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Chapter 4 Steady Electric Currents
Electric current, Electromotive force
Principle of current continuity, Energy dissipation.
1.
2.
3.
4.
5.
6.
Current & Current Density
Electromotive Force
Principle of Current Continuity
Boundary Conditions for Steady Electric Currents
Energy Dissipation in Steady Electric Current Fields
Electrostatic Simulation
1. Current & Current Density
Classification: Conduction current and convection current.
The conduction current is formed by the free electrons (or
holes) in a conductor or the ions in an electrolyte.
The convection current is resulting from the motion of the
electron, the ions, or the other charged particles in vacuum, a
solid, a liquid or a gas.
The amount of charge flowing across a given area per unit time
is called the electric current intensity or electric current, and it is
denoted by I. The unit of electric current is ampere (A)
The relationship between electric current I and electric charge
q is
I
dq
dt
The current density is a vector, and it is denoted as J. The
direction of the current density is the same as the flowing direction
of the positive charges, and the magnitude is the amount of charge
through unit cross-sectional area per unit time.
The relationship between the current element dI across a directed
surface element dS and the current density J is
dI  J  dS
The electric current across the area S is
I   J  dS
S
Which states that the electric current across an area is the flux of the
current density through the area.
In most conducting media, the conduction current density J at a
point is proportional to the electric field intensity E at that point so that
J  E
where  is called the conductivity, and its unit is S/m. A large 
means that the conducting ability of the medium is stronger.
The above equation is called the differential form of the following
Ohm’s law
U  IR
A conductor with infinite is called a perfect electric conductor,
or p.e.c..
In a perfect electric conductor, electric current can be produced
without the influence of an electric field.
There is no steady electric field in a perfect electric conductor.
Otherwise, an infinite current will be generated, and it results in an
infinite energy.
A medium without any conductivity is called a perfect dielectric
or an insulator.
In nature there exists no any p.e.c. or perfect dielectric.
The conductivities of several media
unit in S/m
Media
Conductivities
Media
Conductivities
Silver
6.17  10 7
Sea water
4
Copper
5.80  10 7
Pure water
10 3
Gold
4.10  10 7
Dry soil
10 5
Aluminum
3.54  10 7
Transformer
oil
10 11
Brass
1.57  10 7
Glass
10 12
Iron
10 7
Rubber
10 15
The magnitude of the current density of the convection current is
not proportional to the electric field intensity, and the direction may
be different from that of electric field intensity.
If the charge density is  , and the moving velocity is v, and then
J v
As the polarization of dielectrics, the conducting properties of a
medium can be homogeneous or inhomogeneous, linear or nonlinear,
and isotropic or anisotropic, with same meanings as before.
The above equations are valid only for a linear isotropic medium.
2. Electromotive Force
We first discuss the chemical action inside the impressed source
under open-circuit condition.


E


 Conducting 

 medium


P
N



E

E'
Impressed source





In the impressed source , under the
influence of non-electrostatic force the
positive charges will be moved continuously
to the positive electrode plate P, while the
negative charges to the negative electrode
plate N.
These charges on the plates produce an
electric field E, with the direction pointed to
the plate N from the plate P, and the
electric field E will be stronger with the
increase of the charges on the plates.
The electric force caused by the charges on the plates will resist
the movement of the charges in the source. When the electric force
is equal to the non-electrostatic force, the charges are stopped, and
the charges on the plates will be constant.
Since the non-electrostatic force behaves as the force acting
on the charge, the non-electrostatic force is usually considered
as that produced by an impressed electric field.
P
N



E

E'
Impressed source





This impressed electric field intensity is
still defined as the force acting on unit
positive charge, but it is denoted as E'.
The impressed electric field E' pushes the positive charges to the
positive electrode plate, and the negative charges to the negative
electrode plate, and the direction of is opposite to that of the electric
field E produced by the charges on the plates.
when the impressed electric field is equal but opposite to the
electric field produced by the charges on the plates, and the charges
will be at rest.
P





N



E

E'
Impressed source
If the conducting medium is connected, the
positive charges on the positive electric plate will
be moved to the negative electric plate through
the conducting medium, while the negative
charges on the negative electric plate to the
positive electric plate. In this way, the charges on
the plates will be decreased, and E < E' . The
charges in the source will be moved again.
The impressed source will continuously provide the positive
charges to the positive electric plate, whereas the negative charges to
the negative electric plate, and in view of this a continuous current is
formed.
Consequently, in order to generate the continuous current in the
conducting medium, it must rely on an impressed source.
When it is in dynamical balance, the charges on the plates will be
constant, and they produce a steady electric field in the impressed
source and in the conducting medium.
P





N



E

E'
Impressed source
In the impressed source, E   E  , and there is
a steady current in the circuit consisting of the
impressed source and the conducting medium.
Although the distribution of the charges on the plates is
unchanged, the charges are not at rest. These charges are replaced
without interruption. Hence, they are called sustained charges.
The steady electric field in conducting medium is produced just
by the sustained charge. Once the impressed source is disconnected,
the supply of sustained charge to the conducting medium will vanish.
P
N



E

E'
Impressed source





The line integral of the impressed
electric field along the path from the
negative electric plate N to the positive
electric plate P is defined as the
electromotive force of the impressed source,
and it is denoted as e, i.e.
e
P
N
E   dl
When it is in dynamical balance, E   E  in the impressed
source. Therefore, the above equation can be rewritten as
e  
P
N
E  dl
The steady electrode field caused by the sustained charges on the
plates is also a conservative field, and the line integral of it around a
closed circuit should be zero, i.e.
 E  dl  0
l
P





Consider that in the conducting medium, J  E , we have
J
  dl  0
N



E

E'
Impressed source
l

For homogeneous media, the above equation becomes
 J  dl  0
l
Using Stokes’ theorem, we have
J
   0
 
J  0
In homogeneous conducting media the steady electric current
field is irrotational.
3. Principle of Current Continuity
Assume the density of the sustained charges in the volume V
bound by the closed surface S is , then
q   dV
V
then

S
q

J  dS  
 
dV
V t
t
No impressed source
P





N



E

E'
Impressed source
Because the distribution of the charges in the steady electric
current field is independent of time, i.e.   0 , we find
t
 J  dS  0
S
which states that in the steady electric current field the flux of the
current density through any closed surface will be zero.
If we use a set of curves to describe the current field and let the
tangential direction at a point on the curves be the direction of the
current density at the point, these curves can then be called the
electric current lines.
In this way, the electric current lines must be closed, with no
beginning or end. This result is called the principle of current
continuity.
By using the divergence theorem, we obtain

J 
t
which is called the charge conservation principle in differential
form. Hence, for a steady electric current field, we have
J 0
which states that the steady electric current field is solenoidal.
4. Boundary Conditions for Steady Electric Currents
The integral forms of the equations for steady electric current
field are as follows:

J
l

 dl  0
 J  dS  0
S
And the corresponding differential forms are
J
   0
 
J 0
From the equations in integral form we can find the boundary
condition for the tangential components of the current densities to
be
J 1t J 2 t

1  2
And the normal components are
J1n  J 2n
J 1t
1

J 2t
2
J1n  J 2n
The tangential components of the current densities are
discontinuous, while the normal components are continuous.
Since J  E , we find the boundary conditions for the steady
electric field can be obtained as follows:
E1t  E2t
 1 E1n   2 E2 n
Since there is no the electric field cannot in a perfect electric
conductor, the tangential components of a steady electric current
cannot exist on the surface.
Therefore, when an electric current flows into or out of a perfect
electric conductor, the electric current lines are always perpendicular
to the surface.
5. Energy Dissipation in Steady Electric Current Fields
In a conducting medium, the collision of free electrons with the
atomic lattice will generate thermal energy, and this is an irreversible
energy conversion process. The impressed source has to compensate the
energy dissipation in order to maintain the steady electric current.
dl

J
U
dS
In a steady electric current field, we
construct a small cylinder of length dl and
end face area dS, and assume the two end
faces of the cylinder are equipotential
surfaces.
Under the influence of the electric field, electric charge dq is moved
to the right end face from the left end face in dt , with the corresponding
work done by the electric force as
dW  dqE  dl  Edqdl
The power dissipation P is
P
dW
dq
 E dl  EIdl  EJdSdl
dt
dt
Then the power dissipation per unit volume as
pl  EJ  E 
2
J2

If the direction of J is different from that of E, the above equation
can be written in the following general form
pl  E  J
Which is called the differential form of Joule’s law, and it states that
the power dissipation at a point is equal to the product of the electric
field intensity and the current density at the point.
Suppose the electric potential difference between two end faces
U
is U, then E 
. And we know that J  I . Hence, the power
dl
dS
dissipation per unit volume can be expressed as
pl 
UI
UI

dSdl dV
The total power dissipation in the cylinder is
P  pl dV  UI
which is Joule’s law.
Example 1. A parallel plate capacitor consists of two imperfect
dielectrics in series. Their permittivities are 1 and 2 , the conductivities
are 1 and 2 , and the thickness are d1 and d2, respectively. If the
impressed voltage is U, find the electric field intensities, the electric
energies per unit volume, and the power dissipations per unit volume in
two dielectrics.
Solution: Since no current exists outside the
capacitor, the electric current lines in the
 1 1
d1 capacitor can be considered to be perpendicular
U
d2 to the boundaries. Then we have
 2 2
E11  E2 2
E1d1  E2d2  U
In view of this we find
E1 
2
d1 2  d 2 1
U
E2 
1
d1 2  d 2 1
U
The electric energies per unit volume in two dielectrics, respectively,
are
1
1
we1   1 E12 ,
we 2   2 E22
2
2
The power dissipations per unit volume in two dielectrics, respectively,
are
pl1   1 E12 ,
pl 2   2 E22
Two special cases are worth noting:
If  1  0 , then E1  U ,E 2  0 , w  0 , pl 2  0 .
e2
d1
If  2  0 , then E  0 , we1  0 , pl1  0 , E  U .
1
2
d2
U
 1= 0
d1
E 2= 0
d2
U
E 1= 0
 2= 0
Example 2. A quarter of a flat circular conducting washer is
shown in the figure. Calculate the resistance between two end faces.
Solution: The cylindrical coordinate
system should be selected. Assume the
electric potential difference between two
end faces is U, and let
y
U
t
(r,)
r
0

a
0
b
x
The electric potential 1  0 at   0
The electric potential 2  U at  
π
2
Since the electric potential  is related to the angle , it should
satisfy the following equation
d 2
0
2
d
The general solution is
  C1  C2
Based on the given boundary conditions, we find
2U

π
The current density J in the conducting medium is

y
U
t
J  E    e
(r,)
r
0

a
0
b
x

2U
 e
r
πr
Then the current I flowing into the conducting
π
medium across the end face at   is
2
2Ut b dr 2Ut  b 
2U 



ln  
I   J  dS     e

(

e
t
d
r
)



a
S
S
π
r
π
πr 
a

Consequently, the resistance R between two end faces is
R
U

I
π
b
2 t ln  
a
6. Electrostatic Simulation
Two fields are found to be very similar in source-free region.
Steady Electric Current Field
Electrostatic Field
( E   0)
(   0)
 J  dl  0
 E  dl  0
 J  dS  0
 E  dS  0
 J  0
 E  0
 J  0
 E  0
l
S
l
S
The electric current density J corresponds to the electric field
intensity E, and the electric current lines to the electric field lines.
If the steady electric current field has the same boundary
conditions as that for the electrostatic field, the distribution of
the current density will be the same as that of the electric field
intensity.
Based on this similarity, the solution of the steady electric
current field can be found directly from the results of the
electrostatic field.
In some cases, since the steady electric current field is easy
to be constructed and measured, the electrostatic field can be
investigated based on the steady electric current field with the
same boundary conditions, and this method is called electrostatic
simulation.
The electrostatic field and the steady electric current
field between two electrodes as follows:
P

N
Steady electric
current field
P

N
Electrostatic
field
The calculation of the resistance of conducting media
can be determined based on the results of the corresponding
electrostatic field.
Based on the equations for two fields, we can find the
resistance and conductance between two electrodes as
R

C
G

C

If the capacitance between two electrodes is known, from the
above equations the resistance and the conductance between two
electrodes can be found out.
R

C
G

C

The capacitance of a parallel plate capacitor of plate area S and
S . If the conductivity of the imperfect dielectric
separation d is
C
d
is , the leakage conductance G between two electric plates of the
parallel plate capacitor is
G
 S  S


 d
d
2π
,
ln( b / a)
where b is the inner radius of the outer conductor, and a is the
The capacitance of a coaxial line per unit length is C1 
radius of the internal conductor. If the conductivity of the filled
dielectric is , the leakage conductance per unit length G1 is
G1 
2π
ln( b / a)