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

Taking the divergence operation for both sides of the above

equation gives

In a linear, homogeneous, and isotropic medium, the divergence

of the electric field intensity E is

E

2 E

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

02

which is called Laplace’s equation.

The solution of Poisson’s Equation.

VV

d||

)(

π4

1)(

rr

rr

In infinite free space, the electric charge density

confined to in V produces the electric potential given by

)(r

which is just the solution for Poisson’s Equation in free space.

Applying Green’s function gives the general solution of

Poission’s equation

) ,( rr G

Srrrrrr

rrrr

d)] ,()()() ,([

d) (

) ,()(

GG

VG

S

V

| |π4

1) ,(0 rr

rr

G

For infinite free space, the surface integral in the above equation

will become zero, and Green’s function becomes

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.

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.

1. Dirichet boundary condition: The physical quantities on the

boundaries are specified.

Usually the boundary conditions are classified into three types:

For any mathematical physics equation, the existence, the stability,

and the uniqueness of the solutions need to be investigated.

The uniqueness of the solution is whether the solution is unique or

not for the prescribed 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 existence of the solution is that whether the equation has a

solution or not for the given 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.

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.

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.

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

Dielectric

Conductor

q r

P

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',

r

q

r

q

π4 π4

Considering the electric potential of an infinite conducting plane

is zero, we have .qq

Dielectric

q r

P

h

h

r

qDielectric

The distribution of the electric field lines and the equipotential

surfaces are the same as that of an electric dipole in the upper half-

space.

The electric field lines are perpendicular to the conducting

surface everywhere, which has zero potential.

z

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.

The above equivalence is established only for the upper half-

space for which the source and the boundary condition are both

unchanged.

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.

z

q

For the semi-infinite wedge conducting boundary, the method of

images is also applicable. However, the images can be found only for

conducting wedges with angle given by where n is an integer. In

order to keep the wedge boundary at zero-potential, several image

charges are required.

n

π

/3

/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.

f

qO

（ 2 ） A point charge and a conducting

sphere. 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 poten-

tial on the surface of the sphere is

then given by

r

q

r

q

π4 π4

Requiring that the electric potential at any point on the surface

of the sphere be zero, the image charge must be

qr

rq

P

a

d

r

qr'

The ratio must be constant for any point on the surface of

the sphere to obtain an image charge with a fixed value. r

r

qf

aq

The distance d is

f

ad

2

The electric field intensity outside the sphere can be found out

from q and q' .

If △OPq'~ △OqP , then . Thus the quantity of the

image charge should bef

a

r

r

f

qO

P

a

d

r

qr'

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.

qq

If the image charge q' is put in, then another image charge q" is

needed in order to satisfy the neutrality condition.

f

qO

P

a

d

r

qr'

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.

f

qO

P

a

d

r

qr'

l

（ 3 ） A line charge and a charged conducting

cylinder. P

a

fd

r-lO

An image line charge is put in to represent the charge on

the cylinder and placed parallel to it, at a distance d from the axis.

l

rl

reE

π2

The electric potential with respect to a reference point at a

distance r from the line electric charge is

0r

r

rrE l

r

r

0

ln

π2d

0

The electric field intensity produced by an infinite line charge of

density given byl

If is also taken as the reference point for the electric

potential produced by the image line charge , then and

produce an electric potential at a point P on the surface of the

cylinder given by

l

0r

l l

r

r

r

r llP

00 lnπ2

lnπ2

r

rl lnπ2

Since the conducting cylinder is an equipotential body, in order

to satisfy this boundary condition the ratio must be a constant. r

r

f

ad

2

a

d

f

a

r

r

Let , we find

l

P

a

fd

r-lO

（ 4 ） A point charge and an infinite dielectric

plane.

E

1

1

q

r0

E'

tE

nE

Et

En

0r

q'

2

2

q"0r

nE

tE

E"

1

2

q

et

en

= +

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 permittivity1.

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. We have therefore, 2t1t1t EEE n21n1n DDD

The electric field intensities produced by the point charges q, q',

and q" are, respectively,

To satisfy the boundary condition, we find the image charges

are, respectively,

rr

qeE

21

1 π4 rr

q

eE

21

1 )(π4 rr

q

eE

22

2 )(π4

qq21

21

qq21

22

E

1

1

q

r0

E'

tE

nE

Et

En

0r

q'

2

2

q"0r

nE

tE

E"

1

2

q

et

en

= +

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.

0d

d

d

d12

rr

rr

21 ln CrC We have

Vb

aO

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

Consideringar

U

br

0

0ln

ln

21

21

CbC

UCaCWe have

ba

br

U lnln

ba

U

rrr

r

ln

eeE

We obtain

ba

UC

ln1

babU

Cln

ln2

Solving for the constants C1 and C2 gives

Vb

aO

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 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.

In general, the boundary of the region of interest should conform

with the coordinate system to be used.

3. Method of Separation of Variables in Rectangular Coordinates

02

2

2

2

2

2

zyx

In rectangular coordinate system, Laplace’s Equation for electric

potential is

)()()() , ,( zZyYxXzyx Let

Substituting it into the above equation, and dividing both sides

by X(x)Y(y)Z(z), we have

0d

d1

d

d1

d

d12

2

2

2

2

2

z

Z

Zy

Y

Yx

X

X

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 , and we have222 , , zyx kkk

0d

d 22

2

Xkx

Xx 0

d

d 22

2

Yky

Yy 0

d

d 22

2

Zkz

Zz

The three separation constants are not independent of each other,

and they satisfy the following equation

0222 zyx kkk

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.

xkxk xx BAxX jj ee)( xkDxkCxX xx cossin)( or

where A, B, C, D are the constants to be

determined.

where kx ， ky ， kz are called the separation constants, and they

could be real or imaginary numbers.

The solution of the equation for the variable x can be written as

xx BAxX ee)( xDxCxX coshsinh)( or

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.

The separation constants could be imaginary numbers. If is

an imaginary number, written as , then the equation becomesjxk

xk

The constants in the solutions are also related to the boundary

conditions.

It is very important to select the forms of the solutions, which

depend on the given 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.

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

02

2

2

2

yx

d

x

y = 0

= 0

= 0

O

)()() ,( yYxXyx Using the method of separation of variables, and let

The boundary conditions for the electric potential in the slot

can be expressed as

In order to satisfy the boundary conditions and

, the solution of Y(y) should be selected as 0) ,( dx0)0 ,( x

ykBykAyY yy cossin)(

0) ,0( y 0) ,( y

0)0 ,( x 0) ,( dx

3, 2, 1, ,π

nd

nk y

From the boundary condition , we have = 0 at y = 0

, and the constant B = 0. In order to satisfy , the constant

ky should be

0) ,( dx

0)0 ,( x

d

x

y = 0

= 0

= 0

O

y

d

nAyY

πsin)(We find

Since ， we obtain022 yx kk

d

nkx

πj

The constant kx is an imaginary number, and the solution of

X(x) should be xd

nx

d

n

DCxXππ

ee)(

Since at x = 0 , the constant C = 0, and

xd

n

DxXπ

e)(

y

d

nCyx

xd

n πsine),(

π

Then

Where the constant C = AD .

Since = 0 at x = 0 , and we have

y

d

nC

πsin0

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

y

d

nCyx

n

xd

n

n

πsine),(

1

π

In order to satisfy the boundary condition x = 0, = 0 , and

we have

dyyd

nC

nn

0 ,π

sin1

0

The right side of the above equation is Fourier series. By using the

orthogonality between the terms of a Fourier series, the coefficients Cn

can be found as

n

xd

n

yd

n

nyx

πsine

1

π

4),(

π0

Finally, we find the electric potential in the slot as

5, 3, ,1n

0

d

x

y = 0

= 0

= 0

Electric field lines

Equipotential surfaces

even is if , 0

odd is if ,π

4 0

n

nnCn

nm

nmmxdxnx

0

sinsin

4. Method of Separation of Variables in Cylindrical Coordinates

In cylindrical coordinate system, Laplace’s equation has the form

011

2

2

2

2

2

zrrr

rr

Let )()()(),,( zZrRzr

0d

d

d

d1

d

d

d

d2

22

2

2

z

Z

Z

r

r

Rr

rR

r

And we have

where the second term is a function of the variable only, while the

first and the third are independent of , leading to

22

2

d

d1

k

0d

d 22

2

kor

where k is the separation constant, and it could be real or imaginary

number. The domain of the variable is , in this case the

change of the field with must be a periodic function with the period

of 2.

π20

Let (m is integer), then the solution of the above equation

is

mk

mBmA cossin)(

where A and B are the constants to be determined.

Considering and the above equation for variable , the

previous equation can be rewritten as

mk

0d

d1

d

d

d

d12

2

2

2

z

Z

Zr

m

r

Rr

rRr

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

22

2

d

d1zk

z

Z

Z 0

d

d 22

2

Zkz

Zzor

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

zkDzkCzZ zz cossin)(

where C and D are constants to be determined.

Substituting the equation for the variable z into the previous

equation gives

0)(d

d

d

d 2222

22 Rmrk

r

Rr

r

Rr z

If we let , then the above equation becomes222 xrk z

0)(d

d

d

d 222

2 Rmxx

Rx

x

Rx

which is called a Bessel equation, and its solution is a Bessel function,

given as )(N)(J)( rkFrkErR zmzm

The solution of the above quation should be the linear combination

of products of the solutions R(r) ,() , Z(z).

where E and F are constants to be determined, with being the

first kind of Bessel function of order m, and being the second

kind of Bessel function of order m.

)(J rkzm

)(N rkzm

Since 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 .

)(N rkzm

If the electric fields are independent of z, then we have .

The above equation becomes

0zk

0d

d

d

d 22

22 Rm

r

Rr

r

Rr

The solutions of the above equation are exponential functions,

mm FrErrR )(

If the field is independent of z, and also independent of , then m =

0 . The solution is given by

00 ln)( BrArR

Considering all of the above situations, the solution can be written as

1

10

)cossin(

)cossin(ln),(

mmm

m

mmm

m

mDmCr

mBmArrAr

xE eE 00

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.

x

y

a

E0

O

Example. An infinitely long conducting cylinder of radius a is

placed in a homogeneous electrostatic field . The direction of is

perpendicular to the axis of the conducting cylinder, as shown in the

figure. Find the electric fields intensity inside and outside the cylinder.

0E0E

Solution: Select the cylindrical coordinate

system. Let z-axis be the axis of the cylinder,

and is aligned with the x-axis, so that. 0E

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:

01

ar

r

eE

(a) The tangential component of electric field intensity on the

surface of the cylinder is zero, and has no component in the z-

direction due to symmetry. This leads to the result

(b) The electric potential at infinity should be the

same as that required by the original field as given

by cos) ,( 00 rExE

0

ar

which states that if , the electric potential is independent of

the function , but proportional to r and . Hence, one may

conclude that the coefficients , and m = 0.00 mm CAA

r

sin cos

x

y

a

E0

O

01 EB 201 aED

Based on the given boundary conditions, we find the coefficients

B1 and D1 as

And the electric potential outside the cylinder as

coscos),(2

00 r

aErEr

zrr zr

eeeE

1

sin1cos1 02

2

02

2

Er

aE

r

ar

ee

Hence, the electric potential function now reduces to

coscos),( 11 r

DrBr

x

y

a

E0

O

The electric field intensity outside the cylinder can

be obtained as

x

y

a

E0

Electric field lines

Equipotential surfaces

The electric field lines, the equipotential surfaces outside the

cylinder, and the charges on the surface are shown as follows:

Surface charges

5. Method of Separation of Variables in Spherical Coordinates

In spherical coordinate system, Laplace’s equation becomes

0sin

1sin

sin

112

2

2222

2

rrrr

rr

)()()(),,( rRr Let

0d

d1

d

dsin

d

dsin

d

d

d

dsin2

22

2

r

Rr

rR

We have

For the same reason as before, the solution for should be

mBmA cossin)(

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

d

d

d1 2

nnr

Rr

rR

0)1(d

d2

d

d2

22 Rnn

r

Rr

r

Rr

where n is an integer. This is Euler’s equation, and the general solution

being given by

1)(

n

n

r

DCrrR

0sind

dsin

d

d

sin

1

d

d

d

d12

22

m

r

Rr

rRAnd

Let , then we have xcos

01

)1(d

d)1(

d

d2

22

x

mnn

xx

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 and the second kind of associated Legender’s

functions , where m < n .

)(P xmn

)(Q xmn

If n is an integer, and are polynomials with finite

number of terms, and n was required to be an integer.

)(P xmn

)(Q xmn

0sin

sin)1(d

dsin

d

d 2

m

nnAnd

)(cosP)(P)( mn

mn x

Therefore, we take the following linear combination as the general

solution

)(cosP))(cossin(),,( )1(

0 0

mn

nn

nnm

n

m nm rDrCmBmAr

The field is independent of , so that m = 0 . In this case,

which is called Legendre’s function of the first kind,

and the general solution is

)(P)(P0 xx nn

)(cosP)(),(0

)1( nn

nn

nn rDrCr

From the properties of the second kind of associated

Legender’s functions , we know that as . Thus, if

the region includes or , for which , we only can take

the first kind of associated Legender’s functions as the solution. In

view of this, we have

1x)(Q xmn

0 1x

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.

E0

z

y

0

a

Solution: Select spherical coordinate system,

and let the direction of E0 coincide with the z-

axis, i.e. . Obviously, in this case the

field is rotationally symmetrical to the z-axis,

and independent of .

zE eE 00

0

)1(

0i )(cosP)(cosP),(

nn

nn

nn

nn rDrCr

0

)1(

0o )(cosP)(cosP),(

nn

nn

nn

nn rBrAr

In this way, the solution of the distribution functions of the

electric potential inside and outside the sphere as follows, respectively,

The distribution functions of the electric potentials inside and

outside the sphere should satisfy the following boundary

conditions:

② The electric potential at infinity should

be )(coscos) ,( 100o rPErE

③ The electric potential should be continues at the surface of the

sphere, i.e. ) ,() ,( oi aa

④ The normal derivatives of electric potentials

at the surface should satisfy

arar rr

o0

i

① The electric potential at the center of the sphere should be

finite.

) ,0(i

E0

z

y

0

a

Considering the boundary condition ①, the coefficient Dn

should be zero, hence

0

i )(cos),(n

nn

n PrCr

In order to satisfy the boundary condition ②, all of the coefficients An

except A1 should be zero, and , so that01 EA

)(cos)(cos),( )1(

010o n

n

nn PrBrPEr

Reconsidering the boundary condition ③ , we have

0

)1(10

0

)(cos)(cos)(cosn

nn

nn

nn

n PaBaPEPaC

To satisfy the boundary condition ④, we obtain

0

)2(10

0

1r )(cos)1()(cos)(cos

nn

nn

nn

nn PaBnPEPanC

0r

where

Since the above equations hold for any , the corresponding

coefficients on both sides should be equal. Hence, we find

000 CB

2

1

r

r301

aEB

2

3

r

01

EC )2( ,0 nCB nn

2

3cos

2

3),(

r

0

r

0i z

Er

Er

2

1cos),(

2

30

r

r0o r

aErEr

Finally, we find the electric potentials inside and outside the sphere

are, respectively,

E0

z

y

0

a

Since , we find the electric field intensity inside the sphere isE

zz

E

zeeE

0

00ii 2

3

The field inside the sphere is still uniform, and the field intensity

inside the sphere is less than that outside the sphere.

z

EeE

2

3

0

0i

The electric field intensity inside the

sphere is greater than that outside the

sphere.

zz EE

ee 0r

0

2

3

zz EE

ee 0r

0r

21

3

If there is a spherical air bubble in

an infinite homogeneous dielectric with

the permittivity , then the electric field

intensity isiE