Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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234 The Stationary Current Density Field

(4.9)

This is a consequence of the Helmholtz theorem (which will be discussed in

Appendix A.5)

However, to create stationary currents is possible by using an impressed electric

field such as batteries produce. This results in

(4.10)

i.e., there must be a voltage source which provides an electromotive force (EMF).

Putting things together gives

(4.11)

and therefore

i.e.,

(4.12)

It is then

(4.13)

which may be regarded as the definition for resistance R. R depends on the

geometry and the material, but not on the current. If we separate the resistance into

an external part and an internal one for the voltage source , then

(4.14)

The definition

(4.15)

and eq. (4.13) yields

(4.16)

If we do not integrate over the closed circuit, but just the section outside the voltage

source, then we find for

(4.17)

Fig. 4.3,makes it obvious that the actual integration path is irrelevant, since in any

case

Also irrelevant is the choice of the cross section, since the current

g 0=

V E

ds•

∫

0≠=

g κ EE

+()=

g ds•

∫

κ E ds•

∫

κ E

ds•

∫

---

g ds•

∫

---

g ds•

∫

IR=

VIRIR

+()==

I g dA•

∫

---

g ds•

∫

g dA•

∫

----------------------

---

g ds•

∫

g dA•

∫

----------------------

---

g ds•

∫

Eds•

∫

–==

4.2 Relaxation Time 235

in the stationary case is the same for every cross section. This simplifies for a

conductor with uniform cross section and uniform current density:

(4.18)

which is the familiar formula for the resistance.

This shows that one can create a stationary current density by means of an

EMF. The fact that this is not strictly true for an actual EMF (i.e. after its depletion)

is immaterial. What matters is that it should be possible to maintain the current

density for a time period that is longer than the characteristic times of the system

and the relaxation time in particular, which we will discuss in the next section (the

relaxation time is very short for good conductors).

The unit of resistance in the MKSA system is 1 Ohm ( ):

(4.19)

One finds the unit for conductivity from eq. (4.18)

(4.20)

Good conductors are metals such as Silver and Copper with values

4.2 Relaxation Time

Charges and their related fields decay in a conductive medium. In the pure

electrostatic case, the final state is one where the inside of the conductor is free of

any field, and all charges are located at its surface. Consequently, is a function of

and

Fig. 4.3

I g dA•

∫

---

∫

------------

κA

-------==

1Ω

-------=

1Ω 1m

⋅

----------------------

1Ω 1m⋅

--------------------=

6.17 10

Ωm()

1–

⋅=

5.80 10

Ωm()

1–

,⋅=

r t

ρρr t,()=

236 The Stationary Current Density Field

From

and

follows that

(4.21)

This allows for the introduction of the dimensionless time

(4.22)

which gives time in units of the so-called relaxation time

(4.23)

Thus

(4.24)

Trying

yields

i.e.,

and

For this simplifies to

and then

(4.25)

Consequently, the charges decay with the relaxation time . Depending on the

material, may take on a wide range of values. The following examples shall

illustrate this

g∇•

t∂

∂ρ

+0=

g κE

κD

-------==

D∇• ρ=

---

t∂

∂ρ

+0=

τ t

---

εκ⁄

---------

---== =

---=

τ∂

∂ρ

+0=

ρ h r()f τ()=

f τ()

τ∂

∂

f τ()+0=

f τ() C τ–()exp=

ρ Chr() τ–()exp=

τ 0=

ρ r 0,()Chr()=

ρ r t,() ρr 0,() τ–()exp ρ r 0,() tt

⁄–()exp==

4.3 Boundary Conditions 237

good conductor:

poor conductor: distilled water:

Insulator: molten quartz :

4.3 Boundary Conditions

The continuity equation for charges (eq. (4.2)) applied to the boundary between

two media yields a relation between the normal components of g and the surface

charge (Fig. 4.4). Integrating this equation over a small slice of the volume (i.e., an

area) gives

or when cancelling the dA

(4.26)

It was assumed here that there is no surface current, which means that all

conductivities are finite.

The relation for the stationary case is

and therefore the boundary condition for g

(4.27)

Notice that this does not require for to be zero and, as we shall see soon,

sometimes this has to be different from zero. A consequence of (4.27) is that

(4.28)

Furthermore, it is necessary that

εε

=()

Silver t

1.4 10

19–

s⋅≈

Copper t

1.5 10

19–

s⋅≈







6–

s≈

s10days≈≈

Fig. 4.4

dA–

dA g

∂σ

------

dA+–0=

∂σ

------+–0 =

∂σ

------ 0=

238 The Stationary Current Density Field

(4.29)

Putting the two together yields the law of refraction for g and E lines (Fig. 4.5):

(4.30)

On the other hand, eq. (2.120) with a surface charge on the boundary between the

two media 1 and 2, and letting gives

(4.31)

The two eqs eq. (4.30) and eq. (4.31) are compatible only, if

that is, if

(4.32)

This is a rather peculiar result. It says that a stationary condition without

surface charges on the boundary, can only occur if both relaxation times are

identical. Conversely, if , then the stationary condition is reached only after

reaches the value mandated by eq. (4.32) (if it was initially different from that).

The time it takes to reach equilibrium depends on the geometry of the set-up and

the quantities , and . The law of refraction (4.30) does not apply before

Fig. 4.5

tan

--------------

---------

---------==

tan

--------------

-----

dτ

----- 0=

tan

--------------

-----

--------------+=

-----

--------------

σε

-----

–





-----

–





σ g

-----

–





–() ==

≠

,, κ

4.3 Boundary Conditions 239

the equilibrium is reached. During the transition, eq. (4.31) and the actual values of

have to be used. As an illustrative example, consider that at , a voltage is

abruptly applied to a layered resistor, extending to infinity (Fig. 4.6). For this

voltage is constant. The initial surface charge shall be The following

three formulas apply:

(4.33)

(4.34)

.(4.35)

The first two equations yield and

(4.36)

Substituting this in eq. (4.35) results in the differential equation

(4.37)

where

(4.38)

with the solution

(4.39)

σ t 0=

Fig. 4.6

σ t()

t 0≥

σ 0() 0=

t() bE

t()+ V=

t()– ε

t()+ σ t()=

t()– κ

t()

∂σ t()

∂t

-------------++ 0=

t() E

t()

Vε

bσ t()–

aε

bε

-----------------------------=

t()

Vε

aσ t()+

aε

bε

-----------------------------=











∂σ t()

∂t

-------------

σ t()

----------+ V

–

aε

bε

-----------------------------

aε

bε

aκ

bκ

------------------------=

σ t() V

–

aκ

bκ

-----------------------------

------

–





exp–=

240 The Stationary Current Density Field

After a very long time, this becomes

By (4.36), the related electric fields are

(4.40)

and the related current density is

(4.41)

and

(4.42)

which is illustrated in Fig. 4.7.

The stationary state described by eqs. (4.27), (4.30), and (4.32), is reached only

after the typical time when the relaxation process has subsided. A circuit of

capacitors and resistors as shown in Fig. 4.8 allows one to simulate this transition.

If A is the cross sectional area (assumed to be very large) between the layered

resistors (or layered capacitor) of Fig. 4.6, then one can write

(4.43)

∞

–

aκ

bκ

-----------------------------

1∞

aκ

bκ

------------------------

2∞

aκ

bκ

------------------------











∞

1∞

2∞

aκ

bκ

------------------------

===

σ t() g

∞

-----

-----–





------

–





exp–=

∞

–()1

------

–





exp– .=

Fig. 4.7

∞

–()=

σ t()

---------

;= C

---------=

---------

;= R

---------=











4.3 Boundary Conditions 241

Calculating the circuit of Fig. 4.8 leads to the same results as the field-theoretical

derivation. Without going into the details, let it be noted that the relaxation time

results in the expression

(4.44)

which is again eq. (4.38).

An important special case of the law of refraction (4.30) results from

which occurs when while remains finite. This forces and

the g-lines become perpendicular to the surface (Fig. 4.9). Conversely, if

then the g-lines become parallel to the boundary (Fig. 4.10). Thus, the boundary to

a conductor with infinite conductivity is an equipotential surface, i.e.,

(4.45)

Conversely, the boundary towards an insulator is characterized by

(4.46)

Fig. 4.8

------

------+

-------------------

---------

---------+

---------

---------+

--------------------------

b ε

b κ

------------------------== =

Fig. 4.10

finite

Fig. 4.9

finite

∞→

-----

0→

∞→κ

tan 0=

ϕ const.=

n∂

∂ϕ

242 The Stationary Current Density Field

4.4 The formal analogy between D and g

We have learned that

Furthermore

One concludes that there is complete formal analogy between electrostatic fields D,

on one hand, and the stationary current density fields g, on the other hand. This is

an important observation because it allows to apply most of the results from

Chapters 2 and 3 to current density fields. This is particularly true for the formal

methods developed in Chapter 3, which may now be directly reused to solve

boundary value problems of current density fields by separation or conformal

mapping. Based on eqs. (4.45) and (4.46), there is a twofold application of these

results. First, we obtain a Dirichlet problem when conductors are confined by ideal

conductors. Second, for conductors that are terminated by ideal non-conductors

(insulators), we obtain a Neumann problem. Of course, a combination of the two

gives a mixed boundary value problem.

In the charge-free space In the stationary case

therefore in a uniform space we have

The normal components at the boundary are then

while because of , the tangential components are

This results in the law of refraction

E ϕ∇–=

D εE=

D εϕ∇–=

g κE=

g κϕ .∇–=

D∇• 0= g∇• 0=

∇

ϕ 0=

= g

--------

-------

tan

--------------

-----=

tan

--------------

-----=

4.5 Some Current Density Fields 243

4.5 Some Current Density Fields

4.5.1 Point-like Current Source in Space

A current density emerging from a point-like source inside an infinitely large,

uniform medium is illustrated in Fig. 4.11.

For symmetry reasons it must be

that is

(4.47)

and

(4.48)

All other components of g and E vanish. Of course, the assumption of spherical

symmetry is an idealization because of the current supply. For the potential, we get

(4.49)

The image method allows to analyze the case of a point-like source in a volume

that consists of two half spaces, where each of which has a different conductivity

(Fig. 4.12).

We might try to solve this for region 1 by considering the (real) source I and the

image source I’ in region 2, while the solution for region 2 uses the image source

I’’ located in region 1. One writes:

Fig. 4.11

gAd•

∫

4πr

I==

4πr

------------=

-----

4πκr

---------------==

ϕ r()

4πκr

------------=

4πκ

------------

xa–()

----------------------------------------------

I '

xa+()

-----------------------------------------------+





4πκ

------------

I ''

xa–()

----------------------------------------------



