Chapter 2 Static Electric Fields

Electric Field Intensity 电场强度 , Electric Potential 电势Polarization of Dielectric 介质的极化 , Field EquationsBoundary Conditions 边界条件 , Energy and Force

1. Field Intensity, Flux, and Field Lines2. Equations for Electrostatic Fields in Free Space

3. Electric Potential and Equipotential Surfaces4. Polarization of Dielectrics5. Equations for Electrostatic Fields in Dielectric6. Boundary Conditions for Dielectric Interfaces分界面 7. Boundary Conditions for Dielectric-conductor Interface

8. Capacitance电容9. Energy in Electrostatic Field10. Electric Forces

1. Field Intensity, Flux, and Field Lines

The intensity of the electric field at a point is defined as the force produced by the electric field on a unit positive charge at that point, and is denoted by E

q

FE

Where q is the test charge, and F is the force acting on the charge.

The flux of the electric field intensity through a surface is called

electric flux, and it is denoted as , given by,i.e.

S

SE d

0d lEThe vector equation for electric field lines

Electric field tube

Two parallel charge plates

A negativepoint charge

A positive point charge

Distributions of electric field lines

2. Equations for Electrostatic Fields in Free Space

Physical experiments show that the intensity of an electrostatic

field in free space satisfies the following two equations in integral form

S

q

0

d

SE l

0d lE

Where 0 is the permittivity 介电常数 (or dielectric constant) of free

space and its value is

The left equation is called Gauss’s law, and it shows that the outward

flux of the electric field intensity of an electrostatic field over any closed

surface in free space is equal to the ratio of the charge in the closed

surface to the permittivity of free space.

F/m1036π

1m/F10854187817.8 912

0

The right equation states that the circulation of an electrostatic

field around any closed curve is equal to zero.

0

E

0 E

which shows that the curl of an electrostatic field in free space is zero

everywhere.

Using the divergence theorem, from Gauss’ law we have

It is called the differential form of Gauss’s law. It shows that the

divergence of the electric field intensity of an electrostatic field at a

point in free space is equal to the ratio of the density of the charge at

the point to the permittivity 介电常数 of free space.

From Stokes’ theorem and the above equation, we have

The electrostatic field in free space is a irrotational one.

AE

V

V

V

V

d)(

4π

1)(

d)(

4π

1)(

|rr|

rErA

|rr|

rEr

where

After knowing the divergence and the rotation of the electric

field intensity, one may write, with the aid of the Helmholtz’s

theorem

x

P

z

yr

O

V d )(rrr

rV

V

V

0

d)(

4π

1)(

|rr|

rr

0)( rA

Substituting the electric field equations into the above results

gives

EHence The scalar function is called the

electric potential, and the electric field

intensity at a point in free space is equal

to the negative gradient of the electric

potential at that point.

E

According to the National Standard

of China, the electric potential is denoted

by the Greek small character , i.e.x

P

z

yr

O

V d )(rrr

rV

If the electric charge is distributed on a surface S or on a curve

l, we have

S

S S

0

d|

)(

π4

1)(

|rr

rr

S

S S 3

0

d|

))((

π4

1)(

|rr

rrrrE

l

ld)(

π4

1)(

0 |rr|

rr l

l

l l 3

0

d|

))((

π4

1)(

|rr

rrrrE

Substituting the electric potential expression into this

equation, we haveV

V

dπ4

))(()( 3

0 rr

rrrrE

It is easy to see that the electric field

intensity can be determined directly

from the above equations if the

distribution of the charge is known.

x

P

z

yr

O

V d )(rrr

rV

（ a ） The charge q in the Gauss’ law should be the sum of all

positive and negative charges in the closed surface S.

Summary

（ b ） The electric field lines cannot be closed and intersect each other.

（ c ） The line integral of the electric field intensity along a path

between any two points is independent of the path, and electrostatic

field is a conservative field as the gravitational field.

（ d ） If the distribution of the charge is known, the electric field

intensity can be found based on Gauss’ law, the electric potential, or

the distribution of the charge.

Example 1 Calculate the electric field intensity produced by a point charge.

Solution: The point charge is the charge whose volume is zero.

Because of the symmetry of the point charge, if the point charge is

placed at the origin of a spherical coordinate system, the electric

field intensity must be independent of the angles and .

Construct a sphere of radius r and let the point charge be at the

center of the sphere, then the magnitude of electric field intensity at

all of the points on the surface of the sphere will be equal. If the point

charge is positive, the direction of the electric field intensity is the

same as that of the outward normal to the surface of the sphere.

S

q

0

d

SEApplying Gauss’ law

The left hand side of the above equation is

SS S

ErSES

2n

4dd d eESE

And we have

20π4 r

qE

rr

qeE

20π4

or

S

q

0

d

SE

We also can use the formula for electric potential or electric

field intensity to calculate the electric field intensity produced by

the point charge.

If the point charge is placed at the origin, then . We find

the electric potential produced by the point charge as

r || rr

r

q

0π4)(

r

rr

q

r

qeE

200 π4

1

π4

The electric field intensity E as

rV

r

r

qV

re

erE

20

20 π4

dπ4

)(

If the equation for electric field intensity is used directly, we find

the electric field intensity E as

Example 2 Calculate the electric field intensity produced by an electric dipole.

Solution: For an electric dipole consists of

two point electric charges, we can use the

principle of superposition to calculate the

electric field intensity. Then the electric

potential produced by an electric dipole as

rr

rrq

r

q

r

q

000 π4π4π4

If the distance from the observer is much greater than the separation

l, we can consider that and are parallel with , andre

re re

coslrr 2cos

2cos

2r

lr

lrrr

x

-q

+q

z

yl

r

r-

r+

O

where the direction of the vector l is defined as the direction

pointing to the positive charge from the negative charge. The

product ql is often called the electric moment of the dipole, and

is denoted by p, so that

lp q

)(π4

cosπ4 2

02

0rr

ql

r

qel

We have

20

20 π4

cos

π4 r

p

rr

ep

Then the electric potential of the dipole can be written as

x

-q

+q

z

yl

r

r-

r+

O

sin

11

rrrE r eee 3

03

0 π4

sin

π2

cos

r

p

r

pr

ee

By using the relation , we can find the electric field

intensity of the electric dipole as

E

* The electric potential of the dipole is inversely

proportional to the square of the distance, and the

magnitude of the electric field intensity is inversely

proportional to the third power of the distance.

These properties are very different from that of a point charge.

* In addition, the electric potential and the

electric field intensity are both dependent of the

azimuthal angle .

x

-q

+q

z

yl

r

r-

r+

O

Electric field lines

Electric field lines and equipotential surfaces of

an electric dipole

Electric field linesEquipotential surfaces

Example 3 Assume that an infinitely long charged cylinder of radius

a is placed in free space. The density of the charge is . Calculate the

electric field intensities inside and outside the cylinder.

x

z

y

a

L

S1

Solution: Choose a cylindrical coordinate

system. Since the cylinder is infinitely long,

for any z the environment remain the same.

Hence the field is independent of the

coordinate variable z.

In addition, the cylinder is rotationally symmetrical. Thus the field

must be independent of the angle .

Since it is symmetrical about any plane

of z = constant, the electric field intensity

must be perpendicular to the z-axis, and

coplanar with the radial coordinate r.

Since the direction of the electric field

intensity coincides with the outward normal of

lateral surface S1 of the cylinder everywhere and

is perpendicular to the normal to the upper or the

lower end faces, the left side of the surface

integral becomes

rLESESESSS

π2ddd11

SE

For r < a , the charge , and the electric field intensity

is

Lrq 2π

r

reE

02

Construct a cylindrical Gaussian surface of radius r and length L.

Applying Gauss’ law, we have

S

q

0

d

SE

x

z

y

a

L

S1

rl

reE

0π2

For this example, it is easy to calculate

the electric field intensity using Gauss’ law.

If we directly calculate the electric field

intensity from the distribution of the charge,

it will be very complicated.

When r > a, the charge , we have

Laq 2π

rr

aeE

0

2

π2

π

where can be considered as the charge per unit length, so

that the electric field can be thought of as caused by a line charge

with density . In view of this we can derive the electric field

intensity caused by an infinite line charge with line density as

2πal

2πa

l

x

z

y

a

L

S1

x

z

y

r

2

1

r

O

rr

zdz r

ze

re),

2

π,( zrP

Example 4 Find the electric field intensity caused by a line

charge with length L and charge density .l

Solution: Let the z-axis of a cylindrical

coordinate system coincide with the line

charge and the mid point of the line be

at the origin.

It is impossible to apply Gauss’ law to calculate the field because

the direction of the electric field intensity cannot be found out in

advance. We have to evaluate directly the electric potential and the

electric field intensity.

Since it is rotationally symmetrical,

the field is independent of the angle .

lL

L

dπ4

2

2 3

0 |rr|

rrE l

2

π

Since the field is independent of the angle , for convenience,

let the field point P be placed in the yz-plane, i.e. . Then

d csccsc

sincos

π42

220

2

1

rr

a

a

zr

ee

E lWe have

])cos(cos)sin[(sinπ4 1212

0rz

l

ree

x

z

y

r

2

1

r

O

rr

zdz r

ze

re),

2

π,( zrP

Considering

d cscd

cot

)sincos(csc

csc|

2rz

rzz

r

r

rz eerr

|rr

rl

rl

rreeE

00 π22

π4

This result is the same as that of Example 3.

])cos(cos)sin[(sinπ4 1212

0rz

l

reeE

If the length L , then 1 0, 2 and

x

z

y

r

2

1

r

O

rr

zdz r

ze

re),

2

π,( zrP

3. Electric Potential and Equipotential Surfaces

The physical meaning of the electric potential at a point is

the work W by the electric field force to move a unit charge from the

point to infinity, Hence

The electric potential at a point is in fact the electric potential

difference between the point and infinity. In principle, any point can

be taken as the reference point for the electric potential.

Obviously, the electric potentials for a given point will be different

when difference references are chosen.

q

W ( q is the amount of the charge.)

However, the electric potential difference between any two points

is independent of the reference point.

When charges are confined to a region, infinity is usually taken as

the reference point of the electric potential because the electric potential

at infinity will be zero.

The selection of reference point has no effect on the value of electric

field intensity.

The electric field lines are perpendicular to the equipotential

surfaces everywhere.

Equipotential surfaces are the surfaces on which the

electric potential are equalized, and the equation is

Czyx ),,(

where the constant C is equal to the value of electric potential.

Electric field lines

Equipotential surfaces

E

If the electric potential differences between any adjacent equi-

potential surfaces are constant, then where the equipotential surfaces

are concentrated, there the change of electric potential will be faster,

and the field intensity will be stronger.

In this way, the distribution density of equipotential surfaces can

indicate the intensity of the electric field.

4. Polarization of Dielectrics

The electrons in a conductor are called free electrons, and their

charges are called free charges. the electrons in the dielectric cannot

escape from the binding force and these charges are called bound

charges.

The bound charges in dielectrics will be displaced due to the

electric fields, and this phenomenon is called polarization.

Usually, the polarization of nonpolar molecules is called

displacement polarization, while the polarization of polar

molecules is called orientation polarization.

Based on the distribution of the bound charges in dielectric,

molecules can be classified into two kinds: polar and nonpolar

molecules.

Polar

Molecule

Nonpolar

Molecule

Nonpolar Molecule Polar Molecule

Ea

Polarization of dielectrics

The polarization is formed gradually, as shown in the following

figure.

DielectricTotal field Ea+ EsApplied field Ea

Because the directions of electric dipoles are almost coincided

with that of applied field, the direction of the strongest secondary field

in the middle of a dipole is always in opposite direction of applied field,

that isasa EEE

Secondary field Es

Polarization

In order to measure the polarization intensity, we define the

vector sum of the electric dipole moments per unit volume as the

polarization, denoted by P

Vi

i

N

pP 1

where pi is the electric dipole moment of the i-th electric dipole in volume

V , and N is the number of electric dipoles in volume V . Here, V is

an infinitesimal volume.

Most dielectrics are polarized by an applied electric field, their

polarization P is proportional to the total electric field intensity E in

the dielectrics, so that

EP e0

where e is called the electric susceptibility, and it is usually a positive

real number.

In the above dielectrics, the direction of polarization is the same as

that of the total electric field intensity. When the susceptibility e

remains unchanged for all directions of the applied field, the material

is called an isotropic dielectric.

z

y

x

z

y

x

E

E

E

P

P

P

33e32e31e

23e22e21e

13e12e11e

0

For certain dielectric materials, the susceptibility is dependant

upon the direction of the applied field. The polarization is in a different

direction from the total field. The relationship between the polarization

and the electric field intensity will need to be described in terms of the

following matrix:

This kind of material is called an anisotropic dielectric.

isotropic

anisotropic

Dielectrics whose electric suspectibilities are uniform in space are

called homogeneous dielectrics, otherwise they are called

inhomogeneous dielectrics.

If the value of the electric susceptibility is independent of the

magnitude of the electric field, the dielectric is called a linear dielectric,

otherwise it is a nonlinear dielectrics.

If the electric susceptibility is independent of time, the dielectric

is called a static medium, otherwise, it is called a time-dependent

medium. Homogeneity, linearity, and isotropy stand for three very

different properties that are in general independent of each other.

Properties of dielectrics

homogeneous

linear

time-independent

After a dielectric is polarized, there exists as a result some bound

surface charge on the boundary of the dielectric. if the dielectric is

inhomogeneous, there will be some volume bound charges in the

dielectric. These surface and volume charges are also called polarization

charges. We can show that the electric potential produced by these

polarization charges as

VVS

d

||

)(

π4

1

||

d)(

π4

1)(

0

0 rr

rP

rr

SrPr

The net volume bound charge and the net surface bound

charge are equal but opposite.

The right equation can be rewritten as

S

q

dSP

n)()( erPr S )()( rPr and

5. Equations for Electrostatic Fields in Dielectric

The electrostatic field in dielectrics can be considered as the

electrostatic fields produced by both the free charges and the bound

charges in free space. In this way, in dielectrics the electric flux over

any closed surface S is

where q the free charges ， the bound charges . q

)(1

d 0

qq

S

SE

qS

0 d)( SPE

We have known , then S

q

dSP

Let ， we havePED 0

qS

d SD

D

Where the vector D defined is called electric flux density (or electric

displacement). We can see that the outward flux of electric flux

density over any closed surface is equal to the total net free charges

in the closed surface, and independent of the bound charges.

The above equation is called Gauss’ law for dielectrics.

Using the divergence theorem and considering , we

can find

VVq

d

where is the differential form of Gauss’ law for dielectrics, and it

states that the divergence of the electric flux density at a point is

equal to the density of the net free charge at the point.

Electric flux density lines: The direction of tangent line at a point

on the curve is the direction of electric flux density.

Electric flux density lines start from positive free charges and end

with negative free charges, and they are independent of bound charges.

The polarization of isotropic dielectrics , and EP e0ε

EEED )1( e0e00 εε

Let , then)1( e0

ED εwhere is called the permittivity of the dielectrics (or called dielectric

constant).

Since the electric susceptibility e is usually a positive real number,

the permittivities of most dielectrics are greater than that of free space.

Relative permittivity r is defined as

e0

r 1

The relative permittivities of all dielectrics are usually greater than 1.

Relative permittivities of several dielectrics

Dielectrics r Dielectrics r

Air 1.0 Quartz 3.3

Transformer Oil 2.3 Mica 6.0

Paper 1.3~4.0 Ceramic 5.3~6.5

Plexiglass 2.6~3.5 Pure Water 81

Paraffin Wax 2.1 Resin 3.3

Polythene 2.3 Polystyrene 2.6

For anisotropic dielectrics, the relationship between the electric

flux density and the electric field intensity can be written as

z

y

x

z

y

x

E

E

E

D

D

D

333231

232221

131211

The relationship between the electric flux density and the electric

field intensity is dependent upon the direction of applied electric field.

In addition, the permittivity of a homogeneous dielectric is

independent of the space coordinates.

The permittivity of a static media is independent of time.

The permittivity of linear dielectrics is independent of the magnitude

of the electric field intensity.

For homogeneous dielectrics, since the permittivity is not

a function of the coordinates, and we have

q

S d SE

E

The previous relationship between the electric field, the electric

potential and the free charges still hold as long as the permittivity of

free space is replaced by the permittivity of dielectrics.

6. Boundary Conditions for Dielectric Interfaces

Since different properties of the media, the field quantities will

be discontinuous at the interface between two media. This change is

governed by the boundary conditions for the electrostatic field.

Usually, the boundary conditions include tangential and normal

components.

To derive the relationships between the field quantities at the

boundary, we may let h 0, then two of the linear integrals become

0d d 1

4

3

2 lElE

E2

E1 1

2

et

Tangential Components

1

4

4

3

3

2

2

1 d d d d d lElElElElE

l

Then the circulation of the electric

field intensity around the rectangular

path is given by

Constructing a rectangle about

the point of consideration on the

boundary, the length is l and the

height is h. l

h3

1 2

4

In order to find the relationships between the field quantities at a

point of the boundary, the length l should be small enough so that the

field is constant over l, then

lElE ΔΔd d d t21t

4

3 2

2

1 1 lElElE

2t1t EE

The tangential components of the electric field intensities in both

sides of the interface between two dielectrics are equal. Or say, the

tangential components of the electric field intensities are continuous.

For linear isotropic dielectrics, we have

2

2t

1

t1

DD

Since , we have0d

lE

Constructing a small cylindrical

surface about a point on the boundary. Its

height is h and the end face area is S.

Let h 0 , the flux through the lateral surface will be zero.

And considering S to be very small, one has

S

SDSD 1n2ndSD

The direction of the normal to the boundary is specified as that

from dielectric ① to ②.

hS 1

2

en D2

D1

Normal Components

S

q

dSD

Then the flux of electric flux density

over the cylindrical surface is equal to the

amount of net free charge enclosed by the

surface, that is

SS

qDD

1n2nWe have

where S is the surface density of the free charge at the boundary.

2n1n DD

The normal components of the electric flux densities at the boundary

between two dielectrics are equal. In other words, the normal components

of the electric flux densities are continuous at the boundary.

For linear isotropic dielectrics, we obtain

n221n1 EE

In addition, we can find the relationship between the bound charges

and the normal component of the electric field intensity as

)( n1n20 EES

In the absence of net surface free charge, one has

7. Boundary Conditions for Dielectric-conductor Interface

A state of electrostatic equilibrium.

An electrostatic field cannot exist inside a conductor.

Electrostatic equilibrium

The free charge in conductors must be zero, and they are only

found on the surface of the conductors.

A conductor in electrostatic equilibrium is an equipotential body,

and the surface is an equipotential surface.

Since the electrostatic field cannot exist inside a conductor, but

the tangential component of electric field intensity at any boundary

is continuous, and the electric field intensity must be perpendicular

to the surface of the conductor, we have0n Ee

Dielectric

E, D

Conductor

en

De nS SE nor

Due to , we haven

E

n

S

n

Since there is no electrostatic field inside a conductor, the

polarization is zero. We have

Sn Pe

If an electric field exists near to a

conductor, surface electric charge will

be induced on the surface of the

conductor

S

Electrostatic shielding

If there are no free charges inside of a closed cavity made of

conductor, no electrostatic field exists inside the cavity, although

charges may exist outside.

q = 0

E = 0

A closed cavity made of conductor shields its interior from the

influence of any electrostatic field outside of it, and this effect is called

electrostatic shielding.

If the cavity is grounded, the charges inside it cannot produce any

electrostatic fields outside of it.

If a fictitious closed surface is constructed within the wall of the

cavity, the electric flux over the surface must be zero.

There is either no charge enclosed by this surface, or it contains

positive and negative charges of equal amount. If the former is true, no

electric fields can exist inside the cavity. If it is the latter case, there

could be electric field inside the cavity. However, it is impossible.

The positive and the negative charges

only exist on the surface, and the red

path could be an electric field line.

Then the circulation of the electric field

around the entire closed curve will not be

zero. That is contrary to the basic property

of electrostatic field.

Example. A conducting sphere of radius r1 and with positive

charge q is enclosed by a conducting spherical shell of internal radius r2.

The permittivity of the dielectric between the sphere and the shell is 1,

and the external radius of the shell is r3 . The spherical shell is covered

by a dielectric of 2 with external radius r4. The outer region is vacuum.

r1

r2

r3

r4

0 2

1

Solution: In view of the spherical

symmetry of the structure and the fields,

Gauss’ law can been applied.

Find: (a) The electric field intensities in every region. (b) The free

charges and the bound charges on each surface.

Taking the spherical surfaces as

Gaussian surfaces, it can be seen that the

electric field intensities are perpendicular

to them.

In the regions r < r1 and r2<r <

r3 , E = 0, since electrostatic field

cannot exist in conductors. r1

r2

r3

r4

0 2

1

rr

qeE

21

1 π4

rr

qeE

22

2 π4 For the same reason, in the region r3<r <

r4 ,rr

qeE

20

0 π4 In the region r > r4 ,

In the region r1<r < r2 , due to

S

q

dSD we have

Based on the equations and ， we can

find the free charges and the bound charges on each surface as

follows:

SPenSDen

r =

r1:

21π4 r

qS

011

π4 1r2

1n10

r

qE SS

r =

r4: 01

1π4

)(2r

24

n2n004

r

qEES

0S

r =

r2:

22

2 π4 r

qS

01

1π4 1r

22

21n02

r

qE SS

r =

r3:

23

3 π4 r

qS

011

π4 2r2

332n03

r

qE SS

r1

r2

r3

r4

0 2

1

8. Capacitance

we know that the ratio of the positive charge q on the plate of a

parallel plate capacitor to the electric potential difference U across

the two plates is a constant. This constant is called the capacitance of

the parallel plate capacitor, and

U

qC

The unit of capacitance is F, but it is too big to use.

For instance, an isolated conducting sphere with the radius of

earth has a capacitance of only F. 310708.0

In practice, we often use F ( ) and pF ( ) as

the unit for capacitance.

F10μF1 6 F10pF 1 12

The capacitance of an isolated conductor can be considered as the

capacitance between the isolated conductor and infinity.

To calculate the capacitances between multiple conductors, the

partial capacitance is used. In this case, the electric potential of each

conductor depends not only on its own charge but also on the charges

of the other conductors.q1 q3

qn

q2

nnnjnnjnnnnn

niinjiijiiiii

nnkj

nnhj

CCCCq

CCCCq

CCCCq

CCCCq

)()()(

)()()(

)()()(

)()()(

2211

141

222222212212

111121121111

If the medium is linear, the

potentials of the individual conductors

are proportional to the charges, and

the charges are

where Cii is called the partial self-capacitance of the i-th conductor,

and Cij is called the partial mutual-capacitance between the i-th and

the j-th conductors.

Example. Given that a coaxial line of internal radius a and

external radius b, and the permittivity of the dielectric between the

internal and the external conductors is . Find the capacitance per

unit length between the internal and the external conductors.

Solution: Since the electric field intensity

must be perpendicular to the surface of the

conductor, the direction of the electric field

intensity in the coaxial line must be along the

radial direction. Due to symmetry, Gauss’ law

can be conveniently applied.

a

b

a

b

Suppose the charge per unit length of

internal conductor is q, and construct a

cylindrical surface around the internal

conductor as a Gaussian surface S, then

S

q

d

SE

The electric potential difference U between the internal and the

external conductor is

b

a a

bqrEU

ln

π2d

Hence, the capacitance per unit length for the coaxial line is

a

b

U

qC ln/π2

rr

qeE

π2

Gaussian surface

9. Energy in Electrostatic Field

Under the influence of electrostatic fields, a positively charged

body will be moved along the direction of the electric field. In this

case, work is being done by the field. In order to do this work the

electrostatic field must lose energy, implying that the electrostatic

field has stored energy.

The total energy stored in an electrostatic field can be determined

from the work done to assemble the charges giving rise to the field.

If the charged body is moved into the electrostatic field from

infinity by an applied force, work is being done against the electric

force. This work will be become part of the energy stored in the

field, and the total energy of the electrostatic field will be increased.

First, we calculate the energy of an isolated body with charge Q.

Assuming that the charge Q was moved in from infinity. Since

there was no field in space at the beginning, the applied force did not

need to do the work to move in the first element dq.

qqWQ

d )(

0 e

The electric potential of a charged body will be increased

gradually as more charge is brought in. When the charge is increased

to the final value Q, the total work by the applied force is given by

When the second element dq was moved in, the applied force had

to do the work against the electric field. If the electric potential at the

charge is , then the work by the applied force is dq . Hence, the

increment in the energy of the electric field is dq.

The electric potential of an isolated conductor is equal to the

ratio of the charge q to the capacitance C, i.e.

C

q

We find the energy of the isolated conductor with the charge Q is

C

QW

2

e 2

1

C

QQW ,

2

1eor

The total energy of n charged bodies can be calculated, in the

same way assuming the charges of all charged bodies are increased

by the same ratio from zero. If the dielectric around the bodies is

linear, then the electric potentials will be double when these charges

are doubled.

ddd11

e

n

iii

n

iii QΦqW

When the charges of all bodies are increased to the final values

, the total energy of the system is given bynQQQ ,,, 21

1

0 1

ee dd n

iiiQWW

n

iiiQW

1e 2

1We have

Let the final electric potential of the i-th body be i and the final

charge on it be Qi. At the moment when the charge of the i-th body is

increased to , the electric potential of the charged

body is then . ii 10 , ii Qq

In this way, when the charges of all bodies are increased by the

same ratio , the work by the applied force, and hence the incremental

electrostatic energy of the charged system is

If the charge is distributed in a volume, on a surface, or at a

line, considering , then the total energy of

the body with the distributed charge can be written accordingly as

lSVq lS ddd d

lSVW llS SVd

2

1d

2

1d

2

1e

Where is the electric potential at the element volume dV, the

element surface dS, or the element line dl , respectively.

From the point of view of the field, the electrostatic energy

is distributed in the entire space where the electric field is found.

We now discuss how to calculate the energy density of electro-

static field.

The energy density of electrostatic field is denoted by small

letter character we.

S2

Q2

Q1

S1

V en

en

ne

ne

Suppose the charges on two conductors are Q1 and Q2, and the

surfaces are S1 and S2, respectively.

S

Since the charges are on the

surfaces of the conductors, the total

energy of the system is therefore

21

d 2

1d

2

1e S SS S SSW

21

d 2

1d

2

1e SS

W SDSD

Constructing an infinite spherical surface at infinity, the

electric potential and the field on S will be zero since the charges

are finite and confined. Therefore,

0d

SDS

Where , andnn eDeD S

The above equation for the stored energy can be written as

SDSDSD d 2

1d

2

1d

2

121

e SSS

W SD d 2

1 S

where , which encloses the whole space occupied by the

electrostatic field. By using divergence theorem, we have SSSS 21

VWV

d ) (2

1e D V

Vd ) (

2

1 DD

Since there is no charge in the region V outside the conductors, it

follows that, and , we obtain 0 D E

VWV

d 2

1e

ED

We can see that the energy density of the electrostatic field

is ED2

1ew

For linear isotropic dielectrics, , we haveED

2e

2

1Ew

The energy of an electrostatic field is proportional to the square

of the magnitude of the field intensity. Consequently, the energy does

not obey the principle of superposition.

Reason: When the second charged body is introduced, work needs

to be done against the electric field due to the first charged body, and

this work done becomes the energy stored in the field. This energy is

called mutual energy, while the energy is called self energy when the

charged body is alone.

The total energy is not equal to the sum of their energies when they

exist on their own.

Example. Find the energy of a conducting sphere with radius a

and charge Q. The permittivity of the dielectric around the conductor

is .Solution: Three methods can be used.

(1) The electric potential of a conducting sphere with radius a and

charge Q is

a

Q

π4

a

QQW

π82

1 2

e We find

(2) Since the surface of a conductor is an equipotential surface, and

Sa

QW SS

d π42

1e

a

Q

π8

2

(3) The electric field intensity caused by a conducting sphere with

charge Q is

2 π4 r

QE

a

QrrwW

a π8d sindd

2

2e

π

0

π2

0 e

42

2

e π32 r

Qw

And the energy density is

Integrating over the region outside the conductor gives

Three results above are in agreement.

10. Electric Forces

The magnitude of the electrostatic field intensity at a point is equal

to that of the force experienced per unit positive charge. Hence the

force acting on the point charge isq

EF q

If E is the electric field intensity produced by another point charge

q, then we haverr

qeE

2 π4

Then we find the force acting the point charge due to the point

charge q as

q

rr

qqeF

2 π4

where er is the unit vector directed to the point charge from the point

charge q. The equation is Coulomb’s law, and it is formulated by the

French scientist, Charles Augustin de Coulomb, based on his

experiments.

q

In principle, if the charge distributions of the charged bodies

are known, then the forces among the charges can be calculated

from Coulomb’s law. Nevertheless, for charged systems with

complex charge distributions, it is very difficult to calculate the

electric forces directly using Coulomb’s law.

To calculate the forces among the charged bodies with

distributed charges, the method of virtual work is usually employed.

This method assumes that the charged body undergoes a

certain displacement in the electric field. From the relationships

between the change in energy, the applied force, and the work by

the electric field, the electric force can be calculated. 。

we take a parallel plate capacitor for

example. The charges on the two plates

are assumed to be +q and -q , respectively,

and the separation between them is l. dl

l

–q

+q

As is well know that the force between two plates leads to the

separation, therefore the force calculated by this hypothesis should

be negative.

For convenience in calculation, we

suppose that the differential increment of

the distance between the two plates due to

the electric force is dl .

By definition of work done by a force, we have . lFdd lF

edd WlF

constante

d

d ql

WFAnd we have

where q = constant states that the charges on the plates were unchanged

when they were displaced. This charged system is called a constant

charge system.

The energy of the parallel plate capacitor is 。 For

constant charge system, the charges q are unchanged when the

displacement happens, and only the capacitance C is changed.

C

qW

2

e 2

1

Based on the principle of energy conversation, this work should be

equal to the decrement in the energy of the electric field, so that

And we find the force between two plates of the capacitor to be

The capacitance of the parallel plate capacitor is .l

SC

S

qF

2

2

Here the negative sign in right hand side of the equation states that

the practical direction of the force is pointed to the direction of the

decreasing displacement.

If we suppose the capacitor is always connected with an

impressed source when the plates are displaced, then the electric

potential is constant when the virtual displacement happens. This is a

system of constant electric potential.

The force between two plates of the parallel plate capacitor can be

determined from this hypothesis of constant electric potential, and the

same results will be obtained.

Assume the differential increment of the separation between the

two plates is dl due to the electric force.

qUqqW d2

1d

2

1d

2

1d 21e

where is the electric potential difference between the two

plates. 21 U

Suppose the electric potentials of the positive and the negative plates

are 1 and 2 , respectively, then the differential increment of the

energy of electric field is

Since the capacitance is changed, in order to keep the electric

potential the charge of the positive plate has to have a differential

increment dq, and that of the negative plate -dq .

In order to push the charge dq to the positive electric plate with

the electric potential 1, and to move the charge -dq to the negative

electric plate with the electric potential 2 , the work done by the

impressed source is

e21 d2d)d(d WqUqq

Based on the principle of energy conversation, one part of the

work by the impressed source is provided to the electric force to do

work, and another part of that will be become the differential

increment of the energy stored in the electric field, hence

ee ddd2 WlFW

constante

d

d l

WF

edd WSF constante

d

d qS

WF

Example 1. Calculate the surface tension acting on the electric

plates of a parallel plate capacitor by using method of virtual

work. Solution: Using the virtual work, we suppose that the surface

tension on a plate due to the repellent function of the same charges

is F. The plate will be extended by dS due to this repellent force,

and the work by the electric force is FdS.

Based on the principle of energy conversation, this work is equal

to the decrement of the energy of the electric field, that is

The energy of the parallel plate

capacitor , and we find that the surface

tension is

l

SC

C

qW

,

2

1 2

e

N/m 2 2

2

S

lqF

Consider , we obtain the same results. l

SCCUW

,

2

1 2e

constante

d

d l

WF

If the plates are connected with an impressed source in the

virtual displacement, then the electric potentials are unchanged and

the surface tension F should be

If the variable l is considered as a generalized coordinate variable,

so that l can stand for displacement, surface area, volume, or even the

angle, the forces to change these generalized coordinates are then

called the generalized forces for these generalized coordinates.

constante

ql

WF constant

e

l

WF

If the directions of the generalized forces are still specified as

the direction of the generalized coordinates, then the product of the

generalized force by the generalized coordinate is still equal to the

work. In this way, the above equations can be rewritten as

Where the l represents the coordinate corresponding to the generalized

force.

Generalizedcoordinates

Generalizedforces

Units

Displacement Normal force N

Surface Surface tension N/m

Volume Tension/Pressure N/m2

Angle Torque Nm

Generalized Coordinates & Generalized Forces

The number of generalized coordinates affecting the energy

of the charged system is equal to the number of generalized forces

existing in this system.

For instance, the energy of a parallel plate capacitor is related

not only to the separation between two plates, but also to the

surfaces of the plates. Therefore, there exist two kinds of forces in

the capacitor. One is the force between the two plates, and another

is the surface tension on each plate.

If one generalized coordinate of a charged system is changed,

but no change in the energy of the system results, then there exists

no generalized force for this generalized coordinate in this system.

Example 2. Calculate the tension on a charged bubble made of

soap liquid.

Solution: Assume the charge of the soap bubble is q and the radius

is a. Using constant charge system and let the generalized coordinate l

represent the volume V, then the tension F is

constante

qV

WF

The electric potential of a charged sphere with the radius a

and the charge q isa

q

0π4

And the energy isa

qqW

0

2

e π8

2

1

The volume of the sphere is 3 π3

4aV aaV d π4d 2

We obtain 24

02

2e

2N/m

π32 π4

1

a

q

a

W

aF