Download 3.1. Solutions of Laplace`s Equation in One

Chapter 3. Special Techniques for Calculating Potentials
Given a stationary charge distribution r  we can, in principle, calculate the electric field:
Er  
ˆr
1
4 0
 r  r 'd '
2
where r  r 'r . This integral involves a vector as an integrand and is, in general, difficult to
calculate. In most cases it is easier to evaluate first the electrostatic potential V which is defined
as
Vr  
1
1
4 0
 r r ' d '
since the integrand of the integral is a scalar. The corresponding electric field E can then be
obtained from the gradient of V since
E  V
The electrostatic potential V can only be evaluated analytically for the simplest charge
configurations. In addition, in many electrostatic problems, conductors are involved and the
charge distribution  is not known in advance (only the total charge on each conductor is
known).
A better approach to determine the electrostatic potential is to start with Poisson's equation
2V  

0
Very often we only want to determine the potential in a region where  = 0. In this region
Poisson's equation reduces to Laplace's equation
 V  0
2
There are an infinite number of functions that satisfy Laplace's equation and the appropriate
solution is selected by specifying the appropriate boundary conditions. This Chapter will
concentrate on the various techniques that can be used to calculate the solutions of Laplace's
equation and on the boundary conditions required to uniquely determine a solution.
- 1 -
3.1. Solutions of Laplace's Equation in One-, Two, and Three Dimensions
3.1.1. Laplace's Equation in One Dimension
In one dimension the electrostatic potential V depends on only one variable x.
electrostatic potential V(x) is a solution of the one-dimensional Laplace equation
The
d 2V
2  0
dx
The general solution of this equation is
Vx   sx  b
where s and b are arbitrary constants. These constants are fixed when the value of the potential
is specified at two different positions.
Example
Consider a one-dimensional world with two point conductors located at x = 0 m and at x = 10
m. The conductor at x = 0 m is grounded (V = 0 V) and the conductor at x = 10 m is kept at a
constant potential of 200 V. Determine V.
The boundary conditions for V are
V0  b  0V
and
V10 10s  b  200V
The first boundary condition shows that b = 0 V. The second boundary condition shows that s =
20 V/m. The electrostatic potential for this system of conductors is thus
Vx   20x
The corresponding electric field can be obtained from the gradient of V
Ex   
dV x 
 20 V / m
dx
- 2 -
The boundary conditions used here, can be used to specify the electrostatic potential between x =
0 m and x = 10 m but not in the region x < 0 m and x > 10 m. If the solution obtained here was
the general solution for all x, then V would approach infinity when x approaches infinity and V
would approach minus infinity when x approaches minus infinity. The boundary conditions
therefore provide the information necessary to uniquely define a solution to Laplace's equation,
but they also define the boundary of the region where this solution is valid (in this example 0 m <
x < 10 m).
The following properties are true for any solution of the one-dimensional Laplace equation:
Property 1:
V(x) is the average of V(x + R) and V(x - R) for any R as long as x + R and x - R are located in
the region between the boundary points. This property is easy to proof:
Vx  r   V x  R  sx  R   b  sx  R   b

 sx  b  V x 
2
2
This property immediately suggests a powerful analytical method to determine the solution of
Laplace's equation. If the boundary values of V are
Vx  a  Va
and
Vx  b  Vb
then property 1 can be used to determine the value of the potential at (a + b)/2:
 a  b  1
V x 
 V  Vb 
2  2 a
Next we can determine the value of the potential at x = (3 a + b)/4 and at x = (a + 3 b)/4 :
1 
 3a  b  1 
 a  b  1 3
V x
 V x  a   V x 
  Va  Vb 



2  2 
2  2 2
2 
 1 1
3 
 a  3b  1   a  b 
V x
 V x 
 V x  b   Va  Vb 

2  2  
2 
2 
 2 2
This process can be repeated and V can be calculated in this manner at any point between x =
a and x = b (but not in the region x > b and x < a).
- 3 -
Property 2:
The solution of Laplace's equation can not have local maxima or minima. Extreme values
must occur at the end points (the boundaries). This is a direct consequence of property 1.
Property 2 has an important consequence: a charged particle can not be held in stable
equilibrium by electrostatic forces alone (Earnshaw's Theorem). A particle is in a stable
equilibrium if it is located at a position where the potential has a minimum value. A small
displacement away from the equilibrium position will increase the electrostatic potential of
the particle, and a restoring force will try to move the particle back to its equilibrium
position. However, since there can be no local maxima or minima in the electrostatic
potential, the particle can not be held in stable equilibrium by just electrostatic forces.
3.1.2. Laplace's Equation in Two Dimensions
In two dimensions the electrostatic potential depends on two variables x and y. Laplace's
equation now becomes
 2V  2V
2 
2  0
x
y
This equation does not have a simple analytical solution as the one-dimensional Laplace equation
does. However, the properties of solutions of the one-dimensional Laplace equation are also
valid for solutions of the two-dimensional Laplace equation:
Property 1:
The value of V at a point (x, y) is equal to the average value of V around this point
Vx,y  
1
 VRd
2R Circle
where the path integral is along a circle of arbitrary radius, centered at (x, y) and with radius
R.
Property 2:
V has no local maxima or minima; all extremes occur at the boundaries.
- 4 -
3.1.3. Laplace's Equation in Three Dimensions
In three dimensions the electrostatic potential depends on three variables x, y, and z.
Laplace's equation now becomes
 2V  2V  2V
2 
2 
2  0
x
y
z
This equation does not have a simple analytical solution as the one-dimensional Laplace equation
does. However, the properties of solutions of the one-dimensional Laplace equation are also
valid for solutions of the three-dimensional Laplace equation:
Property 1:
The value of V at a point (x, y, z) is equal to the average value of V around this point
Vx,y,z  
1
2
VR sin d d
2

4R Sphere
where the surface integral is across the surface of a sphere of arbitrary radius, centered at
(x,y,z) and with radius R.
y
P
R
d
q

z
r
x
Figure 3.1. Proof of property 1.
To proof this property of V consider the electrostatic potential generated by a point charge q
located on the z axis, a distance r away from the center of a sphere of radius R (see Figure
3.1). The potential at P, generated by charge q, is equal to
VP 
1
q
4 0 d
- 5 -
where d is the distance between P and q. Using the cosine rule we can express d in terms of
r, R and 
d  r  R  2rRcos
2
2
2
The potential at P due to charge q is therefore equal to
VP 
1
q
4 0
r  R  2rR cos
2
2
The average potential on the surface of the sphere can be obtained by integrating VP across
the surface of the sphere. The average potential is equal to
Vaverage


1
1 1
VP R 2 sin d d 
2

4R Sphere
4 4 0 
q
8 0
r 2  R 2  2rR cos
rR


0
q
r  R  2rR cos
2
2
2 sin d 
q r  R r  R 
1 q


8 0  rR
rR  4 0 r
which is equal to the potential due to q at the center of the sphere. Applying the principle of
superposition it is easy to show that the average potential generated by a collection of point
charges is equal to the net potential they produce at the center of the sphere.
Property 2:
The electrostatic potential V has no local maxima or minima; all extremes occur at the
boundaries.
Example: Problem 3.3
Find the general solution to Laplace's equation in spherical coordinates, for the case where V
depends only on r. Then do the same for cylindrical coordinates.
Laplace's equation in spherical coordinates is given by
1   2 V 
1
 
V 
1
 2V
r

sin


2
2
2
2
2 0
r r  r  r sin  
  r sin  
If V is only a function of r then
V
0

- 6 -
and
V
0

Therefore, Laplace's equation can be rewritten as
1   2 V 
r
0
r 2 r  r 
The solution V of this second-order differential equation must satisfy the following first-order
differential equation:
r
2
V
 a  constant
r
This differential equation can be rewritten as
V a
 2
r r
The general solution of this first-order differential equation is
Vr   
a
b
r
where b is a constant. If V = 0 at infinity then b must be equal to zero, and consequently
Vr   
a
r
Laplace's equation in cylindrical coordinates is
1   V  1  2V  2V
r
 2
2 
2 0
r r  r  r 
z
If V is only a function of r then
V
0

and
- 7 -
V
0
z
Therefore, Laplace's equation can be rewritten as
1   V 
r
0
r r  r 
The solution V of this second-order differential equation must satisfy the following first-order
differential equation:
r
V
 a  constant
r
This differential equation can be rewritten as
V a

r r
The general solution of this first-order differential equation is
Vr   a lnr   b
where b is a constant. The constants a and b are determined by the boundary conditions.
3.1.4. Uniqueness Theorems
Consider a volume (see Figure 3.2) within which the charge density is equal to zero.
Suppose that the value of the electrostatic potential is specified at every point on the surface of
this volume. The first uniqueness theorem states that in this case the solution of Laplace's
equation is uniquely defined.
- 8 -
Boundary
Volume
Figure 3.2. First Uniqueness Theorem
To proof the first uniqueness theorem we will consider what happens when there are two
solutions V1 and V2 of Laplace's equation in the volume shown in Figure 3.2. Since V1 and V2
are solutions of Laplace's equation we know that
 V1  0
2
and
 V2  0
2
Since both V1 and V2 are solutions, they must have the same value on the boundary. Thus V1 =
V2 on the boundary of the volume. Now consider a third function V3, which is the difference
between V1 and V2
V3 V1 V2
The function V3 is also a solution of Laplace's equation. This can be demonstrated easily:
 V3  V1   V2  0
2
2
2
The value of the function V3 is equal to zero on the boundary of the volume since V1 = V2 there.
However, property 2 of any solution of Laplace's equation states that it can have no local maxima
or minima and that the extreme values of the solution must occur at the boundaries. Since V3 is
a solution of Laplace's equation and its value is zero everywhere on the boundary of the volume,
the maximum and minimum value of V3 must be equal to zero. Therefore, V3 must be equal to
zero everywhere. This immediately implies that
- 9 -
V1 V2
everywhere. This proves that there can be no two different functions V1 and V2 that are
solutions of Laplace's equation and satisfy the same boundary conditions. Therefore, the
solution of Laplace's equation is uniquely determined if its value is a specified function on all
boundaries of the region. This also indicates that it does not matter how you come by your
solution: if (a) it is a solution of Laplace's equation, and (b) it has the correct value on the
boundaries, then it is the right and only solution.
q4
Boundary
q3
q2
q1
Figure 3.3. System with conductors.
The first uniqueness theorem can only be applied in those regions that are free of charge and
surrounded by a boundary with a known potential (not necessarily constant). In the laboratory
the boundaries are usually conductors connected to batteries to keep them at a fixed potential. In
many other electrostatic problems we do not know the potential at the boundaries of the system.
Instead we might know the total charge on the various conductors that make up the system (note:
knowing the total charge on a conductor does not imply a knowledge of the charge distribution 
since it is influenced by the presence of the other conductors). In addition to the conductors that
make up the system, there might be a charge distribution  filling the regions between the
conductors (see Figure 3.3). For this type of system the first uniqueness theorem does not apply.
The second uniqueness theorem states that the electric field is uniquely determined if the total
charge on each conductor is given and the charge distribution in the regions between the
conductors is known.
The proof of the second uniqueness theorem is similar to the proof of the first uniqueness
theorem. Suppose that there are two fields E1 and E2 that are solutions of Poisson's equation in
the region between the conductors. Thus
- 10 -
 E1 

0
 E2 

0
and
where  is the charge density at the point where the electric field is evaluated. The surface
integrals of E1 and E2 , evaluated using a surface that is just outside one of the conductors with
charge Qi, are equal to Qi / 0 . Thus

E1  da 
Qi
E2  da 
Qi
Surface
conductor i

Surface
conductor i
0
0
The difference between E1 and E2 , E3  E1  E2, satisfies the following equations:
 E3   E1   E2 


E3  da 
Surface
conductor i

E1  da 
Surface
conductor i
 
 0
0  0
E2  da 
Surface
conductor i
Qi
0

Qi
0
0
Consider the surface integral of E3 , integrated over all surfaces (the surface of all conductors and
the outer surface). Since the potential on the surface of any conductor is constant, the
electrostatic potential associated with E1 and E2 must also be constant on the surface of each
conductor. Therefore, V3 V1 V2 will also be constant on the surface of each conductor. The
surface integral of V3 E3 over the surface of conductor i can be written as

Surface
conductor i

V3 E3  da  V3
E3  da  0
Surface
conductor i
Since the surface integral of V3 E3 over the surface of conductor i is equal to zero, the surface
integral of V3 E3 over all conductor surfaces will also be equal to zero. The surface integral of
V3 E3 over the outer surface will also be equal to zero since V3  0 on this surface. Thus
- 11 -

V3 E3  da  0
All surfaces
The surface integral of V3 E3 can be rewritten using Green's identity as

0 
All surfaces
 

V3 E3  da  

V3 V3  da  
All surfaces
3
Volume
between
conductors
3
2
3
3
3
3
Volume
between
conductors
V  E  E  E d   
3
V  V  V  V d 

3
E3 d  0
2
Volume
between
conductors
where the volume integration is over all space between the conductors and the outer surface.
Since E3 2 is always positive, the volume integral of E3 2 can only be equal to zero if E3 2  0
everywhere. This implies immediately that E1  E2 everywhere, and proves the second
uniqueness theorem.
3.2. Method of Images
Consider a point charge q held as a distance d above an infinite grounded conducting plane
(see Figure 3.4). The electrostatic potential of this system must satisfy the following two
boundary conditions:
Vx,y,0  0
x  

Vx,y,z  0 when y  
z  

A direct calculation of the electrostatic potential can not be carried out since the charge
distribution on the grounded conductor is unknown. Note: the charge distribution on the surface
of a grounded conductor does not need to be zero.
- 12 -
z axis
q
d
Figure 3.4. Method of images.
Consider a second system, consisting of two point charges with charges +q and -q, located at
z = d and z = -d, respectively (see Figure 3.5). The electrostatic potential generated by these two
charges can be calculated directly at any point in space. At a point P = (x, y, 0) on the xy plane
the electrostatic potential is equal to
Vx,y,0 
1 
q


2
4 0  x  y 2  d 2


 0
x 2  y 2  d 2 

q
Figure 3.5. Charge and image charge.
The potential of this system at infinity will approach zero since the potential generated by each
charge will decrease as 1/r with increasing distance r. Therefore, the electrostatic potential
generated by the two charges shown in Figure 3.5 satisfies the same boundary conditions as the
- 13 -
system shown in Figure 3.4. Since the charge distribution in the region z > 0 (bounded by the xy
plane boundary and the boundary at infinity) for the two systems is identical, the corollary of the
first uniqueness theorem states that the electrostatic potential in this region is uniquely defined.
Therefore, if we find any function that satisfies the boundary conditions and Poisson's equation,
it will be the right answer. Consider a point (x, y, z) with z > 0. The electrostatic potential at
this point can be calculated easily for the charge distribution shown in Figure 3.5. It is equal to
1 
q

Vx,y,z  

4 0  x 2  y 2  z  d 2



2 
2
2


x  y  z  d 
q
Since this solution satisfies the boundary conditions, it must be the correct solution in the region
z > 0 for the system shown in Figure 3.4. This technique of using image charges to obtain the
electrostatic potential in some region of space is called the method of images.
The electrostatic potential can be used to calculate the charge distribution on the grounded
conductor. Since the electric field inside the conductor is equal to zero, the boundary condition
for E (see Chapter 2) shows that the electric field right outside the conductor is equal to
EOutside 


nˆ  kˆ
0
0
where  is the surface charge density and nˆ is the unit vector normal to the surface of the
conductor. Expressing the electric field in terms of the electrostatic potential V we can rewrite
this equation as
  0 Ez   0
V
z
z0
Substituting the solution for V in this equation we find

q 
z  d 
 
4  x 2  y 2  z  d 2




d
   q

3/ 2
3/ 2
2

2 x 2  y 2  d 2 
x 2  y 2  z  d 

z0
 
3/ 2
z  d

Only in the last step of this calculation have we substituted z = 0. The induced charge
distribution is negative and the charge density is greatest at (x = 0, y = 0, z = 0). The total charge
on the conductor can be calculated by surface integrating of :
Qtotal 

Surface
2
 da  0
- 14 -


0
 r rdrd
where r 2  x 2  y 2. Substituting the expression for  in the integral we obtain
Qtotal  qd 
1

0
r
2
d

2 3/ 2
qd
rdr 
r 2  d2

0
 1 
 qd 0   q
 d 
As a result of the induced surface charge on the conductor, the point charge q will be
attracted towards the conductor. Since the electrostatic potential generated by the charge imagecharge system is the same as the charge-conductor system in the region where z > 0, the
associated electric field (and consequently the force on point charge q) will also be the same.
The force exerted on point charge q can be obtained immediately by calculating the force exerted
on the point charge by the image charge. This force is equal to
F 
q2
ˆ
2 k
4 0 2d 
1
There is however one important difference between the image-charge system and the real
system. This difference is the total electrostatic energy of the system. The electric field in the
image-charge system is present everywhere, and the magnitude of the electric field at (x, y, z) will
be the same as the magnitude of the electric field at (x, y, -z). On the other hand, in the real
system the electric field will only be non zero in the region with z > 0. Since the electrostatic
energy of a system is proportional to the volume integral of E 2 the electrostatic energy of the
real system will be 1/2 of the electrostatic energy of the image-charge system (only 1/2 of the
total volume has a non-zero electric field in the real system). The electrostatic energy of the
image-charge system is equal to
Wimage 
q2
4 0 2d
1
The electrostatic energy of the real system is therefore equal to
1
1 q2
Wreal  W image  
2
4 0 4d
The electrostatic energy of the real system can also be obtained by calculating the work required
to be done to assemble the system. In order to move the charge q to its final position we will
have to exert a force opposite to the force exerted on it by the grounded conductor. The work
done to move the charge from infinity along the z axis to z = d is equal to
- 15 -
Wreal 
1
d
4 0 
d
q2
1 q2
1 q2
dz



4z 2
4 0 4z 
4 0 4d
which is identical to the result obtained using the electrostatic potential energy of the imagecharge system.
Example: Example 3.2 + Problem 3.7
A point charge q is situated a distance s from the center of a grounded conducting sphere of
radius R (see Figure 3.6).
a) Find the potential everywhere.
b) Find the induced surface charge on the sphere, as function of q. Integrate this to get the total
induced charge.
c) Calculate the electrostatic energy of this configuration.
Figure 3.6. Example 3.2 + Problem 3.7.
a) Consider a system consisting of two charges q and q', located on the z axis at z = s and z = z',
respectively. If the potential produced by this system is identical everywhere to the potential
produced by the system shown in Figure 3.6 then the position of point charge q' must be chosen
such that the potential on the surface of a sphere of radius R, centered at the origin, is equal to
zero (in this case the boundary conditions for the potential generated by both systems are
identical).
We will start with determining the correct position of point charge q'. The electrostatic
potential at P (see Figure 3.7) is equal to
VP  0 
1
q
1
q'

4 0 s  R 4 0 R  z'
This equation can be rewritten as
- 16 -
q' 
q
R  z '
sR
Figure 3.7. Image-charge system.
The electrostatic potential at Q is equal to
VQ  0 
1
q
1
q'

4 0 s  R 4 0 R  z'
This equation can be rewritten as
q' 
q
R  z '
s R
Combining the two expression for q' we obtain
q
q
R  z ' 
R  z '
sR
s R
or
s  RR  z'  s  RR  z'
This equation can be rewritten as
z's  R  z's  R  2sz' Rs  R  Rs  R  2R
The position of the image charge is equal to
R2
z'
s
The value of the image charge is equal to
- 17 -
2
q
q 
R 2 
R


q' 
R  z'  
R    q
s R
s  R 
s 
s
Now consider an arbitrary point P' on the circle. The distance between P' and charge q is d and
the distance between P' and charge q' is equal to d'. Using the cosine rule (see Figure 3.7) we
can express d and d' in terms of R, s, and :
d  R  s  2Rscos
2
2
R4
R2
d' R  z ' 2Rz 'cos  R  2  2R cos
s
s
2
2
2
The electrostatic potential at P' is equal to
VP'


1 
q

40  R 2  s2  2Rs cos





4
2

R
R
R 2  2  2R cos 
s
s


1 q q' 
 
4 0 
d d' 


1 
q
q

 2
 0
4 0  R  s2  2Rs cos
R 2  s2  2Rs cos 
q
R
s
Thus we conclude that the configuration of charge and image charge produces an electrostatic
potential that is zero at any point on a sphere with radius R and centered at the origin.
Therefore, this charge configuration produces an electrostatic potential that satisfies exactly the
same boundary conditions as the potential produced by the charge-sphere system. In the region
outside the sphere, the electrostatic potential is therefore equal to the electrostatic potential
produced by the charge and image charge. Consider an arbitrary point r , , . The distance
between this point and charge q is d and the distance between this point and charge q' is equal to
d'. These distances can be expressed in terms of r, s, and  using the cosine rule:
d  r  s  2rscos
2
2
R4
R2
d' r  z ' 2rz 'cos  r  2  2r
cos
s
s
2
2
2
The electrostatic potential at r , ,  will therefore be equal to
- 18 -
Vr , ,  

1 q q'  1
q
 
{ 2


4 0 d d'  4 0 r  s2  2rs cos
1
4 0
{
q
r  s  2rs cos
2
2

q
R
s
R4
R2
r  2  2r
cos
s
s
}
2
q
rs
2
 R  2rs cos
R 
2
}
b) The surface charge density  on the sphere can be obtained from the boundary conditions of
E
Eoutside  Einside  Eoutside 


nˆ  rˆ
0
0
where we have used the fact that the electric field inside the sphere is zero. This equation can be
rewritten as
  0 Er   0
V
r
Substituting the general expression for V into this equation we obtain

 
 
rs2
 2  s cos
R
q
r  s cos
{ 2

3/ 2 }
4 r  s2  2rs cos 3/ 2 rs2

2
R   R  2rs cos 


q
R  s cos
{ 2
4 R  s2  2Rs cos 3/ 2

r R
s2
 s cos
R
 2
}
s  R 2  2Rs cos 3/ 2

q
s2  R 2
 
4R R 2  s2  2Rs cos 3/ 2
The total charge on the sphere can be obtained by integrating  over the surface of the sphere.
The result is
- 19 -
Q

 R
2

q 2
sin d
2
sin d d   Rs  R 
3/ 2 
0
2
R 2  s2  2Rs cos 
1



q 2
sR
  Rs  R 2  2
2
2
 R  s  2Rs cos



2
2

q s  R  1
1 
R


 q



2 s  R  s
R  s 
s

0
c) To obtain the electrostatic energy of the system we can determine the work it takes to
assemble the system by calculating the path integral of the force that we need to exert in charge q
in order to move it from infinity to its final position (z = s). Charge q will feel an attractive force
exerted by the induced charge on the sphere. The strength of this force is equal to the force on
charge q exerted by the image charge q'. This force is equal to
Fqq' 
1
qq'
ˆ
2 k 
4 0 s  z '
 R 
q  q
 s 
1
sRq2
ˆ
ˆ
2 k 
2 k
4 0  R 2 
4 0 s2  R 2 
s  

s 
1
The force that we must exert on q to move it from infinity to its current position is opposite to
Fqq' . The total work required to move the charge is therefore equal to
s
Rq2
1
Rq2
W   Fqq'  dl 

2 dz 
2
2

4 0  z 2  R 2 
4 0 2z 2  R 2 
8 0 s  R
s
1
s
zRq2
1

Example: Problem 3.10
Two semi-infinite grounded conducting planes meet at right angles. In the region between
them, there is a point charge q, situated as shown in Figure 3.8. Set up the image configuration,
and calculate the potential in this region. What charges do you need, and where should they be
located? What is the force on q? How much work did it take to bring q in from infinity?
Consider the system of four charges shown in Figure 3.9. The electrostatic potential
generated by this charge distribution is zero at every point on the yz plane and at every point on
the xz plane. Therefore, the electrostatic potential generated by this image charge distribution
satisfies the same boundary conditions as the electrostatic potential of the original system. The
potential generated by the image charge distribution in the region where x > 0 and y > 0 will be
identical to the potential of the original system. The potential at a point P = (x, y, z) is equal to
- 20 -
VP

1
q
4 0 x  a 2  y  b  z 2
 
2

1

q
4 0 x  a 2  y  b 2  z 2
q
1
4 0 x  a 2  y  b 2  z 2
 

1

q
4 0 x  a 2  y  b 2  z 2
Figure 3.8. Problem 3.10.
Figure 3.9. Image charges for problem 3.10.
The force exerted on q can be obtained by calculating the force exerted on q by the image
charges. The total force is equal to the vector sum of the forces exerted by each of the three
image charges. The force exerted by the image charge located at (-a, b, 0) is directed along the
negative x axis and is equal to
F1  
q2
ˆ
2 i
4 0 4a
1
- 21 -
The force exerted by the image charge located at (a, -b, 0) is directed along the negative y axis
and is equal to
F2  
q2
ˆ
2 j
4 0 4b
1
The force exerted by the image charge located at (-a, -b, 0) is directed along the vector
connecting (-a, -b, 0) and (a, b, 0) and is equal to
q2
aiˆ  bˆj
1
q2
ˆ ˆ
F3 

2
2
3/ 2 ai  bj
4 0 4a  4b a 2  b 2 16 0 a 2  b 2 

1

The total force on charge q is the vector sum of F1 , F2 and F3 :
Ftot  F1  F2  F3  
  1
 
q2  1
a
b
ˆ
ˆj 

i



 2
 2
2
2 3/ 2 
2
2 3/ 2  
16 0 
a
b
a  b    a  b   

The electrostatic potential energy of the system can, in principle, be obtained by calculating the
path integral of Ftot between infinity and (a, b, 0). However, this is not trivial since the force
Ftot is a rather complex function of a and b. An easier technique is to calculate the
electrostatic potential energy of the system with charge and image charges. The potential energy
of this system is equal to
Wimage 
1 q2 q2



4 0  2a
2b
 q2  1
1 1 
  
 
 2
2
2 
2
80  a  b
a b 
4a  4b
q2
However, in the real system the electric field is only non-zero in the region where x > 0 and y > 0.
Therefore, the total electrostatic potential energy of the real system is only 1/4 of the total
electrostatic potential energy of the image charge system. Thus
1
q2  1
1 1 
Wreal  W image 

 
 2
4
32 0  a  b 2 a b 
- 22 -
3.3. Separation of Variables
3.3.1. Separation of variables: Cartesian coordinates
A powerful technique very frequently used to solve partial differential equations is
separation of variables. In this section we will demonstrate the power of this technique by
discussing several examples.
Example: Example 3.3 (Griffiths)
Two infinite, grounded, metal plates lie parallel to the xz plane, one at y = 0, the other at y = π
(see Figure 3.10). The left end, at x = 0, is closed off with an infinite strip insulated from the
two plates and maintained at a specified potential V0 y . Find the potential inside this "slot".
y
V=0

V(y)
V=0
x
z
Figure 3.10. Example 3.3 (Griffiths).
The electrostatic potential in the slot must satisfy the three-dimensional Laplace equation.
However, since V does not have a z dependence, the three-dimensional Laplace equation reduces
to the two-dimensional Laplace equation:
 2V  2V
2 
2  0
x
y
The boundary conditions for the solution of Laplace's equation are:
1. V(x, y = 0) = 0 (grounded bottom plate).
2. V(x, y = π) = 0 (grounded top plate).
- 23 -
3. V(x = 0, y) = V0(y) (plate at x = 0).
4. V  0 when x  ∞.
These four boundary conditions specify the value of the potential on all boundaries surrounding
the slot and are therefore sufficient to uniquely determine the solution of Laplace's equation
inside the slot. Therefore, if we find one solution of Laplace's equation satisfying these
boundary conditions than it must be the correct one. Consider solutions of the following form:
Vx,y   X x Y y 
If this is a solution of the two-dimensional Laplace equation than we must require that
 Y y 
2
2
 2 X x 




X
x
Y

y


X
x
Y

y


Y

y

 X x 
0
2
2
2
x
y
x
y 2




2
This equation can be rewritten as
1  2 X x 
1  Y y 

0
2
2
Xx  x
Y y  y
2
The first term of the left-hand side of this equation depends only on x while the second term
depends only on y. Therefore, if this equation must hold for all x and y in the slot we must
require that
1  2 X x 
 C1  constant
2
Xx  x
and
1  Y y 
 C1
2
Yy  y
2
The differential equation for X can be rewritten as
 2 Xx 
 C1X x   0
2
x
If C1 is a negative number than this equation can be rewritten as
- 24 -
 2 Xx 
 k 2 X x   0
2
x
where k2 = -C1 . The most general solution of this equation is
Xx   A coskx  B sinkx
However, this function is an oscillatory function and does not satisfy boundary condition # 4,
which requires that V approaches zero when x approaches infinity. We therefore conclude that
C1 can not be a negative number. If C1 is a positive number then the differential equation for X
can be written as
 2 Xx  2
 k X x   0
2
x
The most general solution of this equation is
Xx   Ae  Be
kx
kx
This solution will approach zero when x approaches infinity if A = 0. Thus
Xx   Be
kx
The solution for Y can be obtained by solving the following differential equation:
 Yy 
2
y
2
 k 2Y y   0
The most general solution of this equation is
Yy  C cosky D sinky
Therefore, the general solution for the electrostatic potential V(x,y) is equal to
Vx,y   e
kx
C cosky D sinky
where we have absorbed the constant B into the constants C and D. The constants C and D must
be chosen such that the remaining three boundary conditions (1, 2, and 3) are satisfied. The first
boundary condition requires that V(x, y = 0) = 0:
Vx,y  0  e
kx
C cos0  D sin0 Ce
- 25 -
kx
0
which requires that C = 0. The second boundary condition requires that V(x, y = π) = 0:
Vx,y     De
kx
sink   0
which requires that sin k = 0 . This condition limits the possible values of k to positive
integers:
k = 1, 2, 3, 4, .....
Note: negative values of k are not allowed since exp(-kx) approaches zero at infinity only if k >
0. To satisfy boundary condition # 3 we must require that
Vx  0,y   D sinky V0y 
This last expression suggests that the only time at which we can find a solution of Laplace's
equation that satisfies all four boundary conditions has the form exp – kx sin ky is when
V0 y  happens to have the form sin ky . However, since k can take on an infinite number of
values, there will be an infinite number of solutions of Laplace's equation satisfying boundary
conditions # 1, # 2 and # 4. The most general form of the solution of Laplace's equation will be
a linear superposition of all possible solutions. Thus

Vx,y    Dk e kx sinky
k 1
Boundary condition # 3 can now be written as

Vx  0,y    Dk sinky  V0 y 
k 1
Multiplying both sides by sin(ny) and integrating each side between y = 0 and y = π we obtain

 Dk 0 sinnysinkydy  0 sinnyV0ydy


k 1
The integral on the left-hand side of this equation is equal to zero for all values of k except k = n.
Thus

D 
k 1

k 0

sinny sinkydy   Dk
k 1
- 26 -

2
 kn 

2
Dn
The coefficients Dk can thus be calculated easily:
Dk 

2
 
0
sinkyV0 y dy
The coefficients Dk are called the Fourier coefficients of V0 y . The solution of Laplace's
equation in the slot is therefore equal to

Vx,y    Dk e kx sinky
k 1
where
Dk 
2



0
sinkyV0 y dy
Now consider the special case in which V0y  constant  V0 . In this case the coefficients
Dk are equal to
Dk 
2



0
sinkyV0 y dy  
cosky

2

V0
k
0

0
2 V0



1  cosk   
 k
4 V0

 k
if k is even
if k is odd
The solution of Laplace's equation is thus equal to
Vx,y  

1 kx
e sinky
 k 1, 3, 5, .. k
4V0

Example: Problem 3.12
Find the potential in the infinite slot of Example 3.3 (Griffiths) if the boundary at x = 0
consists to two metal stripes: one, from y = 0 to y = π/2, is held at constant potential V0 , and the
other, from y = π/2 to y = π is at potential  V0 .
The boundary condition at x = 0 is
V0

V0,y   
V
 0
for 0  y   / 2
for  / 2  y  
- 27 -
The Fourier coefficients of the function V0 y  are equal to
Dk
2

sinkyV0 y dy 
2
sinkydy 
2
 

2 V0 
1  2 V0
1  cosk   2cos k  
C

2   k k
 k 
0

V0 
 /2

0

V0 

 /2
sinkydy 
The values for the first four C coefficients are
C1  0
C3  0
C2  4
C4  0
It is easy to see that Ck + 4 = Ck and therefore we conclude that
4 for k  2,6,10,...

Ck  
0 otherwise

The Fourier coefficients Ck are thus equal to
8V0
k

Dk  

0


for k  2,6,10,...
otherwise
The electrostatic potential is thus equal to
Vx,y  

1 kx
e sinky
 k 2, 6,10, ... k
8V0

Example: Problem 3.13
For the infinite slot (Example 3.3 Griffiths) determine the charge density  y  on the strip at
x=0, assuming it is a conductor at constant potential V0 .
The electrostatic potential in the slot is equal to
Vx,y  

1 kx
e sinky
 k 1, 3, 5, .. k
4V0

- 28 -
The charge density at the plate at x = 0 can be obtained using the boundary condition for the
electric field at a boundary:
Ex  0  Ex 0  Ex  0 

nˆ
0
where nˆ is directed along the positive x axis. Since E  V this boundary condition can be
rewritten as
V


x x  0
0
Differentiating V(x,y) with respect to x we obtain
V
4V0 
kx

e sinky
x
 k 1
, 3, 5, ...
At the x = 0 boundary we obtain
V
4V0 

sinky
x x  0
 k 1
, 3, 5, ...
The charge density  on the x = 0 strip is therefore equal to
   0
V
x

4V0 0
x  0



k 1, 3, 5, ...
sinky
Example: Double infinite slots
The slot of example 3.3 in Griffiths and its mirror image at negative x are separated by an
insulating strip at x = 0. If the charge density (y) on the dividing strip is given, determine the
potential in the slot.
The boundary condition at x = 0 requires that
Ex  0  Ex 0  2Ex  0 

nˆ
0
where nˆ is directed along the positive x axis. Here we have used the symmetry of the
configuration which requires that the electric field in the region x < 0 is the mirror image of the
field in the region x > 0. Since E  V this boundary condition can be rewritten as
- 29 -
V
1  y 

x x  0
2 0
We will first determine the potential in the x > 0 region. Following the same procedure as in
Example 3 we obtain for the electrostatic potential

Vx,y    Dk e kx sinky
k 1
where the constants Dk must be chosen such that the boundary condition at x = 0 is satisfied.
This requires that


V
1  y 
kx
  kDk e sinky
  kDk sinky 
x x  0
2 0
k 1
k 1
x  0
Thus
1  y 

 kD sinky  2
k
k 1
0
The constants Dk can be determined by multiplying both sides of this equation with sinmy  and
integrating both sides with respect to y between y = 0 and y = π. The result is


1 



y

sin

my

dy

kD


k 0 sinkysinmydy  mDm
0
2 0
2
k 1
The constants Ck are thus equal to
Dk 
 y sinkydy
 k 

1
0
0
The electrostatic potential is thus equal to
Vx,y  
1
 0

 k   ysinkydye

k 1
1

0
kx

sinky

3.3.2. Separation of variables: spherical coordinates
Consider a spherical symmetric system. If we want to solve Laplace's equation it is natural
to use spherical coordinates. Assuming that the system has azimuthal symmetry ( V /   0)
Laplace's equation reads
- 30 -
1   2 V 
1
 
V 
r

sin

0
2
2
r r  r  r sin  
 
Multiplying both sides by r2 we obtain
  2 V 
1  
V 
r

sin

0
r  r  sin  
 
Consider the possibility that the general solution of this equation is the product of a function
Rr , which depends only on the distance r, and a function  , which depends only on the
angle :
Vr,   Rr   
Substituting this "solution" into Laplace's equation we obtain
 
  2 Rr  Rr  
r
 
r 
r  sin 
Dividing each term of this equation by R r  

  
sin
  0

 
we obtain
1   2 Rr 
1
1  
  
r
 
sin
  0
Rr  r 
r     sin  
 
The first term in this expression depends only on the distance r while the second term depends
only on the angle . This equation can only be true for all r and  if
1   2 Rr 
r
  mm  1  constant
Rr  r 
r 
and
1  
  
sin
  mm  1
  sin  
 
1
Consider a solution for R of the following form:
Rr   Ar
k
where A and k are arbitrary constants. Substituting this expression in the differential equation
for R(r) we obtain
- 31 -
1 
k
2
k 1
k

kr Ar   k k 1r  kk 1  mm 1
k
Ar r
r
Therefore, the constant k must satisfy the following relation:
kk 1  k  k  mm 1
2
This equation gives us the following expression for k
1 

 m
1

2
m


1  1  4mm  1
2 
k

  or
2
2
m  1

The general solution for Rr  is thus given by
Rr   Ar 
B
m
r
m 1
where A and B are arbitrary constants.
The angle dependent part of the solution of Laplace's equation must satisfy the following
equation
 
  
sin
  mm  1  sin  0
 
 
The solutions of this equation are known as the Legendre polynomial Pm cos  .
Legendre polynomials have the following properties:
The
1. if m is even: Pm x   Pm x 
2. if m is odd: Pm x   Pm x 
3. Pm 1 1 for all m
4.
P x Pm x dx 
1 n

1
2
 or
2m  1 nm
P cos Pm cos sind 
1 n

1
2

2m 1 nm
Combining the solutions for Rr  and   we obtain the most general solution of Laplace's
equation in a spherical symmetric system with azimuthal symmetry:

B 

Vr,    Am r m  mm1 Pm cos 
r
m 0
- 32 -
Example: Problem 3.18
The potential at the surface of a sphere is given by
V0   k cos3 
where k is some constant. Find the potential inside and outside the sphere, as well as the surface
charge density    on the sphere. (Assume that there is no charge inside or outside of the
sphere.)
The most general solution of Laplace's equation in spherical coordinates is

B 

Vr,    Am r m  mm1 Pm cos 
r
m 0
First consider the region inside the sphere (r < R). In this region Bm = 0 since otherwise
Vr,  would blow up at r = 0. Thus

Vr ,    Am r m Pm cos 
m 0
The potential at r = R is therefore equal to

VR,    Am R m Pm cos   k cos3 
m 0
Using trigonometric relations we can rewrite cos3  as
8
3
3
cos3   4 cos   3cos  P3 cos   P1cos 
5
5
Substituting this expression in the equation for VR,  we obtain

VR,    Am R m Pm cos  
m 0
8k
3k
P3 cos   P1cos 
5
5
This equation immediately shows that Am = 0 unless m = 1 or m = 3 . If m = 1 or m = 3 then
A1  
3k
5R
- 33 -
A3 
8 k
3
5R
The electrostatic potential inside the sphere is therefore equal to
Vr,  
8k r 3
3k r
P cos 
3 P3 cos  
5 R
5 R 1
Now consider the region outsider the sphere (r > R). In this region Am = 0 since otherwise
Vr,  would blow up at infinity. The solution of Laplace's equation in this region is therefore
equal to

Bm
m 1 Pm  cos 
m 0 r
Vr ,   
The potential at r = R is therefore equal to

Bm
8k
3k
P3 cos   P1cos 
m 1 Pm cos  
5
5
m 0 R
VR,   
The equation immediately shows that Bm = 0 except when m = 1 or m = 3 . If m = 1 or m = 3
then
3 2
B1   kR
5
B3 
8 4
kR
5
The electrostatic potential outside the sphere is thus equal to
Vr,   
3k R 2
8k R 4


P
cos


2
4 P cos 
5 r 1
5 r 3
The charge density on the sphere can be obtained using the boundary conditions for the electric
field at a boundary:
Er R   Er R 
  
rˆ
0
Since E  V this boundary condition can be rewritten as
- 34 -
V
V
  


r r R  r r R
0
The first term on the left-hand side of this equation can be calculated using the electrostatic
potential just obtained:
6k R 2

V
32k R 4


 
P
cos


3 1
5 P3 cos 
r r R   5 r
5 r


r R 
k
6Pcos   32P3cos 
5R 1
In the same manner we obtain
 3k 1

V
24k r 2


 
P1 cos 
3 P3 cos 
r r R  5 R
5 R


r R
k
3P1cos   24P3cos 
5R
Therefore,
V
V
k
  



9P1cos   56P3 cos  
r r R  r r R 5R
0
The charge density on the sphere is thus equal to
   
k 0
9P1cos   56P3cos 
5R
Example: Problem 3.19
Suppose the potential V0   at the surface of a sphere is specified, and there is no charge
inside or outside the sphere. Show that the charge density on the sphere is given by
   
0

2m  1 C
2R 
2
m 0
P cos 
m m
where

Cm   V0  Pm cos sind
0
Most of the solution of this problem is very similar to the solution of Problem 3.18. First
consider the electrostatic potential inside the sphere. The electrostatic potential in this region is
given by
- 35 -

Vr ,    Am r m Pm cos 
m 0
and the boundary condition is

VR,    Am R m Pm cos   V0  
m 0
The coefficients Am can be determined by multiplying both sides of this equation by
Pncos sin and integrating with respect to  between  = 0 and  = π:


0


V0  Pn cos sind   Am R m 0 Pmcos Pn cos sind 
m 0
Thus
Am 
2m 1 1
m
2 R


0
V0  Pm cos sind
In the region outside the sphere the electrostatic potential is given by

Bm
m 1 Pm  cos 
m 0 r
Vr ,   
and the boundary condition is

Bm
m 1 Pm cos   V0  
m 0 R
VR,   
The coefficients Bm are given by
Bm 
2m 1 m 1 
R 0 V0  Pm cos sind
2
The charge density    on the surface of the sphere is equal to

V
V


r r R  r
     0 



r R 

Differentiating Vr,  with respect to r in the region r > R we obtain

V
Bm
  m  1 m  2 Pm cos 
r r R 
R
m 0
- 36 -
2
An R n
2n  1
Differentiating Vr,  with respect to r in the region r < R we obtain

V
m 1
  mAm R Pm cos 
r r R m 0
The charge density is therefore equal to



 m 0 
     0  m  1

Bm


m 1


P
cos


mAm R Pm cos 


m 2 m

R
 m 0

  
B


  0   m  1 mm 2  mAm R m 1 Pm cos 

R

m 0 
Substituting the expressions for Am and Bm into this equation we obtain


2m  1R m 1
2m  1 m 1
  



     0  m  1

m
R
C
P
cos



m 2
m
m m
2R
2R





m 0 


 0  

2
 2m  1 Cm Pm cos  
2R m 0



where

Cm   V0  Pm cos sind
0
Example: Problem 3.23
Solve Laplace's equation by separation of variables in cylindrical coordinates, assuming there
is no dependence on z (cylindrical symmetry). Make sure that you find all solutions to the radial
equation. Does your result accommodate the case of an infinite line charge?
For a system with cylindrical symmetry the electrostatic potential does not depend on z.
This immediately implies that V / z  0. Under this assumption Laplace's equation reads
1   V  1  2V
r
 2
2 0
r r  r  r 
Consider as a possible solution of V:
- 37 -
Vr,   Rr   
Substituting this solution into Laplace's equation we obtain
    Rr  Rr   2  
0
r
 
r r  r  r 2  2
Multiplying each term in this equation by r2 and dividing by Rr    we obtain
r   Rr 
1    
0
r
 
2
Rr  r  r     
2
The first term in this equation depends only on r while the second term in this equation depends
only on . This equation can therefore be only valid for every r and every  if each term is equal
to a constant. Thus we require that
1   Rr 
r
  constant  
Rr  r  r 
and
1    
 
2
  
2
2
First consider the case in which   m  0 .
rewritten as
  
2
 2
The differential equation for   can be
 m2    0
The most general solution of this differential solution is
m   Cm e m  Dm e m
However, in cylindrical coordinates we require that any solution for a given  is equal to the
solution for  + 2π. Obviously this condition is not satisfied for this solution, and we conclude
2
that   m  0. The differential equation for   can be rewritten as
  
2
 2
 m2    0
- 38 -
The most general solution of this differential solution is
m  Cm cosm  Dm sinm 
The condition that m    m   2  requires that m is an integer. Now consider the radial
function Rr . We will first consider the case in which   m2  0. Consider the following
solution for Rr :
Rr   Ar
k
Substituting this solution into the previous differential equation we obtain
r   
1 
1 
1 2 k 1
k 
k 1
k
r

Ar


r

kAr



kAr


= k 2 = m2
k
k 1
k 1
k 1 k Ar


Ar r r
Ar r
Ar r
Ar


Therefore, the constant k can take on the following two values:
k+  m
k-  m
The most general solution for Rr  under the assumption that m2  0 is therefore
Rr   Am r 
m
Bm
m
r
Now consider the solutions for Rr  when m2  0. In this case we require that
  Rr 
r
  0
r  r 
or
r
Rr 
 constant  a 0
r
This equation can be rewritten as
Rr  a 0

r
r
If a0  0 then the solution of this differential equation is
- 39 -
Rr   b0  constant
If a0  0 then the solution of this differential equation is
Rr   a 0 lnr   b0
Combining the solutions obtained for m2  0 with the solutions obtained for m2  0 we
conclude that the most general solution for Rr  is given by

B 

Rr   a 0 lnr   b0   Am r m  mm 
r 
m 1 
Therefore, the most general solution of Laplace's equation for a system with cylindrical
symmetry is


 m Bm 

Vr ,   a 0 lnr   b0    Am r  m Cm cosm  Dm sinm  


r

m 1 

Example: Problem 3.25
A charge density
  a sin5 
is glued over the surface of an infinite cylinder of radius R. Find the potential inside and outside
the cylinder.
The electrostatic potential can be obtained using the general solution of Laplace's equation
for a system with cylindrical symmetry obtained in Problem 3.24. In the region inside the
cylinder the coefficient Bm must be equal to zero since otherwise Vr ,  would blow up at
r  0. For the same reason a0  0 . Thus



Vr,   bin, 0   r m Cin, m cosm  Din, m sinm 
m 1
In the region outside the cylinder the coefficients Am must be equal to zero since otherwise
Vr ,  would blow up at infinity. For the same reason a0  0 . Thus



 1

Vr,   bout, 0    m Cout, m cosm  Dout, m sinm  

m 1 r
- 40 -
Since Vr ,  must approach 0 when r approaches infinity, we must also require that bout, 0 is
equal to 0. The charge density on the surface of the cylinder is equal to
V
V


r r R  r
     0 


r R 

Differentiating Vr ,  in the region r > R and setting r = R we obtain



V
 m

   m 1 Cout, m cosm  Dout, m sinm  
r r R  m 1  R

Differentiating Vr ,  in the region r < R and setting r = R we obtain




V
m 1
  mR Cin, m cosm  Din, m sinm 
r r R m 1
The charge density on the surface of the cylinder is therefore equal to

m 1 

     0  mR m 1Cin, m 
m
m


 m 1

Din, m  m 1 Dout, m sinm 
m 1 Cout, m cosm  mR



R
R

Since the charge density is proportional to sin5  we can conclude immediately that
Cin, m Cout, m  0 for all m and that Din, m  Dout, m  0 for all m except m = 5 . Therefore

     0 5R 4 Din, 5 
5

6 Dout, 5 sin5  a sin5 

R
This requires that
5R 4 Din, 5 
5
a
6 Dout, 5 
R
0
A second relation between Din, 5 and Dout, 5 can be obtained using the condition that the
electrostatic potential is continuous at any boundary. This requires that
Vin R,  bin, 0  R 5 Din, 5 sin5   Vout R, 
Thus
bin, 0  0
- 41 -
Dout, 5
R5
sin5 
and
Dout, 5  R10Din, 5
We now have two equations with two unknown, Din, 5 and Dout, 5 , which can be solved with the
following result:
Din, 5 
a 1
10 0 R 4
and
Dout, 5 
a 6
R
10 0
The electrostatic potential inside the cylinder is thus equal to
a r5
Vin r ,  r Din, 5 sin5  
4 sin5 
10 0 R
5
The electrostatic potential outside the cylinder is thus equal to
Vout r , 
Dout, 5
r
5
a R6
sin5  
5 sin5 
10 0 r
Example: Problem 3.37
A conducting sphere of radius a, at potential V0 , is surrounded by a thin concentric spherical
shell of radius b, over which someone has glued a surface charge
     0 cos
where  0 is a constant.
a) Find the electrostatic potential in each region:
i) r > b
ii) a < r < b
b) Find the induced surface charge    on the conductor.
c) What is the total charge of the system? Check that your answer is consistent with the
behavior of V at large r.
- 42 -
a) The system has spherical symmetry and we can therefore use the most general solution of
Laplace's equation in spherical coordinates:

B 

Vr,    Am r m  mm1 Pm cos 
r
m 0
In the region inside the sphere Bm  0 since otherwise Vr,  would blow up at r = 0.
Therefore

Vr ,    Am r m Pm cos 
m 0
The boundary condition for Vr,  is that it is equal to V0 at r = a:

Va,    Am a m Pm cos   V0 V0 P0 cos 
m 0
This immediately shows that Am  0 for all m except m  0:
A0  V0
The electrostatic potential inside the sphere is thus given by
Vr ar,   V0
which should not come as a surprise.
In the region outside the shell Am  0 since otherwise Vr,  would blow up at infinity.
Thus

Vr b r ,   
m 0
Bout, m
r m 1
Pm cos 
In the region between the sphere and the shell the most general solution for V r, 

Bin, m 

m
Var b r ,    Ain, m r  m 1 Pm cos 

r 
m 0
The boundary condition for Var br,  at r = a is

Bin, m 

m
Var b a,    Ain, m a  m 1 Pm cos   V0  V0 P0 cos 

a 
m 0
- 43 -
is given by
This equation can only be satisfied if
Ain, m a m 
Ain, 0 
Bin, m
a m 1
Bin, 0
a
 0 if m  0
 V0
if m  0
The requirement that the electrostatic potential is continuous at r = b requires that

Bin, m 
Bout, m

m


Ain, m b  m 1 Pm cos   m 1 Pm cos 


b 
m 0
m 0 b

or
Ain, m b m 
Bin, m
b
m 1

Bout, m
b m 1
This condition can be rewritten as
Bout, m  Bin, m  Ain, m b 2m 1
The other boundary condition for the electrostatic potential at r = b is that it must produce the
charge distribution given in the problem. This requires that
  
V
V
  0 


r r b  r


m  1
m 1
P cos  
  0  m  2 Bout, m  Bin, m  mAin, m b

 m
m 0 b
r b 



  0 cos   0 P1cos 
This condition is satisfied if
m 1
m 1
 0 if m 1
m  2 Bout, m  Bin, m  mAin, m b
b


2
0
3 Bout,1  Bin,1  Ain,1 
b
0


if m 1
Substituting the relation between the various coefficients obtained by applying the continuity
condition we obtain
m 1
2m 1
m 1
m 1
 mAin, m b  2m 1Ain, m b  0 if m 1
m  2 Ain, m b
b
- 44 -
2
0
3
3 Ain,1b  Ain,1  3Ain,1 
b
0
if m 1
These equations show that
Ain, m  0 if m 1
Ain,1 
0
3 0
if m 1
Using these values for Ain, m we can show that
Bout, m  Bin, m  0 if m 1
Bout,1  Bin,1 
0 3
b if m 1
3 0
The boundary condition for V at r = a shows that
Bin, m  Ain, m a 2m 1  0 if m  2
Bin,1  Ain,1a 3  

0 3
a if m 1
3 0

Bin, 0  a V0  Ain, 0  aV0
if m  0
These values for Bin, m immediately fix the values for Bout, m :
Bout, m  Bin, m  0 if m  2
Bout,1 
0 3

b  Bin,1  0 b 3  a 3  if m 1
3 0
3 0
Bout, 0  Bin, 0  aV0
if m  0
The potential in the region outside the shell is therefore equal to
Vr b r,  
aV0
 1
P0 cos   0 2 b 3  a 3 P1cos 
r
30 r
The potential in the region between the sphere and the shell is equal to
- 45 -
Var b r ,  
aV0
  a 3 
P0 cos   0 r  2 P1cos 
r
30  r 
b) The charge density on the surface of the sphere can be found by calculating the slope of the
electrostatic potential at this surface:
V
V


r r a  r
     0 

 V 
  V
  0  0  0 cos  0 0   0 cos
r a 
 a 0
 a

c) The total charge on the sphere is equal to

Qa  0

2
0

  0V0

  0 0 cos sin d  4 a0V
 a

   a 2 sin d d  2a 2 2
The total charge on the shell is equal to zero. Therefore the total charge of the system is equal to
Qtotal  4 a 0V0
The electrostatic potential at large distances will therefore be approximately equal to
V
1 Qtotal
1 4 a 0V aV0


4 0 r
4 0
r
r
This is equal to limit of the exact electrostatic potential when r  .
3.4. Multipole Expansions
Consider a given charge distribution . The potential at a point P (see Figure 3.11) is equal
to
VP  
1


4 0 Volume d
d
where d is the distance between P and a infinitesimal segment of the charge distribution. Figure
3.11 shows that d can be written as a function of r, r' and :
 r '2

r'
d 2  r 2  r'2 2rr 'cos  r 2 1     2 cos 
r
r


- 46 -
y
r'

d
r
P
x
Figure 3.11. Charge distribution .
This equation can be rewritten as
1 1
1

d r  r'2

r '
1     2 cos 
r
r


At large distances from the charge distribution r  r' and consequently r'/r 1 . Using the
following expansion for 1 / 1  x :
1
1
3 2 5 3
 1  x  x  x  ....
2
8
16
1  x 
we can rewrite 1/d as
1
d
2

1 
 1 r 'r '
 3 r '2 r'



1
 2cos  
 2cos   .....



 r 
 8 r  r 

r 

 2 r

2
n


1 
1 
 r '
r ' 3 2
 1 r '

1   cos     cos     .....    Pn cos 

r 
r
r
2
2


 r n0 r
Using this expansion of 1/d we can rewrite the electrostatic potential at P as
VP  
1
4 0

r
n0
1
n 1

r 'r 'n Pn cos d
Volume
This expression is valid for all r (not only r  r' ). However, if r  r' then the potential at P
will be dominated by the first non-zero term in this expansion. This expansion is known as the
multipole expansion. In the limit of r  r' only the first terms in the expansion need to be
considered:
- 47 -
VP  
1

1 
1
1
1 

2 3
2
  d  2  r'cosd  3  r '  cos   d  ....
4 0 
r Volume
r Volume
2
2
r Volume


The first term in this expression, proportional to 1/r, is called the monopole term. The second
term in this expression, proportional to 1/r2, is called the dipole term. The third term in this
expression, proportional to 1/r3, is called the quadrupole term.
3.4.1. The monopole term.
If the total charge of the system is non zero then the electrostatic potential at large distances
is dominated by the monopole term:
VP  
1 1
1 Q
d 

4 0 r Volume
4 0 r
where Q is the total charge of the charge distribution.
The electric field associated with the monopole term can be obtained by calculating the
gradient of V P :
EP    V P   
1 Q
1 
ˆ


2 r


4 0
r
4 0 r
Q
3.4.2. The dipole term.
If the total charge of the charge distribution is equal to zero (Q = 0) then the monopole term
in the multipole expansion will be equal to zero. In this case the dipole term will dominate the
electrostatic potential at large distances
VP  
1
1
2
 r 'cos d
4 0 r Volume
Since  is the angle between r and r ' we can rewrite r'cos as
r'cos  rˆ  r '
The electrostatic potential at P can therefore be rewritten as
VP  
rˆ
1 p  rˆ
2    r ' d 
2
4 0 r Volume
4 0 r
1
- 48 -
In this expression p is the dipole moment of the charge distribution which is defined as

p
r ' d
Volume
The electric field associated with the dipole term can be obtained by calculating the gradient of
VP :
ErP   
V P  2p cos 1

3
r
4 0 r
E P   
1 V P  p sin 1

3
r 
4 0 r
E P   
1 V P 
0
r sin 
Example
Consider a system of two point charges shown in Figure 3.12. The total charge of this
system is zero, and therefore the monopole term is equal to zero. The dipole moment of this
system is equal to
p  qr  qr  qr  r   qs
where s is the vector pointing from -q to +q.
The dipole moment of a charge distribution depends on the origin of the coordinate system
chosen. Consider a coordinate system S and a charge distribution . The dipole moment of this
charge distribution is equal to
pS 

 rS d
Volume
A second coordinate system S' is displaced by d with respect to S:
rS '  rS  d
The dipole moment of the charge distribution in S' is equal to
- 49 -
pS ' 


 rS ' d 
Volume

 rS d  d
Volume
 d  pS  d Q
Volume
This equation shows that if the total charge of the system is zero (Q = 0) then the dipole moment
of the charge distribution is independent of the choice of the origin of the coordinate system.
y
-q
s
r-
+q
r+
x
Figure 3.12. Electric dipole moment.
Example: Problem 3.40
A thin insulating rod, running from z = -a to z = +a, carries the following line charges:
 z 
a)   0 cos
2a 
 z 
b)   0 sin
 a 
 z 
c)   0 cos
 a 
In each case, find the leading term in the multipole expansion of the potential.
a) The total charge on the rod is equal to
a
a
4a
 z 
Qtot  a dz  a 0 cos dz 

2a
 0
Since Qtot  0, the monopole term will dominate the electrostatic potential at large distances.
Thus
VP 
1
4a
4 0 
0
1
r
b) The total charge on the rod is equal to zero. Therefore, the electrostatic potential at large
distances will be dominated by the dipole term (if non-zero). The dipole moment of the rod is
equal to
- 50 -
a
a
p   zdz   z0 sin
a
a
2a 2
 z 
dz 

 a 
 0
Since the dipole moment of the rod is not equal to zero, the dipole term will dominate the
electrostatic potential at large distances. Therefore
VP 
2a 2
1
4 0 
0
1
2 cos
r
c) For this charge distribution the total charge is equal to zero and the dipole moment is equal to
zero. The electrostatic potential of this charge distribution is dominated by the quadrupole term.
a
I 2   z dz  
a
2
a
a
4a 3
 z 
z 0 cos
dz  2 0
 a 

2
The electrostatic potential at large distance from the rod will be equal to
VP =
4 a3
1
–
 0 1 1 3 cos2  – 1
2
40
r3 2

Example: Problem 3.27
Four particles (one of charge q, one of charge 3q, and two of charge -2q) are placed as shown
in Figure 3.12, each a distance d from the origin. Find a simple approximate formula for the
electrostatic potential, valid at a point P far from the origin.
The total charge of the system is equal to zero and therefore the monopole term in the
multipole expansion is equal to zero. The dipole moment of this charge distribution is equal to
p   qi ri  2qd ˆj  qd  kˆ  2qd  ˆj  3qd kˆ  2qd kˆ
i
The Cartesian coordinates of P are
x  r sin cos
y  r sin sin 
z  r cos
The scalar product between p and rˆ is therefore
- 51 -
p  rˆ  2qd cos
The electrostatic potential at P is therefore equal to
VP 
p  rˆ
1 2qd cos

2
4 0 r
4 0
r2
1
z
3q
y
-2q
-2q
+q
Figure 3.13. Problem 3.27.
Example: Problem 3.38
A charge Q is distributed uniformly along the z axis from z = -a to z =a. Show that the
electric potential at a point r ,  is given by
Vr ,  

Q 1  1 a 2
1 a 4


1    P2 cos    P4 cos   ....

4 0 r  3 r
5 r

for r > a.
The charge density along this segment of the z axis is equal to

Q
2a
Therefore, the nth moment of the charge distribution is equal to
a
Q a n
Q z n 1
Q a n 1
Q an
n 1
1 
1n 1
I n  a z dz 
z
dz


1

1


2a a
2a n  1 a 2a n  1
2 n 1
a

n
- 52 -



This equation immediately shows that
an
In 
Q
n 1
if n is even
I n  0 if n is odd
The electrostatic potential at P is therefore equal to
Vr,  
1

4 0 
n0 r
1
n 1 I n Pn cos  

Q 1  1 a 2
1 a 4


cos   ....
1

P
cos


P

2
4
4 0 r  3 r 
5 r 

- 53 -