Lehner G. Electromagnetic Field Theory for Engineers and Physicists

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94 Basics of Electrostatics

The associated polarization is

and, as will be shown later (in 3.42)

which means that there are no bound volume charges inside the dielectric.

Nevertheless, there are bound surface charges, specifically

Fig. 2.53 illustrates the behavior of D and E as functions of r. Now we let

approach zero, but shall go to infinity. As a result, one finds the net charge

inside

This means that the charge appears to be reduced as a result of the bound charges

by the factor . The overall field, where now the dielectric fills the entire space, is

Fig. 2.52

4πεr

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

4πε

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

r==

P ε

χE

1–

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

4πr

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

r==

P∇•–

-----

r∂

∂

1–

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

4πr

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





–0== =

1–

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

4πr

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

– for rr

1–

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

4πr

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

for rr











Q' Q

1–

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

4πr

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

4πr

– Q

1–

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

Q–

----===

4πε

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

2.11 A Point Charge inside a Dielectric 95

i.e., the field is reduced relative to vacuum by the same factor , as was the

charge.

2.11.2 Plane Boundaries between two Dielectrics

Let a point charge reside in a space that is filled with two different dielectrics that

are separated by a plane boundary (Fig. 2.54). Material 1 ( ) occupies the half-

space x>0; material 2 ( ) occupies the half-space x<0. One can show that

1. The field in half-space 1 can be expressed as the superposition of the

field of Q and the field of a fictitious charge (image charge) Q’ , located

in half-space 2 at the same distance (a) from the boundary as Q.

2. The field in half-space 2 can be expressed as the field of the fictitious

charge Q’’, located at the same point as Q.

Fig. 2.53

r()

4πε

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

Fig. 2.54

Q’

(Q’’)

96 Basics of Electrostatics

This justifies the following Ansatz.

From this follow the electric fields

The tangential components of E, that is and , as well as the normal

component of , that is , have to be continuous on the boundary x=0. and

, are continuous if

and is continuous if

One may verify that the Ansatz is correct by showing that the appropriate choice of

Q’ and Q’’ fulfills the boundary conditions everywhere on the boundary x=0, i.e.,

for all y and all z, which is not at all self-evident. Calculating Q’ and Q’’ from these

two equations gives

Q’’ always has the same sign as Q has, while Q’ may have either one. In particular

for , we find Q’ = 0 and Q’’ = Q, as expected. In the limit of

4πε

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

xa–()

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

xa+()

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





4πε

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

Q''

xa–()

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





∇–

4πε

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

Qx a–()

xa–()

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

Q' xa+()

xa+()

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

xa–()

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

Q'y

xa+()

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

xa–()

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

Q'z

xa+()

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

⋅==

∇–

4πε

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

Q'' xa–()

xa–()

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

Q''y

xa–()

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

Q''z

xa–()

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

⋅==

D εE

QQ'+

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

Q''

------=

εE

Q' Q''+ Q=

Q' Q

–

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

Q'' Q

2ε

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

= ε

2.12 A Dielectric Sphere in a Uniform Electric Field 97

approaching infinity, one finds Q’ = -Q, the same result as for the image charge at a

conducting plane. As already mentioned, a conductor behaves in many ways like a

dielectric with infinite permittivity. We will return to this topic in a later section.

The field configuration for

is illustrated in Fig. 2.55 and for in Fig. 2.56. The curvature of the force

lines depends on the sign of . For and we have , i.e. and

attract each other, which results in field lines as shown in Fig. 2.55. For

and , however, we have , i.e., and repel each other, which

results in field lines as shown in Fig. 2.56.

2.12 A Dielectric Sphere in a Uniform Electric Field

2.12.1 The Field of a Uniformly Polarized Sphere

In order to solve the problem of a dielectric sphere, it is useful to consider the

electric field of a uniformly polarized sphere. When is the radius of the sphere

and P its uniform polarization, then the overall dipole moment is

(2.123)

One may think of a uniformly polarized sphere as being created by two spheres,

charged with opposite charges that are slightly displaced against each other

(Fig. 2.57). When is its volume charge and d the displacement, then

Outside of the sphere, the field is that of a dipole at the origin, namely

(2.124)

This is still correct for the surface of the sphere, where

Fig. 2.56

Q> 0

Fig. 2.55

Q> 0

Q' ε

< Q 0> Q'0< Q

Q' ε

Q 0> Q'0> QQ'

4πr

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

P==

ρ±

P ρd=

p θcos

4πε

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

θcos

3ε

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

98 Basics of Electrostatics

(2.125)

By means of theorems about the uniqueness of solutions of potentials, which we

will deal with in Chapter 3, it is possible to show that the potential inside the sphere

has to be

(2.126)

There is another way to prove this without using those theorems. The components

of E at the sphere’s outer surface are

(2.127)

which results from (2.63). The surface charge at the sphere’s surface (bound

charges due to polarization) is

(2.128)

This means that decreases by when passing through the surface

from the outside towards the inside of the sphere, while the other components

remain unchanged. At the inside surface of the sphere, one therefore obtains

(2.129)

Consequently the electric field E written in terms of its Cartesian components is

Fig. 2.57

-ρ

+ρ

ϕϕ

θcos

3ε

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

3ε

--------== =

3ε

--------=

2p θcos

4πε

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

2P θcos

3ε

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

p θsin

4πε

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

P θsin

3ε

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











P θcos=

P θcos()ε

⁄

P θcos

3ε

----------------–=

P θsin

3ε

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

0 .=











2.12 A Dielectric Sphere in a Uniform Electric Field 99

(2.130)

So, E has only a z-component which, furthermore, has the same value everywhere

(on the inside surface). The inside of the sphere is free of (bound) volume charges

and we therefore conclude that the whole inside space is filled with the same field.

The related potential is , as was already suggested. Also notice that in

connection with eq. (2.128), a conducting sphere within a uniform electric field

carries a surface charge proportional to (see .eq. (2.94)). Obviously, the field

of the surface charge identically cancels the external field, i.e., it creates a uniform

field inside the sphere. Outside, it creates a dipole field, as we have also seen when

analyzing the conducting sphere. We obtain the same results when we apply these

results to the current case.

Summarizing, we may state that a uniformly polarized sphere (polariza-

tion P) creates an electric dipole field outside and a uniform electric field

inside .

This provides the field of a permanently, uniform polarized sphere (i.e., that of a

uniform electret). The field in its interior is

θsin ϕcos E

θcos ϕcos+0==

θsin ϕsin E

θcos ϕsin+0==

θcos E

θsin–

3ε

--------– .==











ϕ Pz 3ε

⁄=

θcos

P 3ε

⁄()–

DPε

E+ P

---–

---

P===

100 Basics of Electrostatics

Here (Fig. 2.58), D and E point in different directions: they are anti-parallel, not

parallel as usual. Furthermore, a remarkable fact is that (as always in electrostatics)

E is irrotational but not source free, while D is source free but not irrotational.

Besides spherical bodies, only ellipsoids have such simple characteristics.

The general relation between P and E or D and E, respectively, is rather

complicated. In particular, D and E may point in entirely different directions.

Fig. 2.59 illustrates the fields of a uniformly polarized cuboid. On the inside of it,

the fields of D and E have very different shapes. Of course, on the outside, there is

the usual relation between E and D: .

So far in this section, we have not asked what causes polarization. It could be

due to permanent polarization as illustrated in Fig. 2.58 and Fig. 2.59. It could also

be due to an external, uniform field, which then would need to be considered as

well.

Fig. 2.58

Fig. 2.59

D ε

2.12 A Dielectric Sphere in a Uniform Electric Field 101

2.12.2 An Externally Applied Uniform Field as the Cause of Polariza-

tion

When a sphere is exposed to a uniform field, an additional field is created as the

sphere is polarized. This field needs to be superimposed onto the original field. As

we have just seen for uniform polarization, the internal field is uniform as well.

Uniform polarization would thus cause an overall uniform field inside, which on

the other hand causes uniform polarization. We may therefore state that a uniform

electric field uniformly polarizes a dielectric sphere. Using the relations from

Fig. 2.60, one is now able to write

from which one finds

(2.131)

Furthermore,

(2.132)

and

(2.133)

2.12.3 Dielectric Sphere (ε

) and Dielectric Space (ε

)

Let us generalize the previous problem some more. The space outside the sphere

shall now be a dielectric as well. Now, the field outside consists of a, so far,

unknown dipole field and the uniform field , i.e., we may write

Fig. 2.60

a ∞,

3ε

--------– E

a ∞,

χE

3ε

--------------–==

a ∞,

3 χ+

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

a ∞,

2 ε

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

a ∞,

3ε

2 ε

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

a ∞,

3ε

2 ε

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

a ∞,

3ε

1–

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

a ∞,

102 Basics of Electrostatics

(2.134)

C is a constant that is initially undetermined. It depends on the polarization of

both, the outside space as well as the inside of the sphere. Inside, we have the yet

undetermined uniform field , i.e.,

(2.135)

One can show that with the proper choice of C and , all boundary

conditions at the sphere’s surface ( ) can be fulfilled. Only this is the

justification for the Ansatz in (2.134) and (2.135). has to be continuous for

, i.e.,

Furthermore, also for , has to be continuous, i.e.,

Solving those equations gives

(2.136)

and

(2.137)

(2.136) generalizes (2.131) and both coincide for .

The polarization inside is

(2.138)

and uniform in z-direction. The polarization in the outside space is not uniform, but

it is nevertheless divergenceless (source free), so that there are no bound charges in

the outside space:

2C θcos

4πε

------------------- E

a ∞,

θcos+=

θa

C θsin

4πε

----------------- E

a ∞,

θsin–=

ϕa

0 .=











θcos=

θi

θsin–=

ϕi

0 .=











C θsin

4πε

----------------- E

a ∞,

θsin– E

θsin–=

= D

εE

2C θcos

4πε

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

a ∞,

θcos+ ε

θcos=

3ε

2ε

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

a ∞,

C 4πε

–

2ε

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

a ∞,

–()E

3ε

–

2ε

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

a ∞,

== =

ba,

∇•– ε

–()E

∇•–==

2.12 A Dielectric Sphere in a Uniform Electric Field 103

The expression used here for the divergence in spherical coordinates will be

derived later. Bound charges exist solely at the surface of the sphere, namely

Outside of the sphere, this charge brings about a dipole field which corresponds to

the dipole moment

Note eqs. (2.67) and (2.123). The constant C of the previous Ansatz is just the

dipole moment, caused by the polarization of the inside and outside space, which

justifies the Ansatz (2.134) and also illustrates the formal result of (2.137).

Together, the dipole field and, the uniform field bring about the potential in

the outside space

(2.139)

To compare this potential with that of a conducting sphere in a uniform electric

field as described by eq. (2.91) is an interesting exercise. It can be derived from the

just obtained potential by the limit approaching zero. In some way, a

conductor behaves like a dielectric in the limit of infinite permittivity. The law of

refraction (2.121) illustrates this fact. Lines of force have to be perpendicular to the

conductor surface. According to the law of refraction, this is also the case for a

dielectric of infinite permittivity.

in (2.136) also determines ,

(2.140)

–()

-----

r∂

∂

r θsin

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

θ∂

∂

θsin E

θa









–=

bi,

ba,

–()

–()E

θcos ε

–()

4πε

----------------- E

a ∞,





θcos–=

3ε

–

2ε

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

a ∞,

θcos=

4π

------





3ε

–

2ε

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

a ∞,

4πε

–

2ε

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

a ∞,

θcos

-----

–

2ε

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

r–







⁄

3ε

2ε

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

a ∞,