Transcript Document

Chapter 3 Boundary-Value Problems
in Electrostatics
Differential Equations for Electric Potential
Method of Images
Method of Separation of Variables
1.Differential Equations for Electric Potential
2.Method of Image
3.Method of Separation of Variables in Rectangular Coordinates
4.Method of Separation of Variables in Cylindrical Coordinates
5.Method of Separation of Variables in Spherical Coordinates
1. Differential Equations for Electric Potential
The relationship between the electric potential  and the electric
field intensity E is
E  
Taking the divergence operation for both sides of the above
equation gives
  E   2
In a linear, homogeneous, and isotropic medium, the divergence
of the electric field intensity E is
E 


The differential equation for the electric potential is
 2  


which is called Poisson’s equation.
In a source-free region, and the above equation becomes
 2  0
which is called Laplace’s equation.
The solution of Poisson’s Equation.
In infinite free space, the electric charge density  (r ) confined
to in V produces the electric potential given by
1
 (r ) 
4π

 (r )
dV 
V | r  r |
which is just the solution for Poisson’s Equation in free space.
Applying Green’s function G (r , r ) gives the general solution of
Poission’s equation
 ( r )
 (r )   G (r , r )
dV  
V

 [G(r , r ) (r )   (r )G(r , r )]  dS
S
For infinite free space, the surface integral in the above equation
will become zero, and Green’s function becomes
G0 (r , r ) 
1
4π | r  r  |
In the source-free region, the volume integral in the above
equation will be zero. Therefore, the second surface integral is
considered to be the solution of Poisson’s equation in source-free
region, or the integral solution of Laplace’s equation in terms of
Green’s function.
An equation in mathematical physics is to describe the changes
of physical quantities with respect to space and time. For the
specified region and moment, the solution of an equation depends on
the initial condition and the boundary condition, respectively, and
both are also called the solving condition.
Usually the boundary conditions are classified into three types:
1. Dirichet boundary condition: The physical quantities on the
boundaries are specified.
2. Neumann boundary condition: The normal derivatives of the
physical quantities on the boundaries are given.
3. Mixed boundary-value condition: The physical quantities on
some boundaries are given, and the normal derivatives of the physical
quantities are specified on the remaining boundaries.
For any mathematical physics equation, the existence, the stability,
and the uniqueness of the solutions need to be investigated.
The existence of the solution is that whether the equation has a
solution or not for the given condition of the solution.
The stability of the solution refers to whether the solution is
changed substantially when the condition or the solution is changed
slightly.
The uniqueness of the solution is whether the solution is unique or
not for the prescribed condition of the solution.
Electrostatic fields exist in nature, and the existence of the solution
of the differential equations for the electric potential is undoubted.
The stability of Poisson’s and Laplace’s equations have been
proved in mathematics, and the uniqueness of the solution of the
differential equations for the electric potential can be proved also.
In many practical situations, the boundary for the electrostatic field
is on a conducting surface. In such cases, the electric potential on the
boundary is given by the first type of boundary condition, and the
electric charge is given by the second type of boundary condition.
Therefore, the solution for the electrostatic field is unique when the
charge is specified on the surface of the conducting boundary.
For electrostatic fields with conductors as boundaries, the field
may be given uniquely when the electric potential , its normal
derivative, or the charges is given on the conducting boundaries. That
is the uniqueness theorem for solutions to problems on electrostatic
fields.
2. Method of Image
Essence: The effect of the boundary is replaced by one or
several equivalent charges, and the original inhomogeneous region
with a boundary becomes an infinite homogeneous space.
Basis：The principle of uniqueness. Therefore, these charges
should not change the original boundary conditions. These
equivalent charges are at the image positions of the original charges,
and are called image charges, and this method is called the method
of images.
Key：To determine the values and the positions of the image
charges.
Restriction：These image charges may be determined only for
some special boundaries and charges with certain distributions.
（1）A point electric charge and an infinite conducting plane
P
P
r
r
q
Dielectric
Conductor
q
h
h
q
r
Dielectric
Dielectric
The effect of the boundary is replaced by a point charge at the
image position, while the entire space becomes homogeneous with
permittivity , then the source of electric potential at any point P
will be due to the charges q and q',
q
q


4π r 4π r 
Considering the electric potential of an infinite conducting plane
is zero, we have q    q.
The distribution of the electric field lines and the equipotential
surfaces are the same as that of an electric dipole in the upper halfspace.
z


The electric field lines are perpendicular to the conducting
surface everywhere, which has zero potential.
动画
Electric charge conservation ：When a point charge q is above
an infinite conducting plane, the induced opposite charge will be
distributed on the conducting surface, and the magnitude of the
image charge should be equal to the total induced charge.
We can also prove the claim by making use of the relationship
between the density of the charge and the electric field intensity or
the derivative of the electric potential on the conducting surface.
The above equivalence is established only for the upper halfspace for which the source and the boundary condition are both
unchanged.
z


For the semi-infinite wedge conducting boundary, the method of
images is also applicable. However, the images can be found only for
π
conducting wedges with angle given by where n is an integer. In
n
order to keep the wedge boundary at zero-potential, several image
charges are required.

/3
q

/3
q



When an infinite line charge is nearby an infinite conducting
plane, the method of images can be applied as well, based on the
principle of superposition.
（2）A point charge and a conducting sphere.
P
a
O
d
r
r'
q
q
f
To replace the effect of the
boundary of the conducting sphere,
let an image point charge q' be
placed on the line segment between
the point charge q and the center of
the sphere. Then the electric potential on the surface of the sphere is
then given by
q
q


4π r 4π r 
Requiring that the electric potential at any point on the surface
of the sphere be zero, the image charge must be
q  
r
q
r
The ratio r  must be constant for any point on the surface of
r
the sphere to obtain an image charge with a fixed value.
r a
If △OPq'~ △OqP , then

r
f
.
Thus the quantity of the
image charge should be
q  
P
a
O
d
a
q
f
r
r'
q
q
f
The distance d is
a2
d
f
The electric field intensity outside the sphere can be found out
from q and q' .
If the conducting sphere is ungrounded, then the opposite
charges will be induced on the side of the conducting sphere facing
the point charge, while the induced charge on the other side of the
sphere is positive. The total induced charge on the surface of the
conducting sphere should be zero.
If the image charge q' is put in, then another image charge q" is
needed in order to satisfy the neutrality condition.
q   q 
P
a
O
d
r
r'
q
q
f
The image charge must be at the center of the sphere to ensure
that the surface of the sphere is an equipotential surface.
In fact, since the sphere is ungrounded, the electric potential is
non-zero. Since q and q' produce a zero potential on the surface of the
sphere, the second image charge q" is present to produce a certain
electric potential.
P
a
O
d
r
r'
q
q
f
（3）A line charge and a charged conducting cylinder.
P
a
O
r
-l
l
d
f
An image line charge   l is put in to represent the charge on
the cylinder and placed parallel to it, at a distance d from the axis.
The electric field intensity produced by an infinite line charge of
density  l given by
l
E
er
2 π r
The electric potential with respect to a reference point r0 at a
distance r from the line electric charge is
  
r
r0
Edr 
 l  r0 
ln  
2π  r 
If r0 is also taken as the reference point for the electric potential
produced by the image line charge   l , then  l and   l produce an
electric potential at a point P on the surface of the cylinder given by
l  r0  l  r0 
l  r  
P 
ln   
ln   
ln  

2π  r  2π  r  2 π  r 
Since the conducting cylinder is an equipotential body, in order
to satisfy this boundary condition the ratio r  must be a constant.
r

Let r  a  d , we find
P
a
O
r
r
-l
l
d
f
f
a
a2
d
f
（4）A point charge and an infinite dielectric plane.
1
2
E n
q
q
et
en
=
1
1
q'
r0
E'
E t Et

r0
En
q"
E
+
2
2
r0
E n
E t

E"
To find the field in the upper half-space, the image charge q'
can be used to replace the effect of the bound charges on the
interface, and the entire space becomes a homogeneous medium
with permittivity1.
For the lower half-space, the function of the point charge q
and the bound charges on the interface can be replaced by the
image point charge q" at the position of the original point charge
q, and the entire space becomes a homogeneous medium with
permittivity 2.
The fields must satisfy the boundary condition, i.e., the tangential
components of the electric fields are continuous, and the normal
components of the electric flux density are equal on the other side.
E1t  E1t  E2t
We have therefore,
D1n  D1n  D2n
The electric field intensities produced by the point charges q, q',
and q" are, respectively,
E1 
q
4π 1r 2
E1 
er
q
er 
4 π 1 (r ) 2
E 2 
q
e r 
4 π 2 (r ) 2
To satisfy the boundary condition, we find the image charges
are, respectively,
q 
et
en
=
1
1
q'
q  
E n
q
q
1
2
1   2
q
1   2
r0
r0
En
q"
E'
E t Et

E
2 2
q
1   2
+
2
2
r0
E n
E t

E"
Example. Given the radius of the internal conductor of a coaxial
line is a, and its electric potential is U. The external conductor is
grounded, and its internal radius is b. Find the electric potential and
the electric field intensity between the internal and the external
conductors.
Solution: The method of images cannot be
used, and we have to solve the differential
equation for electric potential.
V
O
a
b
The cylindrical coordinate system is selected.
Since the field depends only on the variable r, the
Laplace’s equation for the electric potential only
involves the variable r, as given by
1 d  d 

r
0
r dr  dr 
2
We have
  C1 ln r  C2
We have
  0 rb
  U r a
Considering
C1 ln a  C2  U

C1 ln b  C2  0
Solving for the constants C1 and C2 gives
C1 
V
O
a
U
a
ln  
b
C2  
U ln b
a
ln  
b
b
We obtain

 r 
a
  U ln   ln  
 b 
b

er U
  
E    er  




r
r a


ln  
b
For the given boundary-value problem it is very important to
select the coordinate system in order to determine the integration
constants from the given boundary conditions.
In general, the boundary of the region of interest should conform
with the coordinate system to be used.
In practice, the boundary-value problems for electrostatic fields
are related to three coordinate variables. One efficient method to solve
three-dimensional Laplace’s equation is the method of separation of
variables.
This method reduces a three-dimensional partial differential
equation to three ordinary differential equations, and the method of
separation of variables is established for 11 coordinate systems.
3. Method of Separation of Variables in Rectangular Coordinates
In rectangular coordinate system, Laplace’s Equation for electric
potential is
 2  2  2
 2  2 0
2
x
y
z
Let
 ( x, y, z )  X ( x)Y ( y ) Z ( z )
Substituting it into the above equation, and dividing both sides by
X(x)Y(y)Z(z), we have
1 d 2 X 1 d 2Y 1 d 2 Z


0
2
2
2
X dx
Y dy
Z dz
Where each term involves only one variable. The only way the
equation can be satisfied is to have each term equal to a constant.
Let these constants be  k x2 ,  k y2 ,  k z2 , and we have
d2 X
2

k
X 0
x
2
dx
d 2Y
2

k
yY  0
2
dy
d2Z
2

k
Z 0
z
2
dz
where kx ，ky ，kz are called the separation constants, and they could
be real or imaginary numbers.
The three separation constants are not independent of each other,
and they satisfy the following equation
k x2  k y2  k z2  0
The three-dimensional partial differential equation is separated
to three ordinary differential equations, and the solutions of the
ordinary differential equations are easier to obtain.
The solution of the equation for the variable x can be written as
X ( x)  Ae  jk x x  Be jk x x or
X ( x)  C sin k x x  D cos k x x
where A, B, C, D are the constants to be determined.
The separation constants could be imaginary numbers. If k x is
an imaginary number, written as k x  j , then the equation becomes
X ( x)  Aex  Be x
or
X ( x)  C sinh x  D cosh x
The solutions of the equations for the variables y and z have the
same forms. The product of these solutions gives the solution of the
original partial differential equation.
It is very important to select the forms of the solutions, which
depend on the given boundary conditions.
The constants in the solutions are also related to the boundary
conditions.
Example. Two semi-infinite, grounded conducting planes are parallel
to each other with a separation of d. The finite end is closed by a
conducting plane held at electric potential  0 , and is isolated from the
semi-infinite grounded conducting plane with a small gap. Find the
electric potential in the slot constructed by the three conducting planes.
y
=0
 = 0
d
O
=0
x
Solution: Select rectangular coordinate system. Since the
conducting plane is infinite in the z-direction, the potential in the slot
must be independent of z, and this is a two-dimensional problem. The
Laplace’s Equation for the electric potential becomes
 2  2
 2 0
2
x
y
Using the method of separation of variables, and let
 ( x, y )  X ( x)Y ( y )
The boundary conditions for the electric potential in the slot
can be expressed as
 ( x, 0)  0
 ( x, d )  0
 (0, y)   0
 ( , y )  0
In order to satisfy the boundary conditions  ( x, 0)  0 and
 ( x, d )  0 , the solution of Y(y) should be selected as
Y ( y )  A sin k y y  B cos k y y
From the boundary condition  ( x, 0)  0 , we have  = 0 at y = 0 ,
and the constant B = 0. In order to satisfy  ( x, d )  0 , the constant ky
should be
ky 
nπ
, n  1, 2, 3, 
d
y
=0
 = 0
d
O
=0
x
 nπ 
Y ( y )  A sin  y 
d 
Since k x2  k y2  0，we obtain
nπ
kx  j
d
The constant kx is an imaginary number, and the solution of X(x)
We find
should be
X ( x)  Ce
nπ
x
d
 De

nπ
x
d
Since    at x = 0 , the constant C = 0, and
X ( x)  De
Then
 ( x, y)  Ce
Where the constant C = AD .

nπ
x
d

nπ
x
d
 nπ 
sin 
y
d 
Since  =  0 at x = 0 , and we have
 nπ
 d
 0  C sin 

y

The right side of the above equation is variable, since C and n
are not fixed. To satisfy the requirement at x = 0, one needs to take
the linear combination of the equation as the solution, leading to

 ( x, y)   Cn e
n 1

nπ
x
d
 nπ 
sin 
y
d 
In order to satisfy the boundary condition x = 0,  = 0 , and
we have

 nπ 
y , 0  y  d
 d 
 0   Cn sin 
n 1
The right side of the above equation is Fourier series. By using the
orthogonality between the terms of a Fourier series, the coefficients Cn

 sin nx  sin mxdx  0

can be found as
 40
, if n is odd

Cn   nπ
 0 , if n is even
Finally, we find the electric potential in the slot as
4
 ( x, y)  0
π
y


y

n  1, 3, 5, 
=0
Electric field lines
 = 0
0
nπ
1  d x  nπ
n n e sin  d
m  n
m  n
d
=0
Equipotential surfaces
x
4. Method of Separation of Variables in Cylindrical Coordinates
In cylindrical coordinate system, Laplace’s equation has the form
1     1  2  2
 2 0
r
 2
2
r r  r  r 
z
Let
And we have
 (r ,  , z )  R(r ) ( ) Z ( z )
r d  dR  1 d 2 r 2 d 2 Z

0
r

2
2
R dr  dr   d
Z dz
where the second term is a function of the variable  only, while the
first and the third are independent of , leading to
1 d 2
2


k

 d 2
or
d 2
2

k
 0

2
d
where k is the separation constant, and it could be real or imaginary
number. The domain of the variable  is 0    2π , in this case the
change of the field with  must be a periodic function with the period
of 2.
Let k  m (m is integer), then the solution of the above equation
is
 ( )  A sin m  B cos m
where A and B are the constants to be determined.
Considering k  m and the above equation for variable  , the
previous equation can be rewritten as
 1 d  dR  m 2  1 d 2 Z
0
r
 2 

2
 Rr dr  dr  r  Z dz
The first term of the left side is a function of the variable r only,
and the second is a function of the variable z only, they should be
given by a constant. Let
1 d2Z
2

k
z
Z dz 2
or
d2Z
2

k
zZ  0
2
dz
where the constant kz could be real or imaginary number, so that
trigonometric functions, hyperbolic functions, or exponential functions
can be applied. If kz is a real number, we can take
Z ( z)  C sin k z z  D cos k z z
where C and D are constants to be determined.
Substituting the equation for the variable z into the previous
equation gives
2
d
R
dR
r2 2  r
 (k z2 r 2  m 2 ) R  0
dr
dr
If we let k z2 r 2  x 2, then the above equation becomes
d2R
dR
2
2
x

x

(
x

m
)R  0
dx
dx 
2
which is called a Bessel equation, and its solution is a Bessel function,
given as
R(r )  EJ m (k z r )  FN m (k z r )
where E and F are constants to be determined, with J m (k z r ) being the
first kind of Bessel function of order m, and N m (k z r ) being the second
kind of Bessel function of order m.
Since N m (k z r )  at r = 0 , we can only take the first kind of
Bessel’s function as the solution if the region of the field includes the
point r = 0 .
The solution of the above quation should be the linear combination
of products of the solutions R(r) ,() , Z(z).
If the electric fields are independent of z, then we have k z  0 .
The above equation becomes
d2R
dR
2
r

r

m
R0
2
dr
dr
The solutions of the above equation are exponential functions,
2
R(r )  Er m  Fr  m
If the field is independent of z, and also independent of , then m = 0 .
The solution is given by
R(r )  A0 ln r  B0
Considering all of the above situations, the solution can be written as

 (r ,  )  A0 ln r   r m ( Am sin m  Bm cos m )
m 1

  r  m (C m sin m  Dm cos m )
m 1
Example. An infinitely long conducting cylinder of radius a is
placed in a homogeneous electrostatic field E 0 . The direction of E 0 is
perpendicular to the axis of the conducting cylinder, as shown in the
figure. Find the electric fields intensity inside and outside the cylinder.
Solution: Select the cylindrical coordinate
system. Let z-axis be the axis of the cylinder,
and E 0 is aligned with the x-axis, so that.
y
a
x
O
E0
E0  E0e x
When the conducting cylinder is in an
electrostatic equilibrium state, the electric
field intensity inside the cylinder is zero, with
the cylinder being an equipotential body.
The tangential component of the electric field intensity on the
surface of the cylinder is zero.
Since the electric potential outside the cylinder should be
independent of z, the solution becomes the general form and it should
satisfy the following two boundary conditions:
(a) The tangential component of electric field intensity on the
surface of the cylinder is zero, and has no component in the zdirection due to symmetry. This leads to the result
1 
E   e
r 
y
a
x
O
E0
0
r a


0
r a
(b) The electric potential at infinity should be the
same as that required by the original field as given by
 (,  )   E0 x   E 0 r cos 
which states that if r   , the electric potential is independent of
the function sin  , but proportional to r and cos  . Hence, one may
conclude that the coefficients A0  Am  C m  0 , and m = 0.
Hence, the electric potential function now reduces to
 (r , )  B1r cos  
D1
cos 
r
Based on the given boundary conditions, we find the coefficients
B1 and D1 as
B1   E 0
D1  E 0 a 2
And the electric potential outside the cylinder as
E0 a 2
 (r ,  )   E 0 r cos  
cos 
r
The electric field intensity outside the cylinder can be
y
obtained as
a
x
O
E0
E    er

1 

 e
 ez
r
r 
z
 a2 
 a2 
 er 1  2  E0 cos   e 1  2  E0 sin 
 r 
 r 
The electric field lines, the equipotential surfaces outside the
cylinder, and the charges on the surface are shown as follows:
y




Electric field lines

a


x


Surface charges
Equipotential surfaces
E0
5. Method of Separation of Variables in Spherical Coordinates
In spherical coordinate system, Laplace’s equation becomes
1   2  
1
 
 
1
 2
0
r
 2
 sin 
 2
2
2
2
  r sin  
r r  r  r sin   
Let
 (r ,  ,  )  R(r ) ( ) ( )
We have
sin 2  d  2 dR  sin  d 
d  1 d 2
0
r

 sin 

2
R dr  dr   d 
d   d
For the same reason as before, the solution for  should be
 ( )  A sin m  B cos m
And
1 d  2 dR   1
d 
d 
m2 
r

 sin 

0
2
R dr  dr   sin  d 
d  sin  
The first term of the above equation is the function of r only,
while the second is independent of r. Hence, the first term should be a
constant. It is more convenient to write the constant as n(n+1) so that
1 d  2 dR 
r
  n(n  1)
R dr  dr 
2
d
R
dR
r 2 2  2r
 n(n  1) R  0
dr
dr
where n is an integer. This is Euler’s equation, and the general solution
being given by
R (r )  Cr 
n
D
r n 1
And

d 
d 
m2 
 sin 
   n(n  1) sin  
0
d 
d 
sin  

Let cos  x, then we have

d 
m2 
2 d 
(1  x )
  n(n  1) 
0
2


dx 
dx 
1 x 

The above equation is the associated Legendre’s equation, and the
general solution is the sum of the first kind of associated Legender’s
functions Pnm ( x ) and the second kind of associated Legender’s
functions Q mn ( x) , where m < n .
If n is an integer, Pnm ( x ) and Q mn ( x) are polynomials with finite
number of terms, and n was required to be an integer.
From the properties of the second kind of associated Legender’s
functions , we know that Q mn ( x)   as x  1. Thus, if the region
includes   0 or , for which x  1 , we only can take the first kind
of associated Legender’s functions as the solution. In view of this, we
have
 ( )  Pnm ( x)  Pnm (cos  )
Therefore, we take the following linear combination as the general
solution
n

 (r , ,  )   ( Am sin m  Bm cos m )(Cn r n  Dn r ( n 1) )Pnm (cos  )
m 0 n 0
The field is independent of , so that m = 0 . In this case,
Pn0 ( x)  Pn ( x) which is called Legendre’s function of the first kind,
and the general solution is

 (r , )   (Cn r n  Dn r ( n 1) )Pn (cos  )
n 0
Example. Assume a dielectric sphere with radius a and permittivity
 is placed in free space, and with an originally uniform electrostatic
field E0, as shown in the figure. Find the electric field intensity in the
dielectric sphere.
Solution: Select spherical coordinate system,
and let the direction of E0 coincide with the zaxis, i.e. E0  E0e z . Obviously, in this case the
field is rotationally symmetrical to the z-axis,
and independent of .
y
a
z

0
E0
In this way, the solution of the distribution functions of the electric
potential inside and outside the sphere as follows, respectively,


i (r , )   Cn r Pn (cos  )   Dn r ( n 1) Pn (cos  )
n
n 0
n 0


 o (r ,  )   An r Pn (cos  )   Bn r ( n 1) Pn (cos  )
n
n 0
n 0
The distribution functions of the electric potentials inside and
outside the sphere should satisfy the following boundary conditions:
① The electric potentiali (0,  ) at the center of the sphere should be
finite.
② The electric potential at infinity should
o (,  )   E0 r cos    E0 rP1 (cos  )
be
③ The electric potential should be continues at the surface of the
sphere, i.e.
i (a,  )  o (a,  )
y
④ The normal derivatives of electric potentials
a
z

0
E0
at the surface should satisfy
i

r
r a
o
  0
r
r a
Considering the boundary condition ①, the coefficient Dn
should be zero, hence

i (r , )   Cn r n Pn (cos  )
n 0
In order to satisfy the boundary condition ②, all of the coefficients An
except A1 should be zero, and A1   E 0 , so that

 o (r , )   E0 rP1 (cos  )   Bn r ( n 1) Pn (cos  )
n 0
Reconsidering the boundary condition ③ , we have


 ( n 1)
C
a
P
(cos

)


E
aP
(cos

)

B
a
Pn (cos  )
 n n
 n
0
1
n
n 0
n 0
To satisfy the boundary condition ④, we obtain

   r nCn a
n 0
n 1

Pn (cos  )  E0 P1 (cos  )   (n  1) Bn a ( n  2 ) Pn (cos  )
where  r  
0
n 0
Since the above equations hold for any , the corresponding
coefficients on both sides should be equal. Hence, we find
B0  C0  0
C1  
3 E0
r  2
  r 1 

B1  E0 a 
 r  2 
3
Bn  Cn  0, (n  2)
Finally, we find the electric potentials inside and outside the sphere
are, respectively,
3 E0
3 E0
i (r , )  
r cos   
z
r  2
r  2
 r  1 E0 a 3
o (r , )   E0 r cos  
r  2 r2
Since E   , we find the electric field intensity inside the sphere is
Ei  
3 E0
3 E 0 0
i
e z  E0 e z
ez 
ez 
r  2
z
  2 0
The field inside the sphere is still uniform, and the field intensity
inside the sphere is less than that outside the sphere.
If there is a spherical air bubble in
an infinite homogeneous dielectric with
the permittivity , then the electric field
intensity Ei is
y
a
z

0
E0
Ei 
3E0
3 E
e z  r 0 e z  E0 e z
1  2 r
 0  2
The electric field intensity inside the
sphere is greater than that outside the
sphere.