Unit 5

Current, Conductors andDielectrics

In the previous units we have discussed and defined electric force, electric field inten-

sity, electric flux density and potential. We are now in a position to apply some of the

ideas we have considered to materials, such as conductors, resistors and capacitors,

that engineers, particularly electrical engineers, routinely encounter. We will formal-

ize and generalize some of the ideas that have been met in earlier courses (particularly

circuits courses).

5.1 Current and Current Density

To this point we have examined some of the laws associated with stationary (static)

charges. We now turn our attention to the idea of charge flow and define electric

current. By definition, electric current is the time rate of flow of electric

charge. While in conducting materials the charge carriers are negative (i.e. elec-

trons), in keeping with convention we will consider the flow of positive charges. That

is, instead of thinking of electrons flowing in a particular direction, we will equiva-

lently consider positive charges flowing in the opposite direction. You are familiar

with this from electric circuits in which this kind of electric current was referred to

as conventional current flow.

From the definition, an electric current, 𝐼, due to an amount of charge Δ𝑄 flowing

1

past a point over a time interval Δ𝑡 may be expressed as

𝐼 = limΔ𝑡→0

Δ𝑄

Δ𝑡=𝑑𝑄

𝑑𝑡. (5.1)

The unit for 𝐼 is obviously the ‘coulomb per second’ (C/s) which has the name ampere

(A).

A related quantity, referred to as current density, is the current per unit area

passing in a specified direction through a surface. Thus, current density, symbolized

as 𝐽 , is a vector whose direction is that of the positive charge flow and its unit is the

ampere per square metre (A/m2). If 𝐽 is NOT in the direction of the surface normal,

then the incremental amount of current Δ𝐼 passing through an incremental surface

area Δ𝑆 is simply

Δ𝐼 = 𝐽𝑁Δ𝑆

where 𝐽𝑁 is the component of 𝐽 normal to the surface element in the direction of

charge flow. Considering the situation depicted below, we write the previous expres-

sion as a dot product:

Δ𝐼 = 𝐽 ⋅ Δ�⃗�

where Δ�⃗� is perpendicular to the surface element. In a differential sense we may

write

𝑑𝐼 =

Then, the total current flowing through the extended surface 𝑆 is then found simply

integrating this expression to get

𝐼 =∫𝑆𝐽 ⋅ 𝑑�⃗� . (5.2)

Notice that this is not a closed surface.

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Current Density and Charge Velocity

It is easy to relate the current density 𝐽 to the charge velocity �⃗�. To simplify

matters let’s consider an incremental volume Δ𝑣 formed by the surface element Δ𝑆

and the elemental length Δ𝐿 as shown (note that the result will be general even

though, for simplicity of argument, we are considering a special volume):

If the charge density in the volume is 𝜌𝑣, the incremental amount of charge contained

there is clearly

Δ𝑄 =

and since Δ𝑣 = Δ𝑆Δ𝐿

Δ𝑄 = .

Now let’s assume that in a time interval Δ𝑡 a portion of the charge moves a distance

Δ𝑥, and the associated Δ𝑄 may be written as

Δ𝑄 = .

Noting from equation (5.1) that an incremental amount of current Δ𝐼 may be ex-

pressed as

Δ𝐼 =

we immediately have

Δ𝐼 = 𝜌𝑣Δ𝑆Δ𝑥

Δ𝑡

or

Δ𝐼

Δ𝑆= 𝜌𝑣

Δ𝑥

Δ𝑡

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Taking the limit Δ𝑡 → 0, the fraction on the right represents the 𝑥 component, 𝑣𝑥,

of the velocity �⃗�. Along with this, as Δ𝑆 → 0, the fraction on the left becomes the 𝑥

component, 𝐽𝑥, of 𝐽 . Thus, we may write

𝐽𝑥 = 𝜌𝑣𝑣𝑥 .

If we repeated this analysis for the other components and put all three together, we

would arrive at

𝐽 = 𝜌𝑣 �⃗� (5.3)

The current density in this form is called the convection current density. An example

would be the current density associated with an electron beam in a cathode ray tube

– i.e. a volume of charge is moving. Equation (5.3) indicates clearly that charge

in motion constitutes a current. In passing, we note that soon in this course we

will encounter conduction current density and in Term 6 we will discuss displacement

current density.

5.2 Continuity of Current

The continuity of current is a direct consequence of a very important conservation

law, namely the law of conservation of charge which states that charge cannot

be created or destroyed. Of course, neutral charges may become equal and opposite

charges, which may be recombined or simultaneously destroyed as long as the sum

total remains the same.

Consider a closed surface 𝑆 from which a current 𝐼 flows – in an outward sense.

Let the charge inside the surface be 𝑄𝑖 and the current density be 𝐽 . From equation

(5.2), now written for a closed surface through which the current flows in an outward

direction, we have

𝐼 =∮𝑆𝐽 ⋅ 𝑑�⃗� = −𝑑𝑄𝑖

𝑑𝑡. (5.4)

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The negative on the final member of this equation is demanded by the conservation

of charge. That is, a net current 𝐼 flowing outward from 𝑆 results in a decrease in

𝑄𝑖 inside the surface as time progresses. Equation (5.4) is the integral form of the

continuity of current equation.

We may write equation (5.4) in differential, or point, form as follows:

From the divergence theorem we note that

∮𝑆𝐽 ⋅ 𝑑�⃗� =

Also, we may write 𝑄𝑖 in terms of the volume charge density 𝜌𝑣 as

𝑄𝑖 =

Putting these last two expressions into equation (5.4) we have

∫vol

∇⃗ ⋅ 𝐽𝑑𝑣 = − 𝑑

𝑑𝑡

∫vol

𝜌𝑣𝑑𝑣 .

Remember that 𝑑/𝑑𝑡 is a ‘total’ derivative which, in general, also allows for any change

in the spatial coordinates over time as given by (for a function 𝑓(𝑥, 𝑦, 𝑧, 𝑡))

𝑑𝑓

𝑑𝑡=∂𝑓

∂𝑡+∂𝑓

∂𝑥

𝑑𝑥

𝑑𝑡+∂𝑓

∂𝑦

𝑑𝑦

𝑑𝑡+∂𝑓

∂𝑧

𝑑𝑧

𝑑𝑡.

Now if we agree to keep the surface constant, so that the𝑑𝑥

𝑑𝑡=𝑑𝑦

𝑑𝑡=𝑑𝑧

𝑑𝑡= 0 and

∂𝑑𝑣

∂𝑡= 0, the total derivative in the continuity equation becomes a partial derivative

and may appear inside the integral as

∫vol

∇⃗ ⋅ 𝐽𝑑𝑣 =

Thus, ∫vol

[∇⃗ ⋅ 𝐽 +

∂𝜌𝑣∂𝑡

]𝑑𝑣 = 0 .

Since this last integral is true for any volume (i.e. any integral limits), the integrand

must be zero. That is,

∇⃗ ⋅ 𝐽 = −∂𝜌𝑣∂𝑡

. (5.5)

5

Recalling the definiton of divergence, equation (5.5), the point form of (5.4), indicates

that the charge per second (current) diverging from a small volume per unit volume

is equal to the time rate of decrease of the charge density (charge per unit volume)

at every point. This clearly indicates the prinicple that charge is conserved, or the

current is continuous. Note that if the divergence of the current density is zero –

i.e. the current flowing into a volume through the surface surrounding it equals the

current flowing out, then equation (5.5) is simply Kirchhoff’s current law, perhaps

a little disguised. CAREFULLY CONSIDER THE EXAMPLE ON PAGES 117-118

OF THE TEXT.

Example: An electron beam carrying a total current of −500 𝜇A in the 𝑧 direction

has a current density that is not a function of 𝜌 or 𝜙 in the region 0 ≤ 𝜌 < 10−4 m

and is zero for 𝜌 > 10−4 m. If the electron velocities are given as 𝑣𝑧 = 8 × 107𝑧 m/s,

calculate 𝜌𝑣 at 𝑧 = 1 mm.

6

5.3 Conductors, Dielectrics and Boundary Condi-

tions

5.3.1 Conductors and the Point Form of Ohm’s Law

Please read pages 118-119 of text for a brief description of conductors, semiconductors

and insulators.

In a conductor, electrons, with a charge of −𝑒 experience a force 𝐹 determined

by the electric field �⃗� in the conductor given by

�⃗� = 𝑄�⃗� = −𝑒�⃗�

Of course, these electrons are impeded in their motion as they collide with the crystal

lattice of the material of which the conductor is composed. Thus, (in a manner

roughly analogous to the way in which a falling mass reaches a terminal velocity

as a result of air resistance), the moving electrons will reach some constant average

velocity rather than accelerating continuously. This velocity is referred to as the drift

velocity, �⃗�𝑑, which is to be distinguished from the velocity, 𝑣, in the previous section

where the charges were moving as a ‘beam’ of charge. This velocity, �⃗�𝑑, is associated

with a property of the conductor referred to as the mobility, 𝜇𝑒, which by definition

is related to the drift velocity and the electric field intensity by the equation

�⃗�𝑑 = (5.6)

Because �⃗� has units of m/s and �⃗� has units of V/m, the unit for the mobility must

be m2/(V ⋅ s). Since the mobility is a positive quantity, the negative sign in equation

(5.6) clearly indicates that electron drift velocity is in a direction opposite that of the

electric field intensity. By the way, this drift velocity, for conductors, is quite small –

on the order of cm/s or less.

For convection current density we had that 𝐽 = 𝜌𝑣�⃗�. Using a similar idea here, 𝐽

now becomes the conduction current density and 𝑣 is replaced by the drift velocity,

7

�⃗�𝑑. If the electron charge density, which is a negative quantity, is symbolized by 𝜌𝑒,

then it can replace the 𝜌𝑣 of the convection charge density. Using these substitutions,

𝐽 = (5.7)

To distinguish the convection current density from the conduction current density,

some texts use 𝐽𝜎 for the latter, but in keeping with our text we won’t do that

here. Remember that the conduction current density involves the motion of free or

conduction band electrons. The negative sign in equation (5.7) along with the fact

that 𝜌𝑒 will be a negative quantity means that 𝐽 and �⃗� will be in the same direction.

The quantity (−𝜌𝑒𝜇𝑒) in equation (5.7) is referred to as the conductivity of the

conductor and is symbolized by 𝜎. That is

𝜎 = (5.8)

Given that increasing temperature generally increases thermal activity and thus elec-

tron collisions with the crystalline structure of a conductor, we may reason that

mobility, and consequently the conductivity, will generally decrease with rising tem-

perature. Checking the units, 𝜌𝑒 is in coulombs (C), 𝜇𝑒 is in ‘metres squared per volt

second’ (m2/(V ⋅ s), so that 𝜎 has units of

The unit 1/Ω was originally called the mho – i.e. reciprocal of ohms – but now it is

called the siemens and it is abbreviated as S. Thus, the unit for the conductivity is is

‘siemens per metre’ (S/m). With the conductivity so defined we may now write from

equations (5.7) and (5.8) that

𝐽 = 𝜎�⃗� (5.9)

For reasons consistent with the terminology used earlier, equation (5.8) is referred to

as the point form of Ohm’s Law.

8

Ohm’s Law for an Extended Piece of Conductor

Consider a macroscopic (visible) piece of conductor as shown and labeled below:

For a non-time-varying current (i.e. dc current), 𝐼, we may consider that 𝐽 is

uniform across the cross section 𝑆 of the straight cylindrical wire. This means that

we may write

𝐼 = (5.10)

since 𝐽 is uniform and is parallel to �⃗�. Furthermore, �⃗� can be considered uniform

between the ends of the conductor and

𝑉𝑎𝑏 = (5.11)

where we have used the facts that (1) for displacement vectors, −�⃗�𝑏𝑎 = �⃗�𝑎𝑏 and (2)

here �⃗� ∥ �⃗�𝑎𝑏. We may as well simply write

𝑉 = 𝐸𝐿 .

We now have

from which

𝑉 =𝐿

𝜎𝑆𝐼 .

Now the ratio of the potential diffrence between the conductor ends to the current

entering the more positive end is what, in circuits courses, we have referred to as

resistance. Having used the point form of Ohm’s Law, we have thus arrived at the

common form for conducting objects subjected to uniform fields:

𝑉 = 𝐼𝑅 (5.12)

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where, measured in ohms (Ω),

𝑅 =𝑉

𝐼= . (5.13)

If the fields are not uniform, we must use the intgral forms for 𝑉 and 𝐼 and write

explicitly using equations (5.9), (5.10) and (5.11) that

𝑅 =𝑉𝑎𝑏

𝐼=

− ∫ 𝑎𝑏 �⃗� ⋅ 𝑑�⃗�∫

𝑆 𝜎�⃗� ⋅ 𝑑�⃗� . (5.14)

Again in keeping with the special case outlined above, in equation (5.14) the line

integral for the potential difference is carried out between two equipotential surfaces

in the conductor and the surface integral for the current is carried out on the more

positive of these surfaces.

We shall carry out a simple illustrative example of some of the concepts in this

section and leave a more complicated example for a tutorial.

Example: Exercise D5.3 on page 123 of text.

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5.3.2 Conductors and Electrostatic Boundary Conditions

While we wish to contiue looking at static charges associated with conducting mate-

rials, let’s for a moment do a “thought experiment”’ in which free charge is injected

into such a material.

As soon as the charges are “released” inside the conductor, they will repel each other

until they reach the conducting surface at which position they have nowhere else to

go – assume as a minimum that a perfect insulator surrounds the conductor (in fact,

we will consider the conductor to be in free space for which 𝜖 = 𝜖0). A little more

thought will bring us to the following conclusions:

∙ for a conductor, any net static charge must reside on the surface – on the

boundary between the conductor and the surrounding space.

∙ for the electrostatic case there is no charge flow in the conductor so that there

is no current in the conductor and thus 𝐽 = 0 for this case. Because 𝐽 = 𝜎�⃗�,

this means the the �⃗�-field inside the conductor must be zero also.

We shall shortly have something to say about the electric potential on the surface,

but first we need to consider the fields that can exist outside the conductor due to

the static charges on the surface.

Consider the situation in which a conducting material resides in one half-space

while the other half-space is a vacuum as depicted below:

subscript 𝑡 ≡ tangent to the boundary;

subscript 𝑁 ≡ normal to the boundary;

�̂� ≡ unit normal to the boundary point away from the conductor;

11

𝐸𝑁 and 𝐸𝑡 are the normal and tangential components, respectively, of the �⃗�-field

outside the conductor and 𝐷𝑁 and 𝐷𝑡 are similarly defined as the components of the

corresponding �⃗� field.

Tangential Fields:

Now, there are static charges on the boundary – i.e. the charges are NOT moving.

Therefore, there cannot be a tangential component to the �⃗� depicted, for if there were,

the charges would most certainly move since �⃗�𝑡 = 𝑄�⃗�𝑡 would then be non-zero and

the charges would accelerate along the surface and the static case would no longer

exist. We don’t need to ‘worry’ about the normal at the moment, since the charges

cannot jump off the conductor into free space. Let’s attempt to reach these logical

results a little more formally. With reference to the small rectangular path, 𝐶, on

the sketch above and recalling that for conservative fields (of which the static 𝐸-field

is an example) we have from equation (4.9)

∮𝐶�⃗� ⋅ 𝑑�⃗� = 0.

Applying this integral to each piece of the rectangular path,

Then, since �⃗� = 0 inside the conductor, in keeping with the direction of travel around

the path and labels Δ𝑤 and Δℎ shown on the diagram, we have

12

Notice the (1/2) on the two normal components since half the paths for these com-

ponents are inside the conductor where the field is zero. Thus, we have

𝐸𝑡Δ𝑤 − (𝐸𝑁,at b)Δℎ

2+ (𝐸𝑁,at a)

Δℎ

2= 0

If we allow Δℎ to approach 0 faster than Δ𝑤 (this should not cause us too much

anguish since we are looking for the field just next to the conductor surface where

Δℎ = 0),

and there results

𝐸𝑡 = 0 and, consequently, also 𝐷𝑡 = 0 (5.15)

just as we ‘reasoned’ above, with the latter equation being true because 𝐷 = 𝜖0𝐸.

Normal Fields:

In discussing the normal components of the fields, it is convenient to consider the

�⃗�-field. In keeping with our earlier discussion on gaussian surfaces, consider such a

surface constructed at the conductor/free space boundary as shown by the cylinder

in the above diagram. Remember that inside this gaussian surface, the charge exists

only on the conductor surface. Applying Gauss’s law to this situtation we have

where we recall that �⃗� ⋅ 𝑑�⃗� involves the portion of �⃗� which is perpendicular to the

surfaces, of which there are three here – 𝐷𝑁 is normal to the top and bottom surfaces

of the gaussian cylinder while 𝐷𝑡 is normal to the lateral surface. Allow the cylinder

height Δℎ to approach 0. This immediately eliminates contributions from the sides

(lateral surface). Even without doing this, we note:

– 𝐷𝑡 = 0 above the surface since 𝐸𝑡 = 0; also, 𝐷𝑡 = 0 in the conductor; ⇒– 𝐷𝑁 = 0 in the conductor ⇒

13

Therefore, for the whole gaussian surface

∮𝑆�⃗� ⋅ 𝑑�⃗� =

However,

𝑄 = 𝜌𝑠Δ𝑆

also. Therefore,

and

𝐷𝑁 = 𝜌𝑠 . (5.16)

This immediately tells us that the �⃗� and �⃗� fields are normal to the surface of a

perfect conductor in free space if the conductor carries a surface charge density 𝜌𝑠.

Recall, 𝐷𝑁 = 𝜖0𝐸𝑁 . In summary,

𝐷𝑁 = 𝜖0𝐸𝑁 = 𝜌𝑠 . (5.17)

Potential on the Conductor Surface: We have seen that the potential involves

an integral whose integrand is of the form �⃗� ⋅ 𝑑�⃗�. Since anywhere on the surface

𝐸𝑡 = 0 and 𝐸𝑁 ⊥ 𝑑�⃗�, the dot product will always be zero. Therefore, the conductor

surface is an equipotential surface.

5.3.3 The Method of Images

PAY CAREFUL ATTENTION TO SECTION 5.5 OF THE TEXT.

We will now consider the effect of putting charges in the vicinity of a perfectly

conducting plane, infinite in spatial extent, which according to the previous section

is everywhere at a fixed potential (with the 𝐸-field normal to the plane). For our

discussion we will take the charge to be above the plane and the conducting plane to

be the 𝑥-𝑦 plane. Furthermore, by way of introducing the relevant ideas, we recall

the dipole of Section 4.5 of these notes.

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We know that for the dipole configuration shown, the 𝑥-𝑦 plane is at zero potential.

Thus, from the point of view of analysis in the 𝑧 > 0 half-space, we may as well

consider a positive charge 𝑄 above a conducting plane (the 𝑥-𝑦 plane) maintained at

zero potential. If we develop the fields due to 𝑄 in the 𝑧 > 0 halfspace above the

𝑉 = 0 conducting plane we get the same answers as for the original dipole. However,

since there is no charge in the 𝑧 < 0 halfspace, there will be no field there.

Now consider the problem in reverse. If we have a positive charge configuration

in the 𝑧 > 0 halfspace above a perfectly conducting plane at zero potential, we can

remove the plane, introduce a negative charge configuration (an exact image of the

positive one) below the plane, remove the plane and do the analysis for the positive

and negative charge configurations together. Of course, the fields developed in this

way will only be valid in the half space where the charges actually occur. Consider

the following examples and also note the numerical example on page 130 of the text:

15

These ideas may also be extended to current distributions (i.e. not only charge

distributions) and are applied, for example, to so-called monopole antennas operating

over a ground plane. Such an antenna can be modeled as a dipole antenna (see

depiction), the associated analysis may be done for the dipole and then the field for

the monopole is simply that calculated for the 𝑧 > 0 halfspace for the dipole – also,

we have talked about the 𝑧 halfspaces only as a point of discussion; the perfectly

conducting plane could have any orientation and the half-spaces would have to be

defined relative to it.

Example: An infinite uniform line charge, 𝜌𝐿 = 25 nC/m, lies along the line 𝑥 = 0,

𝑧 = 1 in free space. The surface 𝑧 = 0 is a perfect conductor. (a) Find 𝐸 at 𝑃 (1, 2, 3).

(b) Find 𝑉 at 𝑃 if 𝑉 = 0 at the origin. (c) What is the maximum magnitude of 𝜌𝑆

on the conducting plane. Note: Tutorial 3, problem 1, will help with determining the

proper distances to use for the �⃗� field in part (a), Tutorial 5, problem 1 will be helpful

for part (b), and the results from the boundary conditions for the normal-component

of the �⃗� field will be useful for part (c).

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5.3.4 Semiconductors

Read the first couple of pages of Section 5.3 of the text along with Section 5.6 of the

text in conjunction with this part of the notes.

Substances which are neither good conductors nor yet good insulators, referred to

as semiconductors, form a very important class of ‘electrical’ materials, especially

in the realm of solid state electronics. As noted in the text, what distinguishes the

three classes of materials is the ‘energy gap’ which exists between the valence band

and the conduction band as seen in Figure 5.2 of the text and is roughly depicted

below.

In so-called intrinsic semicinductors, which are pure substances such as germanium

and silicon, the energy required to move electrons into the conduction band (from the

valence band) is relatively small – but not as small as in conductors. When valence

electrons are moved into the conduction band of semiconductors they leave behind

features referred to as holes. In this case, not only can the electrons in the conduction

band form a current, but the holes in the valence band can be considered to move in

the opposite direction – the holes can be thought of as positive charges with the same

masses as electrons. Of course, this hole current really comes about as a result of

valence electrons moving into the holes created by the electrons which have jumped to

the conduction band. While semiconductor materials follow the point form of Ohm’s

law – i.e.

𝐽 = 𝜎�⃗�

17

– the conductivity, 𝜎 is now associated with both electron and hole densities and

mobilities. Thus, in addition to the term already appearing in equation (5.8) as a

result of electrons moving in the conduction band, there is now an added term due

to the hole movement. The result is

𝜎 = −𝜌𝑒𝜇𝑒 + 𝜌ℎ𝜇ℎ (5.18)

where 𝜌ℎ and 𝜇ℎ are the hole density and mobility, respectively, and BOTH are

positive quantities. Remember, 𝜌𝑒 is a negative quantity as it is the electron density.

We note that the size of 𝜌𝑒 is the same as 𝜌ℎ, since for every electron moving into the

conduction band there is a hole is left in the valence band. However, 𝜇ℎ < 𝜇𝑒 because

the electrons in the conduction band move more freely than do those moving into the

holes of the valence band.

Intrinsic semiconductors can become much more conductive as temperature rises

– this is the opposite to the effect which temperature rises have on conductors. While

the mobilities decrease with temperature rise, this is more than compensated for

by the fact that more electrons are able to move into the conduction band as the

semiconductor warms up. (In conductors an increase in temperature causes essentially

no increase in the number of conduction band electrons, but it does cause a decrease

in mobility with an overall effect of a reduction in conductivity).

As noted in the text, by adding small amounts of impurities – a process referred

to as doping – the conductivity of a semiconductor can be greatly increased. Donor

atoms used for the doping process provide extra electrons, in which case the doped

material is refered to as an n-type semiconductor, while acceptor atoms provide extra

holes resulting in a p-type semiconductor. The increase in conductivity occurs in either

case, despite the fact that the mobilities are reduced due to the lattice imperfections

caused by the doping process.

18

5.4 Dielectrics and Boundary Conditions

5.4.1 Polarization, Susceptibility and Permittivity of Dielectrics

Polarization:

In ideal dielectrics there are no free charges such as exist in the conduction band

of a conductor. However, there are bound charges firmly associated with individual

molecules and these may slightly shift their positions if subjected to an electric field.

The shift in the relative positions of the positive and negative charges from their usual

positions as a result of applied fields gives rise to energy storage within the material.

The molecular structure of a dielectric may be polar or nonpolar:

∙ In polar molecules there is a natural separation of positive and negative charges

to form dipoles. Ordinarily these dipoles are randomly oriented within the

material, but the application of an external field may (1) cause the dipoles to

align to some extent in the same direction and (2) cause further separation of

the charges within the dipoles themselves.

∙ In nonpolar molecules there are no dipole moments until a field is applied. Then

the positive and negative charges shift apart in a direction opposite to that of

their mutual attraction and produce a dipole aligned with the field.

Illustration:

Recall Section 4.5 of these notes. Now in the present context, each dipole, whether

19

in a polar or a nonpolar material, can be represented by its dipole moment, �⃗� as

𝑝 = 𝑄𝑑 (5.19)

where 𝑄 is the positive charge of the two bound charges forming the dipole and 𝑑 is

the displacement vector from the positive to the negative charge. The unit of 𝑝 is,

as usual, the C ⋅ m. If there are 𝑛 dipoles per unit volume and we consider a small

volume Δ𝑣, then the total dipole moment, 𝑝total is simply found by summing all dipole

moments 𝑝𝑖 in the volume (remember this must be a vector sum). That is

�⃗�total =𝑛Δ𝑣∑𝑖=1

�⃗�𝑖 (5.20)

Remember 𝑛Δ𝑣 represents the total number of dipoles in Δ𝑣 since 𝑛 is ‘dipoles per

unit volume’. If the dipoles in a dielectric are aligned, 𝑝total could be significant. If

the dipoles are randomly oriented 𝑝total may be essentially zero.

Next, we define the polarization for a dielectric as the total dipole moment per

unit volume. That is,

�⃗� = limΔ𝑣→0

1

Δ𝑣

𝑛Δ𝑣∑𝑖=1

𝑝𝑖 (5.21)

Let us initially consider a nonpolar dielectric in which �⃗� = 0 if no field is applied.

On applying an �⃗�, we examine an elemental surface Δ𝑆 within the dielectric. What

follows should be read in relation to the diagram below:

In view of the diagram, we note that once the field is applied, dipole moments,

each of value �⃗� = 𝑄𝑑, are created. Each charge (i.e. the positive and negative

20

charge) associated with the newly formed dipoles moves a distance of 𝑑 cos 𝜃/2 in the

direction of the surface vector Δ�⃗� away from its undisturbed position. Thus, any

positive charges initially within a distance (𝑑 cos 𝜃)/2 below the surface will cross the

surface in the upward direction and any negative charges initially within a distance

(𝑑 cos 𝜃)/2 above the surface will cross the surface in the downward direction. Since,

after the field is applied, there are 𝑛 dipoles per unit volume in the material, the total

amount of charge crossing Δ�⃗� in the upward direction is therefore

Net bound charge crossing Δ�⃗� upward = (positive charge upward)−(negative charge downward)

or, in symbols,

Δ𝑄𝑏 = [𝑛𝑄(1

2𝑑 ⋅ Δ�⃗�)] − [−𝑛𝑄(

1

2𝑑 ⋅ Δ�⃗�)] .

This gives

We recognize 𝑛𝑄𝑑 as ‘dipole moment per unit volume’ which we have defined as the

polarization, 𝑃 . This allows us to write

Δ𝑄𝑏 = 𝑃 ⋅ Δ�⃗� (5.22)

Next, let’s suppose that the surface element Δ𝑆 within the material is closed so that

charges may pass through it in any direction in a manner similar to the special case

just discussed (i.e. where we have just assumed a element of surface not necessarily

closed). For the closed surface, Δ𝑆 may be interpreted as being outward from the

surface. Thus, if we take equation (5.22) to diffrential form and then integrate over

the surface, the net increase of bound charge within the surface is given by

𝑄𝑏 = (5.23)

Notice that while (5.22) is written for 𝑄𝑏 moving out of the surface, equation (5.23)

is written for 𝑄𝑏 inside the surface, so that the negative sign is required. We note

that equation (5.23)looks similar to Gauss’s law which said that∮𝑆�⃗� ⋅ 𝑑�⃗� = 𝑄encl.

21

Let us next suppose that an �⃗� field is applied to a material that may contain both

bound charge, 𝑄𝑏, and free charge 𝑄 so that the total charge, 𝑄𝑇 , within a closed

surface, 𝑆, is given by

𝑄𝑇 = (5.24)

and from Gauss’s law

𝑄𝑇 = 𝑄encl =∮𝑆𝜖0�⃗� ⋅ 𝑑�⃗� (5.25)

For the free charge, 𝑄, we have from equations (5.23) and (5.24)

𝑄 = 𝑄𝑇 −𝑄𝑏 =∮𝑆(𝜖0�⃗� + 𝑃 ) ⋅ 𝑑�⃗� (5.26)

We may define the factor in the parentheses as the flux density, �⃗�, within the material

and write for the free charge that

𝑄 = (5.27)

where

�⃗� = 𝜖0�⃗� + �⃗� (5.28)

Symbolizing the bound, free and total charge densities as 𝜌𝑏, 𝜌, and 𝜌𝑇 , repsec-

tively, we have

𝑄𝑏 =∫𝑣𝜌𝑏 ; ;

Then from equations (5.23), (5.24) and (5.27) we have, on using the divergence the-

orem (∮𝑆�⃗� ⋅ 𝑑�⃗� =

∫𝑣∇⃗ ⋅ �⃗� 𝑑𝑣),

∇⃗ ⋅ �⃗� = −𝜌𝑏 ; ∇⃗ ⋅ 𝜖0�⃗� = 𝜌𝑇

and for the free charge

∇⃗ ⋅ �⃗� = 𝜌𝑣 (5.29)

22

Susceptibility and Permittivity

Please read pages 140,141, and 142 of the text for more details on the following.

For an isotropic material – one whose electrical properties are the same in all

directions – the polarization, �⃗� , and electric field intensity, �⃗�, are parallel. If, in

addition, �⃗� and 𝑃 are linearly related, we may write

𝑃 ∝ �⃗�

and, in particular,

�⃗� = 𝜒𝑒𝜖0�⃗� (5.30)

where 𝜒𝑒 is a dimensionless constant referred to as the susceptibility. The susceptibil-

ity is indicative of the degree to which a material becomes polarized as it is subjected

to an electric field.

Using equation (5.30) in (5.28) we get

Now we define

𝜖𝑟 = ,

a quantity referred to as the relative permittivity or dielectric constant of the material,

and write

�⃗� = (5.31)

where

𝜖 = 𝜖0𝜖𝑟 (5.32)

is the permittivity of the material.

In a class of subtances referred to as ferroelectric materials, �⃗� and �⃗� are not only

nonlinear, but �⃗� depends on the past history of the material – this is an effect known

generally as hysteresis.

Illustration:

23

Finally, we note in passing that if a dielectric is anisotropic – i.e. it exhibits

different electrical properties in different directions – then the permittivity becomes

a 3 × 3 matrix and equation (5.31) becomes a matrix equation. This will not be

discussed further here.

Example: Problem D6.1 on page 143 of text.

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5.5 Boundary Conditions for Dielectrics

Let us now consider the interface (i.e. boundary) between two ideal dielectric mate-

rials. Let us suppose that there may be a static surface charge density 𝜌𝑆 residing

on the boundary. We shall see that treatment of this boundary, while very simi-

lar in form to our discussion of the boundary between a conductor and free space,

leads to slightly more general results. In the diagram below, both materials are ideal

dielectrics.

↓

↓

Δh

Δw

Tangential Fields:

We are still considering static electric fields, so it is permissible to write for the

25

small rectangular path shown in the figure that

Letting Δℎ approach zero so that the normal components disappear from the inte-

gration, we have

𝐸𝑡1Δ𝑤 − 𝐸𝑡2Δ𝑤 = 0

where we have used the subscript 𝑡 to indicate tangential components. This equation

thus gives

𝐸𝑡1 = 𝐸𝑡2 (5.33)

This means that just above and just below the boundary the tangential electric field

intensities are identical. (Notice that this more general case includes the special case

of a conductor and free space boundary where both of these were equal and zero.)

Now consider the tangential flux density components, 𝐷𝑡1 and 𝐷𝑡2 . Using the

constitutive relationship, �⃗� = 𝜖�⃗�, along with equation (5.33), we have

where 𝜖1 and 𝜖2 are the permittivities of Region 1 and Region 2, respectively. There-

fore,

𝐷𝑡1

𝐷𝑡2

=𝜖1𝜖2

(5.34)

From equations (5.33) and (5.34) we see that the tangential electric field intensity is

continuous across the boundary and that the tangential electric flux density is not.

Normal Fields:

For the normal field consider the cylindrical Gaussian surface as shown above. As

we did for the conductor/free space interface, we again apply Gauss’s law assuming

that somehow a surface charge density 𝜌𝑆 may have been deposited on the boundary.

In that case, with the sides of the surface shrinking to zero more rapidly than the top

26

and bottom surfaces of the cylinder we have

𝐷𝑁1Δ𝑆 −𝐷𝑁2Δ𝑆 = Δ𝑄

where the term on the right side of this equation is simply the elemental charge

enclosed by the surface. Then, since

Δ𝑄 =

we have

𝐷𝑁1Δ𝑆 −𝐷𝑁2Δ𝑆 = 𝜌𝑠Δ𝑆

or

𝐷𝑁1 −𝐷𝑁2 = 𝜌𝑠 (5.35)

As seen previously, this indicates that the normal component of the electric flux

density is discontinuous by an amount equal to the surface charge density. If 𝜌𝑆 = 0,

which one would expect for ideal dielectrics (since there is no free charge within

the dielectric and there will be none ‘naturally’ on the boundary unless there is a

deliberate attempt to place it there), then

𝐷𝑁1 = 𝐷𝑁2 . (5.36)

In this case, the normal �⃗� field is continuous across the boundary.

Using the constitutive relation we can immediately observe the normal �⃗� fields to

be discontinuous across the boundary since

𝜖1𝐸𝑁1 = 𝜖2𝐸𝑁2 . (5.37)

Example: In relation to the regions depicted in the diagram at the beginning of this

section, equations (5.33) through (5.37) may be used to determine the fields in one

region if the fields the other are known. Consider the following situation in which, in

Region 1, �⃗�1 = 𝐷1 ∕ 𝜃1. Let us determine �⃗�2 and �⃗�2 – i.e., determine the magnitudes

and directions of the Region 2 fields in terms of the Region 1 fields and the two

permittivities, 𝜖1 for Region 1 and 𝜖2 for Region 2.

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5.5 Capacitance

We have looked at the boundary conditions for conductor/free space, dielectric/dielectric

and conductor/dielectric interfaces. Some of the concepts encountered in those sec-

tions along with earlier material on Gauss’s law and potential diffrence will be useful

as we further consider capacitance.

In the depiction below, 𝑀1 and 𝑀2 are conductors imbedded in a dielectric.

Let us suppose that 𝑀1 carries a negative charge of −𝑄 on the surface, while 𝑀2

carries a positive charge of the same magnitude. From our earlier deliberations we

know the following:

∙ the static charges lie on the surfaces of the conductors.

∙ the electric fields are normal to the conductor surfaces

∙ the conductor surfaces are equipotential surfaces

∙ the electric flux is directed from 𝑀2 to 𝑀1 (i.e. from positive to negative)

∙ 𝑀2 is at a higher potential than 𝑀1 since ‘positive’ work must be done to move

a positive charge from 𝑀1 to 𝑀2.

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Next, suppose that the potential difference between 𝑀1 and 𝑀2 is

𝑉12 =

Now, we define capacitance as the ratio of the magnitude of the charge on either

conductor to the potential difference between the conductors. That is,

𝐶 =𝑄

𝑉0(5.38)

If the flux density in the dielectric is �⃗� and permittivity is 𝜖 then from Gauss’s law,

if we allow a gaussian surface to enclose, say, the positive charge, then

𝑄 =

Additionally, we may symbolically represent the potential difference from the lower

to higher potential surface as

𝑉0 =

Substituting these last two expressions in equation (5.38) allows us to write the ca-

pacitance as

𝐶 =

∮𝑆 𝜖�⃗� ⋅ 𝑑�⃗�

− ∫ +− �⃗� ⋅ 𝑑�⃗� (5.39)

We note that the capacitance is NOT dependent on the charge on the conductors or

the potential difference between them, but rather on the ratio of these two quantities

– that is, as the charge increases, the potential difference increases proportionately.

We will see, however, that capacitance depends on

∙

∙

∙

29

The unit of capacitance is the farad. Obviously, one coulomb/volt (C/V) is one farad

(F), and one farad is an extremely large capacitance ( you have probably encountered

micro, nano or pico farads).

The Parallel Plate Capacitor:

As an initial step to finding the capacitance of the common parallel plate capacitor,

lets consider two identical parallel planes which are infinite in extent and are separated

by a distance 𝑑. The space between the planes is taken to have a permittivity of 𝜖

and one plate has a charge density of 𝜌𝑠 and the other a charge density of −𝜌𝑠. To

simplify the geometry, without loss of generality, suppose that positive charge density

is in the 𝑧 = 0 plane while the negative charge density is in the 𝑧 = 𝑑 plane.

Applying the results of equation (2.25) – this is a good time to review the example

that led to that equation – to the plates oriented as shown above, the electric field

intensity between these plates is

�⃗� =

Now, in setting up the potential difference integral in the denominator of equation

(5.39), we observe that

𝑑�⃗� =

so that

𝑉0 =

Also, we note that

𝑄 =∮𝑆𝜖�⃗� ⋅ 𝑑�⃗� = 𝜌𝑠𝑆 .

30

where 𝑆 is the surface area of the plate. From equation (5.38) we therefore have

𝐶 =𝑄

𝑉0= (5.40)

There is, of course, one small ‘flaw’ in our development of this equation – we have

assumed the plates to be infinite in area and the 𝜌𝑠 to be finite, making 𝑄 infinite

in the strictest sense. However, if we consider a small portion of each plate and we

assume that the ‘fringing’ fields at the edges of the plates do not significantly affect

the capacitance (which is all that we can do in a course at this level and, besides,

it’s not a bad thing to do), equation (5.40) is perfectly fine. Thus, we see that the

capacitance indeed does not depend on 𝑄 or 𝑉0, but on the physical charateristics of

the device as noted above.

Energy in the Capacitor:

Finally, we may find the energy (𝑊𝐸) stored in the capacitor by adapting equation

(4.37) to read

𝑊𝐸 =1

2

∫vol

𝜖𝐸2𝑑𝑣 =

From, equation (5.40) and our result giving

𝑉 20 =

𝜌2𝑠𝑑2

𝜖2and 𝐶 = 𝑄/𝑉0 and 𝑉0 = 𝑄𝐶

we observe that

𝑊𝐸 =1

2𝐶𝑉 2

0 = (5.41)

We shall consider at least one other capacitive example in a tutorial.

31