chapter 2 static electric fields electric field intensity 电场强度, electric potential 电势...
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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
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