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Electrodynamics
REN Xincheng, Postdoctoral , Professor
Tel ： 2331505; 13310918078
Email:[email protected]
The most simple case of electromagnetic field is discussed in this chapter——
the charge is resting, namely, electrostatic field (coulomb field), the
corresponding electric field does not vary with time. The main question of
investigation is:
Electric chargeThe space is given Distribution Solve electrostatic ( )
Dielectric and conductorE x
TreatmentTreatment method: : The characteristics of electrostatic field (Conservative
force field) is used, the electric potential is introduced (Because there are
many convenience in scalar calculation ), electrostatic field is obtained by the
potential.
Chapter 2. Electrostatic Field Chapter 2. Electrostatic Field (Coulomb Field)(Coulomb Field)
Main content:Main content:
2 ） Secondly, several methods for solving electrostatic field is introduced mainly——separation of variables 、 method of images and Green function method;
3 ） Finally, the expanded form of electric field in the distance when the charge distribution is in a small region, the concept of the electric multipole moment is introduced, which have important applications in problems of atomic physics, nuclear physics and electromagnetic radiation.
1 ） Firstly, the scalar potential is introduced, the potential differential equations and boundary conditions are obtained;
§2.1 Electrostatic potential and it’s §2.1 Electrostatic potential and it’s differential equation differential equation 11 、、 The introduction of scalar potentialThe introduction of scalar potential
2
12 1( ) ( ) ( ) (2.1.1)
P
P PP P E dl P E dl
)2.1.2(E
ldEd
In electromagnetics, the concept of electrostatic potential (scalar potential) is i
ntroduced by the results that electrostatic field is a conservative force field, th
e integral relationship of electric potential and electric field (Taken as the refe
rence point zero at infinity) is:
Attention: If the distribution of charge is in infinite area, the point of limited distant
should be taken as the reference point zero of potential. The differential relationship of electric potential and electric field is:
The electric potential is introduced using the characteristics of electrostatic field.
EED
)4.1.2(0)3.1.2(
Illustration ：
2 ） The distribution of electric potential can be obtained by (2.1.1) in the con
dition that field is known, The distribution of field can be obtained by (2.1.2) i
n the condition that electric potential is known.
1 ） Here, the negative sign is introduced, in order to consistent with the pote
ntial defined in electromagnetics. This field points to the direction of decreasi
ng electric potential. So, the field points to the direction of decreasing of elect
ric potential;
Point charge (The reference point zero of electric potential is taken at
infinity)
superposition principle3
0 0 0
1( )
4 4 4
Qr Q dqE x
r r r
3) In general case, the relationship of the charge and electric field is with
interaction and confinement (the inner of medium and conductor is such), the
distribution of charge and field are determined simultaneously (When
electrostatic equilibrium).
For this reason, in general, the problem of interaction between the field and the
field, the charge and the field near it is need to research. The equation of
differential form describes the local nature of the field, the equation of
differential form should be adopted to solve the general problem.
22 、、 The differential equationsThe differential equations of of PotentialPotential
The basic equations describing electrostatic field.
DEE
0
/ EED
2
Poisson equation
For homogeneous, isotropic and linear medium
Which can be
Illustration ：
1)This equation applies to homogeneous dielectric;
2 ） Although the free charge appears in equation only, but the potential is also related to the bound charge, theεin equation is reflected that the medium influence on the electric potential. This can be illustrated by changing of the form of the Poisson equation.
fPEPED
00 )(
0
2
0
PfPfE
)6(0)()5(
12
12
EEnDDn ）（
33 、、 TheThe boundary value relationboundary value relation of of When there are two different medium in the space, and the boundary value relati
on of field is satisfied in their interface:
Accordingly, the boundary value relation of the the scalar potential can be obtained.
1 2
2 12 1
(7)
(8)n n
0 ldEd
21 1介质
2介质
12
1E2E
ld
E
Obtaining (8) by (5) is obvious. To applied to the boundary surface are:
So there It also can be obtained from (6).
Special caseSpecial case ：1 ） For dielectric ； 2 ） For conductor and dielectric
The basic problem of electrodynamics is that solving the boundary value problem of corresponding problem, which solving the Poisson equation that satisfied the boundary value relations and the boundary conditions. Of course, in some cases, applications Coulomb's law directly is more convenient, then Coulomb's law can solve the problem as a starting point.
V
dVDE2
1 W
2
1~,
1~
,
rE
r
DE
1
E D2
1
2
w
But w
Attenti on：
V
dV2
1
44 、、 The energy of electrostatic fieldThe energy of electrostatic field
Example 1. solving the electric potential of homogeneous field (E0 P55)
xEldEPP
000)()(
Here, you can not choose infinity point as the reference point zero, and taken coordinate origin as the reference point.
Example 2. The linear charge density of infinitely long and charged straight wire uniformly isτ, solving the potential. (P56)
school assignment (P93) ： 1
§2.2 Uniqueness theorem of electrostatic §2.2 Uniqueness theorem of electrostatic boundary value problemsboundary value problems
The practical significance of uniqueness theoremuniqueness theorem ：
1 ） It solve what the conditions are met, the solving of the electrostatic field to be the only;
2 ） For many practical problems, some analysis need to be given according to given the conditions frequently, try solution need to be proposed. The uniqueness theorem can judged right or wrong of try solution;
3 ） It is the basis for Electric Image.
11 、 、 Uniqueness theorem of electrostatic bounUniqueness theorem of electrostatic boun
dary value problemsdary value problems TTheoremheorem 1 1 ： To the region V of uniform partition, namely, the
inner of V can be divided into a number of uniform area Vi, the permittivity of any uniform area is εi. The distribution of free charg
e of inner of V is known, then the boundary S on V is given
S|
Sn|
1) Electric potential ； or 2） normal derivative of Electri
c potential； Then, electric field within the V has a unique and definite solution.
The proving of theorem (Using negative approach):
' ''and '''
Suppose there are two solving satisfy the conditions that required by theorem, let
So2
2
2
' /0 (In each homogeneous region)
'' /
11 ,V
iiV ,S( ) ( )
i j
i i j j
region i and j
n n
regi on
0|''|'| SSS
' ''| | | 0S S Sor
n n n
In the boundary of two homogeneous region
In the boundary S of the whole region, there
Consider surface integral of interface Si of uniform region of
section i, such as the following
iS
iii Sd
( )i
i i iVdV
ii V iiiV ii dVdV 22)(
iV
ii dV2)(
i
V iii
S iiiii
dVSd 2)(
Sum of all the partitions Vi ,there is:
Since in interface of Vi and Vj with V the to meet the
jjjiiijjii
ji
nnnn
)()()()(
( )i jand dS dS is both normal of outside
00)( 2 i
V iii
dV
0|0|
SS n或
Therefore, in the integral of the above formula of the left, the integral of inner
interface cancel out each other, while the integral of outer interface, have:
, so the integral of the interface S is also zero. So
have:
Then Φ=constant.namely, up to a constant difference, But, the a
dditional constant of the electric potential doesn’t influence on the electric fiel
d, which proves the uniqueness theorem.
' ''and
22、、 The uniqueness theorem with conductorsThe uniqueness theorem with conductors Conductor is different from the dielectric, the distribution of conductor surfa
ce charge and electric field of outside conductor are interdependent, and ther
efore can not be known in advance. In this case, what conditions to be attach
ed to the conductor, the distribution of electric field within the region was un
iquely identified, which have the following theorem:
TTheoremheorem 2: 2:： There is conductor in the region V, suppose if exc
epting conductor, the distribution of charge in V is known
On the boundary S of V, is known. In addition, if the potential
of the conductor surface or the total charge carried by conductor is known,
then the distribution of electric field of the region V has a unique solution.
| |S Sorn
1conductor'V
S
CertificationCertification ： We prove Theorem 2 based on Theorem 1, the region re
moving the conductor is denoted by V ', in V', only the insulating medium (As shown below right figure, for simplicity, assuming there is only a homogeneous medium).
According to Theorem 1, Theorem 2, it is self-
evident that the establishment of the case of
conductor given the potential, the following is only
required to prove the second case, namely, the case
that the total charge carried the conductor is known.
The conditions that the electric quantity carried by conductor is known can be expressed as:
/ (known to a conductor of arbitrary)d Q
,
0 (to a conductor of arbitrary)d
)conductorgiven any For (0 d
0
' and '' '''
If two solutions satisfy the conditions required by Theorem 2, then the difference between the two solutions must satisfy the following formula:
Consider equipotentiality of conductor, we will have to:
Then the argumentation that is exactly same argument of Theorem 1 is used on
t he region of V’, we can conclude:
Thus, the Theorem 2 is proved.
Example(P62) ： As shown, the interspace of two concentric conducting spher
e is filled with two types of medium, the capacitance ratio of left part is ε1, the
capacitance ratio of right part is ε2, assuming the electrification in inter shell is
Q, outer shell grounded. located within the spherical shell carrying charge Q, outer shell is grounding. Requesting the electric field and charge distribution on the shell.
34 r
rQE
1 21E
11DQ
Explanation:
Method: Based on uniqueness theorem, and consider th
e results of solution of single medium, propose trial
solution, and then further verify that if it satisfies the co
nditions required uniqueness theorem.
In the case that single medium,the field
E is spherically symmetric radial, that is
Here, the conditions required by Uniqueness Theorem are :
nn
tt
DDEE
21
21
3r
rAE
AdSEdSEQSS
)(2 21221121
1 ） E near the conductor must be perpendicular to the conductor surface (equipotential surface);
2 ） The interface between two medium satisfy the boundary value relation:
021 nn DD
The first condition requires E is radial, i.e. along the radial direction, this time, ,It is naturally satisfied.
The second condition requires E is spherically symmetric, based on the solution of single medium, it can be assumed:
Where A is the undetermined coefficient, it is determined charge Q carried by the inter spherical.
So)(2 21
Q
A
1 2 21 2
,2 )
(
Along the radialdirectionQ
E E Dbut unsymmetrir
cal
221
22222
221
11111
)(2
)(2
a
QED
a
QED
rr
rr
It satisfies all the conditions of requirement of uniqueness theorem ， so it is the only correct solution.
Here, the charge distribution on the conductor surface is solved.
(P63)
Using boundary relations
Illustration: It is difficult to obtain the general solution.
§2.3 §2.3 Laplace equationLaplace equation Separation of Separation of variablesvariables
一、一、 Separation of variablesSeparation of variables
02
In many practical problems, the electrostatic field is determined by the
energized conductor. The characteristics of these problems is that the charge
is only in some of the conductor surface, no other charge distribution in space.
If we discuss the region V with boundary of these conductors, then the
Poisson equation reduces to Laplace equation.
The problem inverts to solve Laplace equation satisfied the boundary conditions. The general solution of Laplace equation can be obtained by separation of variables. The basic thought of separation of variables is studied in Mathematical Methods of Physics.
The corresponding coordinate system is selected based on the shape of discussed area before application separation of variables firstly, the specific form of Laplace equation is wrote in the corresponding coordinates, the general solution is obtained by using separation of variables, the undetermined coefficient is determined through the edge conditions (boundary value relations and the boundary conditions), the solution of electrostatic problem is obtained.
The case of Spherical coordinates The case of Spherical coordinates ： The general solution of Laplace equations have been given in Appendix II （ Have a symmetric axis and taken as the polar axis ） ,it is:
)0)(sinsin
1)(( 2
r
rr
拉氏方程为
)(cosnP nn ba , is Legendre polynomials (function) ， is pending con
stant ， determined by the boundary conditions 。 The following few example
s illustrate undetermined coefficients determined by boundary conditions.
Example 1.(P.48) An inner and outer diameter, and Conducti
ng sphere ， ChargedQ ， Concentrically surrounds a radius Conducto
r Ball （ ） . Seeking space at each point and the potential of this c
onductor ball induced charge
1R2R 3R
1 2R R
n
nnnn
n Pr
brar )(cos)(),(
1
Example 2.(P.49) Capacitance rateεdielectric sphere placed in a uniform external electric fieldE0， Seeking potential.
cos2
3
cos
2cos
00
02
2
300
0
001
RE
R
RERE
The general solution。 UseR→0， R→∞andR= R0Undetermined coefficients determined boundary value relationship.
Final results
Example 3 .(P.52) Examples of cylindrical coordinates： Conductor charged potential wedgeV， Analysis of the electric field near its corners（ Approximate） .
Solution：Using cylindrical coordinates， zaxisAlong the edges can be seen as infinitely long
Example 3.(P.51) Conductor radius sphere placed in a uniform field， Seeking potential and surface charge density on the conductors.
2
2 2
1 10( )r
r r r r
Operation(P93)： 2、 7、 18 Accompanied with a steady current of the electric field is
also the same as the electrostatic field EJ
22 2
2
22
20
d R dRr r R
dr dr
d
d
通解
0 0 0 0A B( ln )( ) ( )( cos sin )r C D A r B r C D
0 2 0, ,V r ; 有限
101 1
1
1 22
sin sin ( , , , )n rn n n
n
nV A r V A r n
Wedge space outside（Wedge space outside）
§2.4 §2.4 Mirror ActMirror Act Scope of Review of separation of variables:
Scope mirror law: Only one or a few in the region point charge. Border area is a c
onductor or dielectric interface. And the boundary region has a relatively simple s
hape.
Based on the imaging method: Uniqueness Theorem
The basic idea of the mirror law: Conceived by one or several point charge to repl
ace the conductor equivalently (Or media) surface charge distribution(Free charg
e and bound charge), Field strength generated by it and Conductor (Or media) eq
uivalent surface field intensity distribution generated in the charge.These imagina
ry point charge is called the elephant charge, this method is called Mirror Act, als
o called Electric Image.
So original charge and like charge Co-excitation field is what we are asking the electric field
Note Note Uniqueness theorem requirements: 1 ） Like charge must appear in the region in question so poisson
equation in order to ensure that the original unchanged; 2 ） Ensure that the boundary conditions remain unchanged.We
took advantage of the boundary conditions are usually identifie
d as the location and size of the charge. Example 1(P71) There is a point charge Q near infinite ground pl
ane conductor plate, Seeking in the field of space. Solution ： Analysis from the physical conductor surface induc
ed charge and static equilibrium conditions: conductor surface i
s equipotential surface (potential was0), or field lines and conduc
tor flat vertical.
In about a position facing said infinite imagine an elephant charge-Q .the above conditions can be achieved. The space (The half-space) electric field(Potential) distribution
Q
Q
r
'ra
a
)'
(4
1
0 r
Q
r
Q
))(
)((
4
1
222
2220
azyx
Qazyx
Q
)
'
'(
4
133
0 r
rQ
r
rQE
DiscussDiscuss ： 1 ） When the top was filled capacitorsεis homogeneous medium，
this time to change the continuous discontinuous；
)'
(4
1
r
Q
r
Q
21
2
21
21
2''
'
1
2
Qa
r P
2 ） When the two intersect in a uniform dielectric infinite plane。needing discuss zoning, Electric field were obtained in the first
half and a lower half-space space (Situation), With boundary
conditions determine the coefficients.The required result is
Example 2(P72) ： Vacuum has a radius is R0 grounded conducting
sphere ， From the center of the sphere is a（ a>R0 ） there is a po
int charge Q ， Find the potential at each point of space.
cos444
'
4 2211
1raar
Q
r
Q
r
Q
22 4
''
0R
aQ
b'Q
P
0
r
'r
R solution ： first useGeometric me
thods; Second use Algebraic methods. Results
Discuss ： 1) Conducting sphere surface charge distribution 。 Use of bound
ary relations
)'
(4
1 0
0 ar
QR
r
Q
)cos2
/
cos2(
4
122
0
220 RbbR
aQR
RaaR
Q
Qa
RQ
a
Rb 0
20 ',
0|00 RRRn
0|]
cos2
/
cos2
1[
4 22
0
220
0 RRRbbR
aR
RaaRR
Q
2/30
220
0
20
2
)cos2(4
aRaRR
RaQ
0
20 sin2 dRq
)(cos)cos2(4
20
2/30
220
0
20
220
daRaRR
RaQR
'0 QQa
R
Qq Visible induced charge as a charge equal to the conductor
surface. But Q Description field lines emitted by the surface of a part of the conductor termination, the rest ends at
infinity.
2 ） Conductor ball is not grounded →In the center of the
sphere,the charge increases .'Q
]'
)(
'[
4 20
20 a
ba
QQF
]/1
)/(
/[
4 20
220
0
0
2
a
aR
aRa
aRQF
3 ） Conductor ball is not grounded ， Originally charged Q0 → In the center of the sphere, the charge increases . 4 ） Discuss force issues.Let’s talk about the situation that Conductor
ball is not grounded and originally charged Q0 . Charge Q suffered
electrostatic forces (Including the induced charge on the surface of the conductor and the original charge distribution) .All of these should be charged as a force on the Q, i.e.
If Q0 =Q ， then have
0' QQ
Be seen, electricity of the charged body are the same size and equal p
ositive and negative. However, the force can be repellent smokable. This is determined by Induced charge distribution on the conductor changes with distance. Among them,
Value of 0.618 is important. That is the golden thread. 1/1.618=0.618 。
；， repulsion is0618.1 0 FRaWhen
；， unstressed0618.1 0 isFRaWhen
nalgravitatio is0618.1 0 FRaWhen ，
TaskTask(P95) ： 8 、 9 、 11
§2.5 §2.5 Green's function methodGreen's function method
Corresponding to the first boundary value problem or second boundary value
problem
)||( SS n 或
Object of study ： Given charge distribution ρ inside V , potential
of each point on the boundary of V or the potential of normal
derivative .Find the potential distribution within
the points of V.
Processing method ： solve the problem based mirror image method by means of boundary value problem of point charge
.Point charge is the limit of a continuous charge distribution.Green formula associated with the solution between Boundary Value Problems of point charge and issues to be discussed in this section .To this end we first introduce a point charge density δ function and the Green's function.
1. Representation of point charge
density δ function Point charge is limit of large charge density and small volume of charged
body(Provided that total charge is unchanged). Units point charge
density at the origin is expressed as a δ function that
000
)(xx
x
) chargepoint origin the(1)( includeVdVxV
And
In mathematics, δ function is a generalized function. It can be see
n as limit of certain continuous function .
'x
To Unit point charge of ， its charge density is
''0
)'()(xxxx
xxx
)'(1)'( xincludeVdVxxV
)0()()( fdVxxfV
)'()'()( xfdVxxxfV
)(xf'x
In 's neighborhood of a continuous function have
)(xf
Nature of the δ functionδ ： For the origin neighborhood of the continuous function
2. Green's function It described boundary value problem in mathematical
methods. For the boundary V of an area S ， If the function in V
)()()( 2 xfxsatisfyx
problem value
boundary first thecalled is)(|
)()(1
2
xf
xfxThenS
problem alue v
boundary second thecalled is)(|
)()(
2
2
xfn
xfx
S
)'()( xxx 'x If one Unit point charge is in ， The charge density is
Then the potential ψ in the volume V included of satisfies the Poisson equation
'x
)'(1
)(0
2 xxx
problem alueboundary vfirst 0|
)'(1
)(0
2
isxxx
S
problem alueboundary v second1|
)'(1
)(
0
0
2
is
Sn
xxx
S
Similarly, we said
Respectively, the corresponding solution are referred to the first boundary value problem for the Green's function or the second boundary value problem for the Green's function.
)()'(1
)',(0
2 xtoactxxxxG
)',( xxG
'x
)',( xxG Green's function is generally denoted by .In which, represents
the point where the point charge ， x represents the observation point (field point), meet
Generally speaking ， Green's function is obtained by mirror image method . In the previous section, a few examples ， a point charge Q be replaced by 1 (That is replaced by a point charge) Green's function is obtained in corresponding region.
Example(1) Unbounded space of Green's function
(2) Green's function on the half-space
22200 )'()'()'(4
1
4
1)',(
zzyyxxrxxG
)'(41
41
proveCan 22 xxr
dVrV
2220 )'()'()'(
1(
4
1
zzyyxx
)'
11(
4
1)',(
0 rrxxG
))'()'()'(
1222 zzyyxx
)'(1
)',(0
2 xxxxGSo
(3) Green's function outside the sphere of space
)'
'/1(
4
1)',( 0
0 r
RR
rxxG
cos'2'
1(
4
122
0 RRRR
)
cos'2)'
(
1
20
2
0
RRRR
RR
z r
'P
xR
P'' xR
b
3. 3. solution of Green's formula solution of Green's formula and boundary value problemand boundary value problem To contact the Green‘s function and general Boundary Value Prob
lems through Green formula. Green formula is:
SV
dSnn
dV )()( 22
0
2 1
)',()( xxGx
In which,Ψ and any two functions within the region V.Now we
take meet
get
Then Green formula change to
V
dVxxGxxxxG )]',()()()',([ 22
S
dSn
xxGx
n
xxxG )
)',()(
)()',((
getcanxxxxGUse )'(1
)',(1
0
2
0
2
及
])'()()',([1
0 VV
dVxxxdVxxG
S
dSn
xxGx
n
xxxG )
)',()(
)()',((
)'(x
Customarily, we use for field variables ， for source variabl
e
. For this reason we swap them
'x
x
VdVxxGxx '),'()'()(
S
dSn
xxGx
n
xxxG ')
'
),'()'(
'
)'(),'((
0
0|)',( SxxG
SV
dSn
xxGxdVxxGxx ')
'
),'()'('),'()'()( 0
Sn
xxGS
0
1|
'
),'(
To first boundary value problem
S| )(x Thus given G and , can be obtained within the region V
To second boundary value
problem
Then
SSVdS
n
xxxGdVxxGxx
''
)'(),'('),'()'()( 0
ExplanationExplanation ： Generally,Green's function’s acquisition is very difficult.
Only those case that boundary surfaces are simple, we can calculate the Green's function.
But if we seek a boundary surface of Green's function ， then we get the solution to this boundary surface corresponding with Boundary Problem. This is the thing once and for all.
ExampleExample(P83) TaskTask(P97) ： 19
§2.6 §2.6 Electric multipole momentsElectric multipole moments The charge distribution at a given electric potential distribution in vacuo
VdV
r
xx '
4
)'()(
0
r
x'x
o
xR
R
QdV
R
xx
V00 4
'4
)'()(
If the problem is in a small area in the far field (Charge distribution is in a small area ， The point we examined away from
the area located the charge). And if the required accuracy is not h
igh, we can consider the charge as a point charge focusing on th
e origin.(That is, for a point charge approximation)r=R
This is approximately the lowest potential level of approximation. To
improve the accuracy of the issues discussed, we must consider
the potential level of correction, two other multi-level correction
correction. This can be clearly seen from discussed below electric
potential of the multipole expansion .
1. The multipole expansion of electric potential Far-field problems of distribution charges in a small regional can
be used as potential Taylor expansion.(Actually Taylor expansion to 1 / r )
0')'( xxxf 在 'x
In the neighborhood of an arbitrary continuous function the Taylor expansion of is(Select the
coordinate origin in the source region):
3
1,0'|)'(
''''
!2
1
jix
jiji xxf
xxxx
3
10'0' |)'(
''|)'()'(
ix
iix xxf
xxxxfxxf
)(:''!2
1)(')( xfxxxfxxf
xRrxx
xxfletIf
1
'
1)'(
0' of odneighborho in theThen, x
x
xxx
xxxxr
1:''
!2
11'
1
'
11
Then the potential expression expands to
R
xxR
xR
1:''
!2
11'
1
V
dVR
xxR
xR
xx ')1
:''!2
11'
1)('(
4
1)(
0
system theof quadrupole electric the''')'(3 VdVxxxD
system theof charge total the')'( VdVxQLet
system theof charge total the'')'( VdVxxp
)1
:6
11(
4
1)(
0
RD
Rp
R
Qx
As can be seen, the potential of the lowest order approximation is approximation of point charge (is equivalent to the potential as the full charge is concentrated at the coordinate origin in space ). This is said at the beginning of this section). Here the electrical potential of multipole moments will be discussed separately.
R
Qx
0
)0(
4
1)(
300
)1(
4
1
4
1)(
R
Rp
Rpx
2. 2. Potential of Electric multipole momentsPotential of Electric multipole moments
1 ） The firstIs the lowest order
approximation
2 ） The second
is first lever correction term. It represents one potential that located in the coordinate origin and the electric dipole moment
of the dipole is p in the distance. Be defined as the electric dipole moment. If the charge symmetry about the origin system, the electric dipole moment of system is zero. Therefore, only the charge on the origin of the system of asymmetric electric dipole moment is not zero. The total charge is zero and the easiest charge system of non-zero electric dipole moment is a pair of positive and negative point charge. Such charge-dipole potential system
became the main items of the potential. Assuming that there is a
point charge + Q on the set point ， a point charge -Q on th
e set point ,then 'x
'x
)charge. negative theto charge positive from vector a is ('2 lQlxQp
zeQlp If this point charge located at Z-axis ， then
Now, by point charge superimposing potential we illustrate that the charge system at a certain approximation is dipole potential.
P
x
z
yo
Q
Q2/l
2/l
Rr
r
)11
(4 0
rr
Q
cos2
cos2so field,far a toDue l
Rr
lRr
)cos
2
1
cos2
1(
4 0
lR
lR
Q
This result is the electric dipole potential. However, the process of appr
oximate calculation shows that , Such potential positive and negative c
harges on the system also contains potential correction advanced tha
n the potential of the electric dipole. But the electric dipole potentials pl
ay a major role.
30
202
220 44
cos
cos4
cos
4 R
Rp
R
Ql
lR
lQ
RDx
1:
24
1)(
0
)2(
3 ） The third exhibition style is potential second correction
It represents the electric quadrupole potential at the origin. Let us
first discuss the nature of the electric quadrupole moment,then di
scuss the significance of electric quadrupole moments.
VdVxxxD ''')'(3
)3,2,1,(''')'(3 jidVxxxDV jiij
322331132112332211 DDDDDDDDD 、、、、、
corresponding Component formula
jiij DD Visibility, electric quadrupole moment is symmetric tensor consisting of nine components.
Therefore, it is up to six independent components.
Can prove, in fact, only five independent components
For this reason the definition V
dVlrxxxD ')'''3)('(' 2
RD
RD
1:'
1:
0332211 DDD
D Its contribution to the potential equal to ’s contribution to the potential that
By the new definition of the electric quadrupole moment, it's easy to get
So electric quadrupole moment tensor only have five independent components.
Here let's further analysis of electric quadrupole moments of the physical meaning of the various components.
pp 和
To a pair of positive charge and a negative charge system at Z-axis. It can be seen consisting of a pair of electric dipoles
Figure the positive charge is located at , the negative charge is located at ,By definition
ba
00 pQ，
][3 222233 QbQaQaQbD
plababQabQ 6))((6)(6 22
P
x
z
yoQ
Q
Rr
r
Q
Q]11
[4
121
0
r
pr
p
)11
(4
1
0
rrz
p
In which l = a + b is the center distance between the upper and lower electric dipole.
Potential superimposed by a pair of reverse electric dipole, the potential for the system is
If the charge distribution of a system is spherically symmetric, then
)(4
1)
cos(
4
13
02
0 R
z
zpl
R
l
zp
RzD
Rzpl
1
6
1
4
11
4
12
2
330
2
2
0
')'('3
1')'('')'('')'(' 2222 dVxrdVxzdVxydVxx
2211 ,DD
33D This is the electric quadrupole potential above the sub-system. Th
erefore, the simplest charge system with component is composition of the two pairs of positive and negative charge at the Z-axis. Similarly, the simplest charge system with
component is composition of the two pairs of positive and negative charge at the x-axis and y-axis
So charge distribution system of spherically symmetric, neither have the
electric dipole moment nor the electric quadrupole moment and electric m
ultipole moment. Only charge distribution deviate spherically symmetric,
the system exists electric multipole moments. Electric multipole moment
s of existence marked deviation from the spherically symmetric charge di
stribution. Therefore, the shape of charge distribution can make certain in
ferences through a multi-polar potential measurements in the far field.
0Thus 332211 DDD
0Obviously 231312 DDD
Tip: In fact, to prove 01
:')')'( 2 RdVlrx
V
V
jiij dV
Rxxlrx '
1')'( 2左
V
jiij dV
Rxxrx '
1')'( 2
V
ii
dVRxx
rx '1
')'( 2
0'1
')'( 022 R
VdV
Rrx
3. Energy of charge system in the external field3. Energy of charge system in the external field
Then
V e dVxxW )()(
)0(!2
1)0()0()( e
jijie
iiee xx
xxx
xx
V e
jijie
iie dV
xxxx
xxxW ])0(
!2
1)0()0()[(
)0(:
6
1)0()0( eee DpQ
e)(x
Let potential of the external electric field is .In the external electric field ,the energy of System with the charge distribution (Interaction energy) ：
)(xe
If the charge distribution is in a small area,taking any point in small
area as the coordinate origin point ， Expand on the origin of
In the external field, energy is when charge of the envisaged syste
m focused on the origin .
is energy at the origin of the system of electric dipole in external field.
)0()0(eQW
)0()0()1(ee EppW
)0()]0([)1(ee EpEpWF
sin]cos[
)1(
ee pEpEW
L
eEpL
Generally
Furthermore, in the system by electric dipole in an external electri
c field the force and moment were
is the energy electric quadrupole in external field. Visible only in the non-uniform
external field quadrupole energy was not zero.
)0(:6
1)2(eEDW
Example(P90) ： Uniformly charged elongated spheroid semi-major axis is a, short axle is b, with a total charge Q, seeking its electric quadrupole potential and distant.
TaskTask （ P96 ）： 16
