Chapter 4 Steady Electric Currents

Electric current, Electromotive force

Principle of current continuity, Energy dissipation.

1. Current & Current Density

2. Electromotive Force

3. Principle of Current Continuity

4. Boundary Conditions for Steady Electric Currents

5. Energy Dissipation in Steady Electric Current Fields

6. Electrostatic Simulation

1. Current & Current Density

Classification: Conduction current and convention 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

t

qI

d

d

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

SJ dd I

The electric current across the area S is

S

I

d SJ

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 EJ

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 lawIRU

A conductor with infinite is called a perfect electric conductor,

or p.e.c..

A medium without any conductivity is called a perfect

dielectric or an insulator.

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. Ot

herwise, an infinite current will be generated, and it results in an infi

nite energy.

In nature there exists no any p.e.c. or perfect dielectric.

The conductivities of several mediaunit in S/m

Media Conductivities Media Conductivities

Silver Sea water

Copper Pure water

Gold Dry soil

AluminumTransformer

oil

Brass Glass

Iron Rubber

71017.6

71080.5 310

71010.4 510

71054.3 1110

71057.1 1210

710 1510

4

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.

vJ

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.

If the charge density is , and the moving velocity is v, and then

The above equations are valid only for a linear isotropic medium.

EConducting medium

2. Electromotive Force

We first discuss the chemical action inside the impressed source

under open-circuit condition.

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 t

o the plate N from the plate P, and the elect

ric field E will be stronger with the increase

of the charges on the plates.

P N

E

Impressed sourceE'

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.

This impressed electric field intensity is still defined as the

force acting on unit positive charge, but it is denoted as E'.

Since the non-electrostatic force behaves as the force acting

on the charge, the non-electrostatic force is usually considered a

s that produced by an impressed electric field.

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.

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.

when the impressed electric field is equal but opposite to the ele

ctric field produced by the charges on the plates, and the charges wi

ll be at rest.

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.

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.

In the impressed source, , and there is a steady current

in the circuit consisting of the impressed source and the conducting

medium.

EE

Consequently, in order to generate the continuous current in the c

onducting medium, it must rely on an impressed source.

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, t

he supply of sustained charge to the conducting medium will vanish.

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.

lE d

P

Ne

When it is in dynamical balance, in the impressed

source. Therefore, the above equation can be rewritten as

EE

lE d

P

Ne

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.

l

0d lE

For homogeneous media, the above equation becomes

l

0d lJ

Consider that in the conducting medium, , we haveEJ

l

0d lJ

Using Stokes’ theorem, we have

0

J

0 J

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

V

Vq

d

Vtt

qS V

dd

SJthen

Because the distribution of the charges in the steady electric

current field is independent of time, i.e. , we find0

t

S

0d SJ

which states that in the steady electric current field the flux of the

current density through any closed surface will be zero.

In this way, the electric current lines must be closed, with no be

ginning or end. This result is called the principle of current continu

ity.

If we use a set of curves to describe the current field and let the ta

ngential direction at a point on the curves be the direction of the curr

ent density at the point, these curves can then be called the electric cu

rrent lines.

By using the divergence theorem, we obtain

t

J

which is called the charge conservation principle in differential

form. Hence, for a steady electric current field, we have

0 J

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:

l

0d lJ

S

0d SJ

0

J

0 J

And the corresponding differential forms are

From the equations in integral form we can find the boundary

condition for the tangential components of the current densities to

be

2

t2

1

t1

JJ

And the normal components are 2n1n JJ

The tangential components of the current densities are

discontinuous, while the normal components are continuous.

Since , we find the boundary conditions for the

steady electric field can be obtained as follows:

EJ

n221n1

2tt1

EE

EE

2

t2

1

t1

JJ

2n1n JJ

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.

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.

dl

U

J dS

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

lqEqW ddddd lE

The power dissipation P is

lSEJlEIlt

qE

t

WP dddd

d

d

d

d

Then the power dissipation per unit volume as

22 J

EEJpl

If the direction of J is different from that of E, the above equation

can be written in the following general form

JE lp

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

is U, then . And we know that . Hence, the power

dissipation per unit volume can be expressed asl

UE

d

S

IJ

d

V

UI

lS

UIpl ddd

The total power dissipation in the cylinder is

UIVpP l d

which is Joule’s law.

Example 1. A parallel plate capacitor consists of two imperfect die

lectrics 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 un

it 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

capacitor can be considered to be perpendicular

to the boundaries. Then we have

2211 EE UdEdE 2211

In view of this we find

Udd

E1221

21

Udd

E1221

12

1 1

2 2

d1

d2

U

The electric energies per unit volume in two dielectrics, respectively,

are 2222e

2111e 2

1 ,

2

1EwEw

The power dissipations per unit volume in two dielectrics, respectively,

are 2222

2111 , EpEp ll

Two special cases are worth noting:

If , then , , , .02 01 E 0e1 w 01 lp2

2 d

UE

If , then , , , .01 1

1 d

UE 02 E 02e w 02 lp

d1

d2

1= 0

E 2= 0

UE 1= 0

2= 0

U

Example 2. A quarter of a flat circular conducting washer is

shown in the figure. Calculate the resistance between two end faces.

U

y

x

t

ab

r

0

(r,)

0

Solution: The cylindrical coordinate

system should be selected. Assume the

electric potential difference between two

end faces is U, and let

Since the electric potential is related to the angle , it should

satisfy the following equation

0d

d2

2

The general solution is 21 CC

The electric potential at 01 0

The electric potential atU22

π

Based on the given boundary conditions, we find

π

2U

r

U

r π

2 eeEJ

The current density J in the conducting medium is

Then the current I flowing into the conducting medium across the

end face at is2

π

)d (π

2d rt

r

UI

SS

eeSJ

a

bUt

r

rUt b

aln

π

2d

π

2

Consequently, the resistance R between two end faces is

ab

tI

UR

ln 2

π

6. Electrostatic Simulation

Two fields are found to be very similar in source-free region.

Steady Electric Current Field

)0( E

Electrostatic Field

)0(

0d

llJ 0d

l

lE

0d

SSJ 0d

S

SE

0 J 0 E

0 J 0 E

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.

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.

Based on this similarity, the solution of the steady electric

current field can be found directly from the results of the

electrostatic field.

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.

CR CG

Based on the equations for two fields, we can find the

resistance and conductance between two electrodes as

If the capacitance between two electrodes is known, from the

above equations the resistance and the conductance between two

electrodes can be found out.

The capacitance of a coaxial line per unit length is ,

where b is the inner radius of the outer conductor, and a is the

radius of the internal conductor. If the conductivity of the filled

dielectric is , the leakage conductance per unit length G1 is

)/ln(

π21 ab

C

)/ln(

π21 ab

G

d

S

d

SG

The capacitance of a parallel plate capacitor of plate area S and

separation d is . If the conductivity of the imperfect dielectric

is , the leakage conductance G between two electric plates of the

parallel plate capacitor is

d

SC

CR CG