Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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54 Basics of Electrostatics

This is even true for rotationally symmetric charge distributions of all kinds, not

only for the case of the two point charges discussed here.

If there are more than two charges, the configurations can become rather

complicated. Then, knowing the location of the stagnation points, is a particularly

useful means to understand the structure of such a field.

For future purposes, we investigate here the equipotential surface for

the special case of two opposite charges, where

Fig. 2.7

Fig. 2.8

separatrix

------ 1 . 5=

ϕ 0=

2.4 The Field Generated by two Point Charges 55

Taking the square gives

which is the equation for a sphere. Its characteristics are illustrated in Fig. 2.10. As

before, we assumed . The distances of the charge from the center of the

sphere are

(2.52)

(2.53)

From this, we find

Fig. 2.9

separatrix

4πε

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

---+





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

---–





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

---+





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

---–





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

dx–

----- y

+++





----- y

++++





---





–

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

–







–()

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

---+ d

–

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

---– d

–

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

56 Basics of Electrostatics

(2.54)

and

(2.55)

that is, the product of the two distances equals the square of the radius of the

sphere. We will use this result when we discuss image charges.

Also interesting is the case of opposite charges having the same magnitude.

that is

Based on eq. (2.51), the stagnation point has now moved to infinity. All force

lines that originate at (if is positive) end at . This results in a field as

illustrated in Fig. 2.11, and is called a dipole field. One can assign a dipole moment

to these charges (Fig. 2.12). The dipole moment is a vector quantity which points

from the positive to the negative charge, whose magnitude is

where

(2.56)

Fig. 2.10 0

---

–

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

⋅=

–

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

–

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

–

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

----

-------=

–()

---------------------------- r

Q==

–=

Qd Qr

–=

p Q r

–()=

2.5 The Ideal Dipole 57

(2.57)

. (2.58)

If we let Q approach infinity and d approach zero in such a manner that p

remains finite, then we have created an ideal dipole. It will be discussed in detail in

the next section.

2.5 The Ideal Dipole

2.5.1 The Ideal Dipole and its Potential

Consider a charge be at location and a charge located at . The

dipole moment is (see Fig. 2.12)

Imagine increasing Q and decreasing at the same time, in such a way as to

keep fixed. The related potential for the charges is

In terms of the Taylor expansion:

The gradient operator is marked with the index to express the fact that the

derivatives are with respect to the components of , The potential is now

Fig. 2.12

-Q

- r

Fig. 2.11

p Qd==

d r

–=

Q– r

+Q r

p Qdr

4πε

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

+()–

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

–

----------------–=

+()–

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

–

---------------- dx

∂

∂ 1

–

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

+ dy

∂

∂ 1

–

---------------- dz

∂

∂ 1

–

------------- …++

–

---------------- dr

∇

–

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

…+• +=

58 Basics of Electrostatics

having used the fact that

Finally we get for the potential

(2.59)

where is the observation point and the location of the dipole . Using the

angle between and as illustrated in Fig. 2.13, we write

(2.60)

The dipole field shall be discussed in more detail. It is rotationally symmetric

around the axis parallel to the orientation of . We choose that to be the z-axis of a

Cartesian coordinate system (Fig. 2.14). For this situation one obtains

with

Consequently

4πε

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

∇

–

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

• =

4πε

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

∇

–

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

• ,– =

∇

–

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

∇

–

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

–=

p ∇

–

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

•

4πε

---------------------------------–

prr

–()•

4πε

–

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

θ prr

–

p θcos

4πε

–

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

Fig. 2.14

Fig. 2.13

–

p θcos

4πε

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

4πε

++()

32/

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

θcos

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

2.5 The Ideal Dipole 59

(2.61)

Because of the rotational symmetry, it is sufficient to calculate the field in a plane,

e.g., for the x-z-plane (y=0) (see Fig. 2.15).

(2.62)

If we transform to spherical coordinates ( ) then the azimuthal component

vanishes. The remaining components are:

(2.63)

All lines of force pass through the origin. This may initially seem surprising, but is

quite plausible if we consider Fig. 2.15 as having emerged as the limit from

Fig. 2.11.

x∂

∂ϕ

–

3pxz

4πε

++()

52/

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

y∂

∂ϕ

–

3pyz

4πε

++()

52/

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

z∂

∂ϕ

–

4πε

++()

32/

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

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

1–















Fig. 2.15

3pxz

4πε

+()

52/

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

3p θcos θsin

4πε

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

p 3 θcos

1–()

4πε

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











r θϕ,,

θsin E

θcos+

2p θcos

4πε

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

θcos E

θsin–

p θsin

4πε

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











60 Basics of Electrostatics

Oftentimes, one deals not with individual dipoles, but rather a collection of

dipoles distributed over a volume, surface, or a line, more or less densely filled

with dipoles. Like for potentials for volume, surface, or line charges, where one

employs the superposition principle to integrate over the potentials of point

charges, one similarly makes use of the superposition of the potentials (2.59) of the

“point dipole”.

2.5.2 Volume Distribution of Dipoles

If the dipoles are distributed within a volume, the resulting volume density is

defined by the quantity

This is the polarization, which turns out to be an important quantity. This

distribution generates a potential

(2.64)

This expression leads to interesting consequences. We start by considering the

following integral

where we have used the vector formula

Thus

(2.65)

Comparison of this equation with equations (2.20) and (2.26) reveals that it is

possible to think of a volume distribution of dipoles as the result of superposition

of a volume charge distribution and a surface charge distribution, namely by

(2.66)

dτ

------=

Pr'() ∇

rr'–

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

•τ'd

4πε

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

∫

–=

Pr'() ∇

r '

rr'–

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

•τ' d

4πε

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

∫











4πε

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

∇

Pr'()

rr'–

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

•τ'd

∫

4πε

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

∇

r '

Pr'()•

rr'–

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

τ'd

∫

4πε

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

Pr'() ∇

r '

rr'–

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

•τ'd

∫

4πε

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

Pr'() dA'•

rr'–

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

∫

fa()∇• f a∇• a ∇f•+=

4πε

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

∇

Pr'()•

rr'–

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

τ'd

∫

–

4πε

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

Pr'() dA'•

rr'–

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

∫

ρ r'() Pr'() ∇•–=

2.5 The Ideal Dipole 61

and

(2.67)

This important result can also be made plausible, by considering this example.

Consider the disk shown in Fig. 2.16, filled with uniform polarization .

The charges of the dipoles inside the volume cancel each other. Only at the surface,

there will be a net, bound charge. The net charge at the top is a positive surface

charge and the one at the bottom is negative. One may think of this as the result of

two disks of uniform volume charge which are slightly displaced against each other

(Fig. 2.17). If the volume charges are and , and the displacement is d, then

we find for the polarization and the surface charges are . The

reason is that is perpendicular to the surface of the disk.

The general case is illustrated in Fig. 2.18, where the surface charge is given by

which is also the result previously obtained in a more formal manner. If the

polarization is not uniform, then the charges inside do not cancel entirely and there

remains a net volume charge. Fig. 2.19 illustrates this case. It shows a volume with

σ r'()

Pr'() dA'•

dA'

--------------------------- Pr'() n• ==

Fig. 2.16

dipoles

+ρ -ρ

P ρd= ρd± P±=

Fig. 2.17

+ + + +

- - - -

+ + + +

- - - -

+ρ

-ρ

σ -P=

σ +P=

σ r'()

ρddA γcos⋅⋅

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

Pr'() dA•

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

62 Basics of Electrostatics

dipoles inside. At the end of each dipole vector, there is a positive and at its

beginning a negative charge. We have

i.e., the overall flux of the polarization over the surface is equivalent to the

negative of the charge in the volume (a vector that points outward represents a

negative charge inside). Conversely,

and comparison mandates that

Fig. 2.18

+ + + + +

+ +

+ + + + +

+ + +

dA γcos

+ + + + +

+ +

+ + + + +

+ + + +

Fig. 2.19

P dA•

∫

Q– ρτd

∫

–==

P dA•

∫

P∇• τd

∫

ρ P∇•–=

2.5 The Ideal Dipole 63

2.5.3 Surface Distributions of Dipoles (Dipole Layers)

If we place dipoles on a surface, one obtains a so-called double layer or dipole

layer. The name stems from the fact that it is equivalent to two layers of opposite

charges. As shown in Fig. 2.20, let point in the direction of . We define the

surface density of the dipole moment

(2.68)

Using (2.60), we find for the potential

(2.69)

With the solid angle element

(2.70)

p dA'

Fig. 2.20

dA'

origin

field point

rr'–

dipole layer

dA'

--------=

τ r'() θcos

4πε

rr'–

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

A'd

∫

Fig. 2.21

1 unit length

Ω

dΩ

dA’

dΩ

θ A'dcos

rr'–

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