiences an electric eld E the negative cloud
~ ~ ~
moves opposite toE while the positively charged nucleus moves in the direction ofE.IfE is large enough it
can ionize the atom by pulling the electrons away from the nucleus. For moderately sized electric elds the
atom achieves an equilibrium position where the positive and negative charges are o set. In this situation
the atom is said to be polarized and have a dipole momentp: ~
~
De nition: When a pair of charges +q and −q are separated by a distance 2d the electric dipole
~
moment is de ned by p~=2dq,where p~ has dimensions of [C m].
~ ~
In the special case where d has the same direction as E and the material is symmetric we say that p~
~ ~
is proportional to E and write p~ = E,where is called the atomic polarizability. If in a material subject
to an electric el d their results many such dipoles throughout the material then the dielectric is said to be
~ ~
polarized. The vector quantityP is introduced to represent this e ect. The vector P is called the polarization
2
vector having units of [C=m ], and represents an average dipole moment per unit volume of material. The
vectorsP andE are related through the displacement vectorD such that
i i i
P =D − E : (2:6:42)
i i 0 i
For an anisotropic material (crystal)
j j
D = E and P = E (2:6:43)
i j i j
i i
j j
where is called the dielectric tensor and is called the electric susceptibility tensor. Consequently,
i i
j j j j j j j
P = E = E − E =( − )E so that = − : (2:6:44)
i j j 0 i 0 j 0
i i i i i i i
A dielectric material is called homogeneous if the electric force and displacement vector are the same for any
two points within the medium. This requires that the electric force and displacement vectors be constant
j
parallel vector elds. It is left as an exercise to show that the condition for homogeneity is that =0:
i;k
A dielectric material is called isotropic if the electric force vector and displacement vector have the same
j
i i
direction. This requires that = where is the Kronecker delta. The term = K is called the
0 e
i j j
2
−12 2
dielectric constant of the medium. The constant =8:85(10) coul =N m is the permittivity of free
0

space and the quantityk = is called the relative dielectric constant (relative to ). For free spacek =1:
e 0 e

0
j j
Similarly for an isotropic material we have = where is called the electric susceptibility. For a
0 e e
i i
~ ~ ~
linear medium the vectorsP, D and E are related by
D = E +P = E + E = (1 + )E = K E = E (2:6:45)
i 0 i i 0 i 0 e i 0 e i 0 e i i337
where K =1+ is the relative dielectric constant. The equation (2.6.45) are constitutive equations for
e e
dielectric materials.
The eect of polarization is to produce regions of bound charges within the material and bound
b
surface charges together with free charges which are not a result of the polarization. Within dielectrics
b f
~ ~
we haver P = for bound volume charges and P b e = for bound surface charges, where b e is a
b n b n
unit normal to the bounding surface of the volume. In these circumstances the expression for the potential
function is written
ZZZ ZZ
1 1
b b
V = d + d (2:6:46)
4 r 4 r
0 0
V S
and the Gauss law becomes

~ ~ ~ ~
r E = = + =−r P + or r( E +P)= : (2:6:47)
0 b f f 0 f
~ ~ ~
Since D = E +P the Gauss law can also be written in the form
0
i
~
r D = or D = : (2:6:48)
f f
;i
When no confusion arises we replace by : In integral form the Gauss law for dielectrics is written
f
ZZ
~
D n ^d =Q (2:6:49)
fe
S
where Q is the total free charge density within the enclosing surface.
fe
Magnetostatics
~ ~
A stationary charge generates an electric el d E while a moving charge generates a magnetic eld B:
Magnetic eld lines associated with a steady current moving in a wire form closed loops as illustrated in the
gure 2.6-5.
Figure 2.6-5. Magnetic eld lines.
The direction of the magnetic force is determined by the right hand rule where the thumb of the right
hand points in the direction of the current flow and the ngers of the right hand curl around in the direction
~ ~
of the magnetic eld B: The force on a test chargeQ moving with velocity V in a magnetic eld is
~ ~ ~
F =Q(V B): (2:6:50)
m
The total electromagnetic force acting onQ is the electric force plus the magnetic force and is
h i
~ ~ ~ ~
F =Q E+(V B) (2:6:51)338

which is known as the Lorentz force law. The magnetic force due to a line charge density moving along
acurve C is the line integral
Z Z

~ ~ ~ ~ ~
F = ds(V B)= I Bds: (2:6:52)
mag
C C
Similarly, for a moving surface charge density moving on a surface
ZZ ZZ

~ ~ ~ ~ ~
F = d (V B)= K Bd (2:6:53)
mag
S S
and for a moving volume charge density
ZZZ ZZ

~ ~ ~ ~ ~
F = d (V B)= J Bd (2:6:54)
mag
V V

~ ~ ~ ~ ~ ~
where the quantities I = V , K = V and J = V are respectively the current, the current per unit
length, and current per unit area.
A conductor is any material where the charge is free to move. The flow of charge is governed by Ohm’s
law. Ohm’s law states that the current density vector J is a linear function of the electric intensity or
i
J = E ,where is the conductivity tensor of the material. For homogeneous, isotropic conductors
i im m im
= so that J = E where is the conductivity and 1= is called the resistivity.
im im i i

Surround a charge density with an arbitrary simple closed surface S having volume V and calculate
the flux of the current density across the surface. We nd by the divergence theorem
ZZ ZZZ
~ ~
J n ^d = r Jd : (2:6:55)
S V
If charge is to be conserved, the current flow out of the volume through the surface must equal the loss due
to the time rate of change of charge within the surface which implies
ZZ ZZZ ZZZ ZZZ

d @

~ ~
J n ^d = r Jd =− d =− d (2:6:56)
dt @t
S V V V
or
ZZZ

@
~
r J + d =0: (2:6:57)
@t
V
This implies that for an arbitrary volume we must have

@
~
r J =− : (2:6:58)
@t
Note that equation (2.6.58) has the same form as the continuity equation (2.3.73) for mass conservation and
so it is also called a continuity equation for charge conservation. For magnetostatics there exists steady line

@
~
currents or stationary current so = 0. This requires thatr J =0:
@t339
Figure 2.6-6. Magnetic e ld around wire.
Biot-Savart Law
The Biot-Savart law for magnetostatics describes the magnetic e ld at a point P due to a steady line
current moving along a curveC and is
Z
~
I b e
0 r
~
B(P)= ds (2:6:59)
2
4 r
C
with units [N=amp m] and where the integration is in the direction of the current flow. In the Biot-Savart
−7 2
~
law we have the constant =4 10 N=amp which is called the permeability of free space,I =Ib e is
0 t
the current flowing in the direction of the unit tangent vector b e to the curveC; b e is a unit vector directed
t r
from a point on the curve C toward the point P and r is the distance from a point on the curve to the
~
general point P: Note that for a steady current to exist along the curve the magnitude of I must be the
same everywhere along the curve. Hence, this term can be brought out in front of the integral. For surface
~ ~
currentsK and volume currentsJ the Biot-Savart law is written
ZZ
~
b
K e
0 r
~
B(P)= d
2
4 r
S
ZZZ
~
J b e
0 r
~
and B(P)= d :
2
4 r
V
EXAMPLE 2.6-5.
~ ~
Calculate the magnetic eld B a distance h perpendicular to a wire carrying a constant currentI:
Solution: The magnetic e ld circles around the wire. For the geometry of the gu re 2.6-6, the magnetic
eld points out of the page. We can write
~
I b e =Ib e b e =Ie ^ sin
r t r
where e ^ is a unit vector tangent to the circle of radius h which encircles the wire and cuts the wire perpen-
dicularly.340
For this problem the Biot-Savart law is
Z
I e ^
0
~
B(P)= ds:
2
4 r
In terms of we n d from the geometry of gu re 2.6-6
s h
2
tan = with ds =h sec d and cos = :
h r
Therefore,
Z

2 2
Ie ^ sin h sec
0
~
B(P)= d :
2 2
h = cos

1
But, = = 2+ so that sin =cos and consequently
Z

2
Ie ^ Ie ^
0 0
~
B(P)= cos d = (sin − sin ):
2 1
4 h 4 h

1
Ie ^
0
~
For a long straight wire !− = 2and ! = 2 to give the magnetic eld B(P)= :
1 2
2 h
For volume currents the Biot-Savart law is
ZZZ
~
b
J e
0 r
~
B(P)= d (2:6:60)
2
4 r
V
and consequently (see exercises)
~
r B =0: (2:6:61)


~
Recall the divergence of an electric eld is r E = is known as the Gauss’s law for electric elds and so

0
~ ~
in analogy the divergencer B = 0 is sometimes referred to as Gauss’s law for magnetic elds. If r B =0,
~ ~ ~ ~
then there exists a vector eld A such that B =r A: The vector eld A is called the vector potential of
~ ~ ~ ~ ~
B: Note thatr B =r (r A)=0: Also the vector potential A is not unique since B is also derivable
~
from the vector potentialA +r where is an arbitrary continuous and dieren tiable scalar.
Ampere’s Law
Ampere’s law is associated with the work done in moving around a simple closed path. For example,
~
consider the previous example 2.6-5. In this example the integral of B around a circular path of radius h
which is centered at some point on the wire can be associated with the work done in moving around this
path. The summation of force times distance is
Z Z Z
I
0
~ ~
B d~r = B e ^ds = ds = I (2:6:62)
0
2 h
C C C
Z
where now d~r = e ^ds is a tangent vector to the circle encircling the wire and ds =2 h is the distance
C
around this circle. The equation (2.6.62) holds not only for circles, but for any simple closed curve around
the wire. Using the Stoke’s theorem we have
Z ZZ ZZ
~ ~ ~
B d~r = (r B) b e d = I = J b e d (2:6:63)
n 0 0 n
C S S341
ZZ
~
where J b e d is the total flux (current) passing through the surface which is created by encircling
n
S
some curve about the wire. Equating like terms in equation (2.6.63) gives the di erential form of Ampere’s
law
~ ~
r B = J: (2:6:64)
0
Magnetostatics in Materials
Similar to what happens when charges are introduced into materials we have magnetic e lds whenever
there are moving charges within materials. For example, when electrons move around an atom tiny current
loops are formed. These current loops create what are called magnetic dipole moments m ~ throughout the
~
material. When a magnetic eld B is applied to a material medium there is a net alignment of the magnetic
~ ~
dipoles. The quantity M; called the magnetization vector is introduced. Here M is associated with a
dielectric medium and has the units [amp/m] and represents an average magnetic dipole moment per unit
~ ~
volume and is analogous to the polarization vector P used in electrostatics. The magnetization vector M
~
acts a lot like the previous polarization vector in that it produces bound volume currents J and surface
b
~ ~ ~ ~ ~
currentsK wherer M =J is a volume current density throughout some volume andM b e =K is a
b b n b
surface current on the boundary of this volume.
~
@E
From electrostatics note that the time derivative of has the same units as current density. The
0
@t
~
@E
~ ~ ~ ~ ~
total current in a magnetized material is then J =J +J + whereJ is the bound current,J is the
t b f 0 b f
@t
~
@E
free current and is the induced current. Ampere’s law, equation (2.6.64), in magnetized materials then
0
@t
becomes
~ ~
@E @E
~ ~ ~ ~ ~
r B = J = (J +J + )= J + (2:6:65)
0 t 0 b f 0 0 0 0
@t @t
~
@E
~ ~ ~
where J = J +J : The term is referred to as a displacement current or as a Maxwell correction to
b f 0
@t
the eld equation. This term implies that a changing electric eld induces a magnetic eld.
~
An auxiliary magnet e ld H de ned by
1
H = B −M (2:6:66)
i i i

0
~ ~
is introduced which relates the magnetic force vector B and magnetization vectorM: This is another con-
stitutive equation which describes material properties. For an anisotropic material (crystal)
j j
B = H and M = H (2:6:67)
i j i j
i i
j j
where is called the magnetic permeability tensor and is called the magnetic permeability tensor. Both
i i
of these quantities are dimensionless. For an isotropic material
j j
= where = k : (2:6:68)
0 m
i i
−7 2
Here =4 10 N=amp is the permeability of free space and k = is the relative permeability
0 m

0
j j
coe cient. Similarly, for an isotropic material we have = where is called the magnetic sus-
m m
i i
ceptibility coe cient and is dimensionless. The magnetic susceptibility coe cient has positive values for342
materials called paramagnets and negative values for materials called diamagnets. For a linear medium the
~ ~ ~
quantities B, M and H are related by
B = (H +M )= H + H = (1 + )H = k H = H (2:6:69)
i 0 i i 0 i 0 m i 0 m i 0 m i i
where = k = (1 + ) is called the permeability of the material.
0 m 0 m
~
Note: The auxiliary magnetic vector H for magnetostatics in materials plays a role similar to the
~
displacement vector D for electrostatics in materials. Be careful in using electromagnetic equations from
~ ~ ~
dieren t texts as many authors interchange the roles of B and H. Some authors call H the magnetic eld.
1
~
However, the quantity B should be the fundamental quantity.
Electrodynamics
~
@P
~
In the nonstatic case of electrodynamics there is an additional quantityJ = called the polarization
p
@t
current which satis es
~
@P @ @
b
~ ~
r J =r = r P =− (2:6:70)
p
@t @t @t
and the current density has three parts
~
@P
~ ~ ~ ~ ~ ~
J =J +J +J =r M +J + (2:6:71)
b f p f
@t
consisting of bound, free and polarization currents.
Faraday’s law states that a changing magnetic e ld creates an electric el d. In particular, the electro-
magnetic force induced in a closed loop circuitC is proportional to the rate of change of flux of the magnetic
eld associated with any surface S connected with C: Faraday’s law states
Z ZZ
@
~ ~
E d~r =− B b e d :
n
@t
C S
Using the Stoke’s theorem, we nd
ZZ ZZ
~
@B
~
(r E) b e d =− b e d :
n n
@t
S S
The above equation must hold for an arbitrary surface and loop. Equating like terms we obtain the dieren tial
form of Faraday’s law
~
@B
~
r E =− : (2:6:72)
@t
This is the r st electromagnetic eld equation of Maxwell.
Ampere’s law, equation (2.6.65), written in terms of the total current from equation (2.6.71) , becomes
~ ~
@P @E
~ ~ ~
r B = (r M +J + )+ (2:6:73)
0 f 0 0
@t @t
which can also be written as
1 @
~ ~ ~ ~ ~
r ( B−M)=J + (P + E)
f 0
@t
0
1
D.J. Grit hs, Introduction to Electrodynamics, Prentice Hall, 1981. P.232.343
or
~
@D
~ ~
r H =J + : (2:6:74)
f
@t
This is Maxwell’s second electromagnetic eld equation.
To the equations (2.6.74) and (2.6.73) we add the Gauss’s law for magnetization, equation (2.6.61) and
Gauss’s law for electrostatics, equation (2.6.48). These four equations produce the Maxwell’s equations of
electrodynamics and are now summarized. The general form of Maxwell’s equations involve the quantities
E ; Electric force vector, [E ]=Newton=coulomb
i i
2
B ; Magnetic force vector, [B ]=Weber=m
i i
H; Auxilary magnetic force vector, [H ] = ampere=m
i i
2
D ; Displacement vector, [D ] = coulomb=m
i i
2
J; Free current density, [J ] = ampere=m
i i
2
P ; Polarization vector, [P ] = coulomb=m
i i
M ; Magnetization vector, [M ] = ampere=m
i i
for i=1; 2; 3: There are also the quantities
3
%; representing the free charge density, with units [%] = coulomb=m
2 2
; Permittivity of free space, [ ] = farads=m or coulomb =Newton m :
0 0
2
; Permeability of free space, [ ] = henrys=morkg m=coulomb
0 0
In addition, there arises the material parameters:
i
; magnetic permeability tensor, which is dimensionless
j
i
; dielectric tensor, which is dimensionless
j
i
; electric susceptibility tensor, which is dimensionless
j
i
; magnetic susceptibility tensor, which is dimensionless
j
These parameters are used to express variations in the electric el d E and magnetic eld B when
i i
acting in a material medium. In particular,P;D;M and H are de ned from the equations
i i i i
j j
i i
D = E = E +P = +
i j 0 i i 0
j j
i i
j
i i i
B = H = H + M; = ( + )
i j 0 i 0 i 0
j j j
i
j j
P = E ; and M = H for i=1; 2; 3:
i j i j
i i
The above quantities obey the following laws:
Faraday’s Law
This law states the line integral of the electromagnetic force around a loop is proportional
to the rate of flux of magnetic induction through the loop. This gives rise to the rst electromagnetic eld
equation:
i
~
@B @B
ijk
~
r E =− or E =− : (2:6:75)
k;j
@t @t344
Ampere’s Law
This law states the line integral of the magnetic force vector around a closed loop is
proportional to the sum of the current through the loop and the rate of flux of the displacement vector
through the loop. This produces the second electromagnetic eld equation:
i
~
@D @D
ijk i
~ ~
r H =J + or H =J + : (2:6:76)
f k;j
f
@t @t
Gauss’s Law for Electricity
This law states that the flux of the electric force vector through a closed
surface is proportional to the total charge enclosed by the surface. This results in the third electromagnetic
eld equation:

1 @ p
i i
~
r D = or D = or gD = : (2:6:77)
f f p f
;i
i
g@x
Gauss’s Law for Magnetism
This law states the magnetic flux through any closed volume is zero. This
produces the fourth electromagnetic eld equation:

1 @ p
i i
~
r B =0 or B =0 or p gB =0: (2:6:78)
;i
i
g@x
When no confusion arises it is convenient to drop the subscript f from the above Maxwell equations.
Special expanded forms of the above Maxwell equations are given on the pages 176 to 179.
Electromagnetic Stress and Energy
Let V denote the volume of some simple closed surface S: Let us calculate the rate at which electro-
magnetic energy is lost from this volume. This represents the energy flow per unit volume. Begin with the
rst two Maxwell’s equations in Cartesian form
@B
i
E =− (2:6:79)
ijk k;j
@t
@D
i
H =J + : (2:6:80)
ijk k;j i
@t
Now multiply equation (2.6.79) byH and equation (2.6.80) byE . This gives two terms with dimensions of
i i
energy per unit volume per unit of time which we write
@B
i
E H =− H (2:6:81)
ijk k;j i i
@t
@D
i
H E =J E + E : (2:6:82)
ijk k;j i i i i
@t
Subtracting equation (2.6.82) from equation (2.6.81) we nd
@D @B
i i
(E H −H E )=−J E − E − H
ijk k;j i k;j i i i i i
@t @t
@D @B
i i
[(E H ) −E H +H E ]=−J E − E − H
ijk k i ;j k i;j i;j k i i i i
@t @t
Observe that (E H ) is the same as (E H ) so that the above simpli es to
jki k i ;j ijk j k ;i
@D @B
i i
(E H ) +J E =− E − H: (2:6:83)
ijk j k ;i i i i i
@t @t345
Now integrate equation (2.6.83) over a volume and apply Gauss’s divergence theorem to obtain
ZZ ZZZ ZZZ
@D @B
i i
E H n d + JE d =− ( E + H )d : (2:6:84)
ijk j k i i i i i
@t @t
S V V
The rst term in equation (2.6.84) represents the outward flow of energy across the surface enclosing the
volume. The second term in equation (2.6.84) represents the loss by Joule heating and the right-hand side
is the rate of decrease of stored electric and magnetic energy. The equation (2.6.84) is known as Poynting’s
theorem and can be written in the vector form
ZZ ZZZ
~ ~
@D @B
~ ~ ~ ~ ~ ~
(E H) n ^d = (−E −H −E J)d : (2:6:85)
@t @t
S V
For later use we de ne the quantity
2
~ ~ ~
S = E H or S =E H [Watts=m](2:6:86)
i ijk j k
as Poynting’s energy flux vector and note that S is perpendicular to both E and H and represents units
i i i
of energy density per unit time which crosses a unit surface area within the electromagnetic eld.
Electromagnetic Stress Tensor
Instead of calculating energy flow per unit volume, let us calculate force per unit volume. Consider a
region containing charges and currents but is free from dielectrics and magnetic materials. To obtain terms
with units of force per unit volume we take the cross product of equation (2.6.79) with D and the cross
i
product of equation (2.6.80) with B and subtract to obtain
i

@D @B
i s
− (E D +H B )= JB + B + D
irs ijk k;j s k;j s ris i s ris s i
@t @t
which simpli es using the e− identity to
@
−( − )(E D +H B )= J B + (D B )
rj sk rk sj k;j s k;j s ris i s ris i s
@t
which further simpli es to
@
−E D +E D −H B +H B = JB + ( D B ): (2:6:87)
s;r s r;s s s;r s r;s s ris i s ris i s
@t
Observe that the rst two terms in the equation (2.6.87) can be written
E D −E D =E D − E E
r;s s s;r s r;s s 0 s;r s
1
=(E D ) −E D − ( E E )
r s ;s r s;s 0 s s ;r
2
1
=(E D ) − E − (E D )
r s ;s r j j sr ;s
2
1
=(E D − E D ) − E
r s j j rs ;s r
2
which can be expressed in the form
E
E D −E D =T − E
r;s s s;r s r
rs;s346
where
1
E
T =E D − E D (2:6:88)
r s j j rs
rs
2
is called the electric stress tensor. In matrix form the stress tensor is written
2 3
1
E D − E D E D E D
1 1 j j 1 2 1 3
2
E 1
4 5
T = E D E D − E D E D : (2:6:89)
2 1 2 2 j j 2 3
rs
2
1
E D E D E D − E D
3 1 3 2 3 3 j j
2
By performing similar calculations we can transform the third and fourth terms in the equation (2.6.87) and
obtain
M
H B −H B =T (2:6:90)
r;s s s;r s
rs;s
where
1
M
T =H B − H B (2:6:91)
r S j j rs
rs
2
is the magnetic stress tensor. In matrix form the magnetic stress tensor is written
2 3
1
B H − B H B H B H
1 1 j j 1 2 1 3
2
M
1
4 5
T = B H B H − B H B H : (2:6:92)
2 1 2 2 j j 2 3
rs
2
1
B H B H B H − B H
3 1 3 2 3 3 j j
2
The total electromagnetic stress tensor is
E M
T =T +T : (2:6:93)
rs
rs rs
Then the equation (2.6.87) can be written in the form
@
T − E = JB + ( D B )
rs;s r ris i s ris i s
@t
or
@
E + J B =T − ( D B ): (2:6:94)
r ris i S rs;s ris i s
@t
For free space D = E and B = H so that the last term of equation (2.6.94) can be written in terms
i 0 i i 0 i
of the Poynting vector as
@S @
r
= ( D B ): (2:6:95)
0 0 ris i s
@t @t
Now integrate the equation (2.6.94) over the volume to obtain the total electromagnetic force
ZZZ ZZZ ZZZ ZZZ
@S
r
E d + JB d = T d − d :
r ris i s rs;s 0 0
@t
V V V V
Applying the divergence theorem of Gauss gives
ZZZ ZZZ ZZ ZZZ
@S
r
E d + JB d = T n d − d : (2:6:96)
r ris i s rs s 0 0
@t
V V S V
The left side of the equation (2.6.96) represents the forces acting on charges and currents contained within
the volume element. If the electric and magnetic elds do not vary with time, then the last term on the
right is zero. In this case the forces can be expressed as an integral of the electromagnetic stress tensor.347
EXERCISE 2.6
I 1. Find the eld lines and equipotential curves associated with a positive charge q located at (−a;0) and
a positive chargeq located at (a;0): The eld lines are illustrated in the g ure 2.6-7.
Figure 2.6-7. Lines of electric force between two charges of the same sign.
I 2. Calculate the lines of force and equipotential curves associated with the electric eld
~ ~
b b
E =E(x;y)=2y e +2x e . Sketch the lines of force and equipotential curves. Put arrows on the lines of
1 2
force to show direction of the eld lines.
I 3. A right circular cone is de ned by
x =u sin cos ; y =u sin sin ; z =u cos
0 0 0
A
with 0 2 andu 0: Show the solid angle subtended by this cone is Ω = =2 (1− cos ):
2 0
r
I 4. A charge +q is located at the point (0;a) and a charge−q is located at the point (0;−a): Show that
1 −2aq
~ ~
the electric forceE at the position (x;0); wherex>a is E = b e :
2
2 2 3=2
4 (a +x )
0
2 2 2
I 5. Let the circle x +y = a carry a line charge . Show the electric eld at the point (0 ; 0;z)is

1 az(2 )b e
3
~
E = :
2 2 3=2
4
(a +z )
0
I 6. Use superposition to nd the electric eld associated with two in nite parallel plane sheets each

carrying an equal but opposite sign surface charge density . Find the eld between the planes and outside


of each plane. Hint: Fields are of magnitude and perpendicular to plates.
2
0
ZZZ
~
J b e
0 r
~ ~ ~
I 7. For a volume currentJ the Biot-Savart law givesB = d : Show thatr B =0:
2
4 r
V
~r ~r
~
b
Hint: Let e = and considerr (J ): Then use numbers 13 and 10 of the appendix C. Also note that
r
3
r r
~ ~
r J = 0 because J does not depend upon position.348
I 8. A homogeneous dielectric is de ned by D and E having parallel vector elds. Show that for a
i i
j
homogeneous dielectric =0:
i;k
I 9. Show that for a homogeneous, isotropic dielectric medium that is a constant.
I 10. Show that for a homogeneous, isotropic linear dielectric in Cartesian coordinates

e
P = :
i;i f
1+
e
I 11. Verify the Maxwell’s equations in Gaussian units for a charge free isotropic homogeneous dielectric.
~ ~
1@B @H
1 ~
r E =− =−
~ ~
r E = r D=0
c @t c @t

~ ~
1@D 4 @E 4
~ ~
r B = rH =0
~ ~ ~
r H = + J = + E
c @t c c @t c
I 12. Verify the Maxwell’s equations in Gaussian units for an isotropic homogeneous dielectric with a
charge.
~
1@B
~
~ r E =−
r D =4
c @t
~
~
4 1@D
r B =0
~ ~
r H = J +
c c @t
I 13. For a volume charge in an element of volumed located at a point ( ; ; ) Coulombs law is
ZZZ
1
~
E(x;y;z)= b e d
r
2
4 r
0
V
2 2 2 2
(a) Show that r =(x− ) +(y− ) +(z− ) :
1
(b) Show that b e = ((x− )b e +(y− )b e +(z− )b e ):
r 1 2 3
r
(c) Show that
ZZZ ZZZ
1 (x− )b e +(y− )b e +(z− )b e 1 b e
1 2 3 r
~
E(x;y;z)= d d d = r d d d
2 2 2 3=2 2
4 [(x− ) +(y− ) +(z− ) ] 4 r
0 0
V V
ZZZ
1 ( ; ; )
~
(d) Show that the potential function forE isV = d d d
2 2 2 1=2
4 [(x− ) +(y− ) +(z− ) ]
0
V
~
(e) Show that E =−rV:

2
(f) Show that r V = − Hint: Note that the integrand is zero everywhere except at the point where

( ; ; )=(x;y;z): Consider the integral split into two regions. One region being a small sphere
about the point (x;y;z) in the limit as the radius of this sphere approaches zero. Observe the identity

b e b e
r r
r = −r( ; ; ) enables one to employ the Gauss divergence theorem to obtain a
(x;y;z)
2 2
r r
ZZ
b e
r
surface integral. Use a mean value theorem to show− ndS ^ = 4 sincen ^ =−b e :
r
2
4 r 4
0 0
S

I 14. Show that for a point charge in space =q (x−x ) (y−y ) (z−z ); where is the Dirac delta
0 0 0
function, the equation (2.6.5) can be reduced to the equation (2.6.1).
I 15.
1 ~ r
~
(a) Show the electric eld E = b e is irrotational. Here b e = is a unit vector in the direction of r:
2 r r
r r
~
(b) Find the potential functionV such thatE =−rV which satis es V(r )=0 for r > 0:
0 0349
I 16.
~ ~ ~
(a) If E is a conservative electric eld such that E =−rV, then show that E is irrotational and satis es
~ ~
r E =curlE =0:
~ ~ ~ ~
(b) Ifr E =curlE = 0, show that E is conservative. (i.e. ShowE =−rV:)
Hint: The work done on a test charge Q = 1 along the straight line segments from (x ;y ;z )to
0 0 0
(x;y ;z ) and then from (x;y ;z )to (x;y;z ) and nally from ( x;y;z )to(x;y;z) can be written
0 0 0 0 0 0
Z Z Z
x y z
V =V(x;y;z)=− E (x;y ;z )dx− E (x;y;z )dy− E (x;y;z)dz:
1 0 0 2 0 3
x y z
0 0 0
Now note that
Z
z
@V @E (x;y;z)
3
=−E (x;y;z )− dz
2 0
@y @y
z
0
@E @E @V
3 2
~
and fromr E = 0 we nd = , which implies =−E (x;y;z): Similar results are obtained
2
@y @z @y
@V @V
~
for and : Hence show−rV =E:
@x @z
I 17.
~ ~ ~ ~
(a) Show that ifr B = 0, then there exists some vector eld A such that B =r A.
~ ~
The vector eld A is called the vector potential ofB:
Z
1
~ ~
Hint: Let A(x;y;z)= sB(sx;sy;sz) ~rds where~r =xb e +yb e +zb e
1 2 3
0
Z
1
dB
i
2
and integrate s ds by parts.
ds
0
~
(b) Show thatr (r A)=0:
I 18. Use Faraday’s law and Ampere’s law to show

i
@ @E
im j jm i i
g (E ) −g E =− J +
;m 0 0
;j ;mj
@t @t
~ ~
I 19. Assume thatJ = E where is the conductivity. Show that for = 0 Maxwell’s equations produce
2
~ ~
@E @ E
2
~
+ =r E
0 0 0
2
@t @t
2
~ ~
@B @ B
2
~
and + =r B:
0 0 0
2
@t @t
~ ~
Here both E andB satisfy the same equation which is known as the telegrapher’s equation.
I 20. Show that Maxwell’s equations (2.6.75) through (2.6.78) for the electric eld under electrostatic
conditions reduce to
~
r E =0
~
r D =
f

f
2
~ ~
NowE is irrotational so that E =−rV: Show thatr V =− :
350
I 21. Show that Maxwell’s equations (2.6.75) through (2.6.78) for the magnetic eld under magnetostatic
~ ~ ~ ~ ~
conditions reduce tor H = J andr B =0: The divergence of B being zero implies B can be derived
~ ~ ~ ~
from a vector potential functionA such thatB =r A.HereA is not unique, see problem 24. If we select
~ ~
A such thatr A = 0 then show for a homogeneous, isotropic material, free of any permanent magnets, that
2
~ ~
r A =− J:
I 22. Show that under nonsteady state conditions of electrodynamics the Faraday law from Maxwell’s
~
equations (2.6.75) through (2.6.78) does not allow one to set E = −rV. Why is this? Observe that
~ ~ ~ ~
r B =0 so we can write B =r A for some vector potential A. Using this vector potential show that
!
~
@A
~
Faraday’s law can be writtenr E + = 0. This shows that the quantity inside the parenthesis is
@t
~
@A
~
conservative and so we can writeE + =−rV for some scalar potentialV. The representation
@t
~
@A
~
E =−rV−
@t
~
@A
is a more general representation of the electric potential. Observe that for steady state conditions =0
@t
so that this potential representation reduces to the previous one for electrostatics.
~
@A
~
I 23. Using the potential formulationE =−rV− derived in problem 22, show that in a vacuum
@t
~
@r A
2
(a) Gauss law can be writtenr V + =−
@t
0
(b) Ampere’s law can be written

2
~
@V @ A
~ ~
r r A = J− r −
0 0 0 0 0
2
@t @t
(c) Show the result in part (b) can also be expressed in the form
!

~
@A @V
2
~ ~ ~
r A− −r r A + =− J
0 0 0 0 0
@t @t
I 24. The Maxwell equations in a vacuum have the form
~ ~
@B @D
~ ~ ~ ~ ~
r E =− r H = + V r D = r B =0
@t @t
2
~ ~ ~ ~
where D = E; B = H with and constants satisfying =1=c where c is the speed of light.
0 0 0 0 0 0
~
@A
~ ~ ~ ~
Introduce the vector potential A and scalar potential V de ned by B =r A and E = − −rV:
@t
Note that the vector potential is not unique. For example, given as a scalar potential we can write
~ ~ ~
B =r A =r (A +r ); since the curl of a gradient is zero. Therefore, it is customary to impose some
~ ~
kind of additional requirement on the potentials. These additional conditions are such that E and B are
1 @V
~ ~
not changed. One such condition is that A andV satisfyr A + =0: This relation is known as the
2
c @t
~
Lorentz relation or Lorentz gauge. Find the Maxwell’s equations in a vacuum in terms ofA andV and show
that

2 2
1 @ 1 @
2 2
~ ~
r − V =− and r − A =− V:
0
2 2 2 2
c @t c @t
0351
~ ~
I 25. In a vacuum show thatE and B satisfy
2 2
~ ~
1 @ E 1 @ B
2 2
~ ~ ~ ~
r E = r B = r E =0 rB =0
2 2 2 2
c @t c @t
I 26.
(a) Show that the wave equations in problem 25 have solutions in the form of waves traveling in the
x- direction given by
i(kx !t) i(kx !t)
~ ~ ~ ~ ~ ~
E =E(x;t)=E e and B =B(x;t)=B e
0 0
i(kx !t)
~ ~
where E and B are constants. Note that wave functions of the form u =Ae are called plane
0 0
2
harmonic waves. Sometimes they are called monochromatic waves. Here i =−1 is an imaginary unit.
Euler’s identity shows that the real and imaginary parts of these type wave functions have the form
Acos(kx !t)and Asin(kx !t):
These represent plane waves. The constantA is the amplitude of the wave ,! is the angular frequency,
andk=2 is called the wave number. The motion is a simple harmonic motion both in time and space.
That is, at a xed point x the motion is simple harmonic in time and at a xed time t, the motion is
harmonic in space. By examining each term in the sine and cosine terms we nd that x has dimensions of
length,k has dimension of reciprocal length,t has dimensions of time and! has dimensions of reciprocal
time or angular velocity. The quantityc =!=k is the wave velocity. The value =2 =k has dimension
of length and is called the wavelength and 1= is called the wave number. The wave number represents
the number of waves per unit of distance along the x-axis. The period of the wave is T = =c =2 =!
and the frequency is f =1=T: The frequency represents the number of waves which pass a xed point
in a unit of time.
(b) Show that ! =2 f
(c) Show that c =f
(d) Is the wave motion u = sin(kx−!t) + sin(kx +!t) a traveling wave? Explain.
2
1 @
2
(e) Show that in general the wave equationr = have solutions in the form of waves traveling in
2 2
c @t
either the +x or−x direction given by
= (x;t)=f(x +ct)+g(x−ct)
where f and g are arbitrary twice dieren tiable functions.
(f) Assume a plane electromagnetic wave is moving in the +x direction. Show that the electric eld is in
the xy−plane and the magnetic eld is in the xz−plane.
Hint: Assume solutionsE =g (x−ct);E =g (x−ct);E =g (x−ct);B =g (x−ct);
x 1 y 2 z 3 x 4
B = g (x−ct);B = g (x−ct)where g ,i =1;:::;6 are arbitrary functions. Then show that E
y 5 z 6 i x
~
does not satisfyr E = 0 which implies g must be independent of x and so not a wave function. Do
1
~ ~ ~
the same for the components of B. Since bothr E =r B =0 then E = B = 0. Such waves
x x
are called transverse waves because the electric and magnetic elds are perpendicular to the direction
~ ~
of propagation. Faraday’s law implies that the E and B waves must be in phase and be mutually
perpendicular to each other.