1

Chapter 3. Boundary-Value Problems in Electrostatics:

Spherical and Cylindrical Geometries

3.1 Laplace Equation in Spherical Coordinates

The spherical coordinate system is probably the most useful of all coordinate systems in study

of electrostatics, particularly at the microscopic level. In spherical coordinates (𝑟, 𝜃, 𝜙), the

Laplace equation reads:

1

𝑟2𝜕

𝜕𝑟(𝑟2

𝜕Φ

𝜕𝑟) +

1

𝑟2 sin 𝜃

𝜕

𝜕𝜃(sin𝜃

𝜕Φ

𝜕𝜃) +

1

𝑟2 sin2 𝜃

𝜕2Φ

𝜕𝜙2= 0

Try separation of variables Φ(𝑟, 𝜃, 𝜙) = 𝑅(𝑟)𝑌(𝜃, 𝜙). Then Eq. 3.1 becomes

1

𝑅[𝑑

𝑑𝑟(𝑟2

𝑑𝑅

𝑑𝑟)] = −

1

𝑌[1

sin 𝜃

𝜕

𝜕𝜃(sin𝜃

𝜕𝑌

𝜕𝜃) +

1

sin2 𝜃

𝜕2𝑌

𝜕𝜙2]

Since each side must be equal to a constant, 𝜆 = 𝑙(𝑙 + 1), we get two equations.

Radial equation

𝑑

𝑑𝑟(𝑟2

𝑑

𝑑𝑟𝑅) − 𝑙(𝑙 + 1)𝑅 = 0

Introducing 𝑈(𝑟) ≡ 𝑟𝑅(𝑟), we obtain

𝑑2𝑈

𝑑𝑟2−𝑙(𝑙 + 1)

𝑟2𝑈 = 0

The general solution of this equation is

𝑈(𝑟) = 𝐴𝑟𝑙+1 + 𝐵𝑟−1 → 𝑅(𝑟) = 𝐴𝑟𝑙 +𝐵

𝑟𝑙+1

Angular equation

1

sin 𝜃

𝜕

𝜕𝜃(sin𝜃

𝜕𝑌

𝜕𝜃) +

1

sin2 𝜃

𝜕2𝑌

𝜕𝜙2= −𝑙(𝑙 + 1)𝑌

We solve Eq. 3.6 using the separation of variable method again: 𝑌(𝜃, 𝜙) = 𝑃(𝜃)𝑄(𝜙). Then we

obtain two ordinary differential equations for 𝜃 and 𝜙 :

(3.1)

(3.2)

(3.3)

(3.4)

(3.5)

(3.6)

2

1

sin𝜃

𝑑

𝑑𝜃(sin 𝜃

𝑑𝑃

𝑑𝜃) + [𝑙(𝑙 + 1) −

𝑚2

sin2 𝜃] 𝑃 = 0

𝜕2𝑄

𝜕𝜙2+𝑚2𝑄 = 0

Equation 3.8 has solutions

𝑄𝑚(𝜙) = 𝑒±𝑖𝑚𝜙

If the full azimuthal range (0 ≤ 𝜙 ≤ 2𝜋) is allowed, 𝑄(𝜙) = 𝑄(𝜙 + 2𝜋) so that 𝑚 must be an

integer. The functions 𝑄𝑚 form a complete set of orthogonal functions on the interval 0 ≤ 𝜙 ≤

2𝜋, which is nothing but a basis for Fourier series. The orthogonality relation is

∫ 𝑄𝑚(𝜙)𝑄𝑚′∗ (𝜙)

2𝜋

0𝑑𝜙 = ∫ 𝑒𝑖(𝑚−𝑚

′)𝜙2𝜋

0𝑑𝜙 = 2𝜋 𝛿𝑚𝑚′

Simultaneously, the completeness relation is

1

2𝜋∑ 𝑄𝑚(𝜙)𝑄𝑚

∗ (𝜙′)

∞

𝑚=−∞

=1

2𝜋∑ 𝑒𝑖𝑚(𝜙−𝜙

′)

∞

𝑚=−∞

= 𝛿(𝜙 − 𝜙′)

Associated Legendre Polynomials

Equation 3.7 can be written in terms of 𝑥 ≡ cos 𝜃 :

𝑑

𝑑𝑥[(1 − 𝑥2)

𝑑𝑃(𝑥)

𝑑𝑥] + [𝑙(𝑙 + 1) −

𝑚2

1 − 𝑥2] 𝑃(𝑥) = 0

This is the differential equation for the associated Legendre polynomials. Physically acceptable

solution (i.e., −1 ≤ 𝑥 ≤ 1) is obtained only if 𝑙 is a positive integer or 0. The solution is then

associated Legendre polynomial 𝑃𝑙𝑚(𝑥) where 𝑙 = 0,1,2, … and −𝑙 ≤ 𝑚 ≤ 𝑙.

𝑃𝑙𝑚(𝑥) has the form, (1 − 𝑥2)

|𝑚|

2 times a polynomial of order 𝑙 − |𝑚|, which can be obtained by

Rodrigues formula,

𝑃𝑙𝑚(𝑥) =

(−1)𝑚

2𝑙𝑙!(1 − 𝑥2)

|𝑚|

2𝑑𝑚+𝑙

𝑑𝑥𝑚+𝑙(𝑥2 − 1)𝑙 for 𝑥 ≥ 0

and the parity relation

𝑃𝑙𝑚(−𝑥) = (−1)𝑙−|𝑚|𝑃𝑙

𝑚(𝑥)

The orthogonal relation of the functions 𝑃𝑙𝑚 is expressed as

(3.7)

(3.8)

(3.9)

(3.12)

(3.13)

(3.14)

(3.10)

(3.11)

3

∫ 𝑃𝑙𝑚(𝑥)𝑃𝑙′

𝑚(𝑥)1

−1𝑑𝑥 = (

2

2𝑙+1 )(𝑙+|𝑚|)!

(𝑙−|𝑚|)! 𝛿𝑙𝑙′

The functions 𝑃𝑙(𝑥) for 𝑚 = 0 are called Legendre polynomials which are the solutions when

the problem has azimuthal symmetry so that Φ is independent of 𝜙. The first few 𝑃𝑙 are

𝑃0(𝑥) = 1, 𝑃1(𝑥) = 𝑥, 𝑃2(𝑥) =1

2(3𝑥2 − 1), …

The Legendre polynomials form a complete orthogonal set of functions on the interval −1 ≤

𝑥 ≤ 1. The orthogonality condition can be written as

∫ 𝑃𝑙(𝑥)𝑃𝑙′(𝑥)1

−1

𝑑𝑥 = 2

2𝑙 + 1 𝛿𝑙𝑙′

and the completeness relation is expressed as

∑(2𝑙 + 1

2)𝑃𝑙(𝑥)𝑃𝑙(𝑥

′)

∞

𝑙=0

= 𝛿(𝑥 − 𝑥′)

Spherical Harmonics

The angular function can be written as 𝑌𝑙𝑚(𝜃, 𝜙) = 𝑁𝑙𝑚𝑃𝑙𝑚(cos 𝜃)𝑒𝑖𝑚𝜙, where 𝑁𝑙𝑚 is

normalization constant. The normalized angular functions,

𝑌𝑙𝑚(𝜃, 𝜙) = √(2𝑙+1)

4𝜋

(𝑙−𝑚)!

(𝑙+𝑚)! 𝑃𝑙𝑚(cos 𝜃)𝑒

𝑖𝑚𝜙 for 𝑚 ≥ 0

𝑌𝑙,−𝑚(𝜃, 𝜙) = (−1)𝑚𝑌𝑙𝑚

∗ (𝜃, 𝜙)

are called spherical harmonics.

𝑌𝑙𝑚 is either even or odd, depending on 𝑙, i.e., 𝑌𝑙𝑚(−𝐱) = (−1)𝑙𝑌𝑙𝑚(𝐱):

In the spherical coordinate, 𝐱 → −𝐱 corresponds to 𝑟 → 𝑟, 𝜃 → 𝜋 − 𝜃, 𝜙 → 𝜋 + 𝜙, therefore

𝑌𝑙𝑚(𝜋 − 𝜃, 𝜋 + 𝜙) = √(2𝑙+1)

4𝜋

(𝑙−𝑚)!

(𝑙+𝑚)! 𝑃𝑙𝑚(cos(𝜋 − 𝜃))𝑒

𝑖𝑚(𝜙+𝜋)

= √(2𝑙+1)

4𝜋

(𝑙−𝑚)!

(𝑙+𝑚)! (−1)𝑙−𝑚𝑃𝑙𝑚(cos 𝜃)(−1)

𝑚𝑒𝑖𝜙

= (−1)𝑙𝑌𝑙𝑚(𝜃, 𝜙)

The first few spherical harmonics are

(3.19)

(3.15)

(3.20)

(3.21)

(3.16)

(3.17)

(3.18)

4

𝑌00 =1

√4𝜋, 𝑌1±1 = ∓√

3

8𝜋sin 𝜃 𝑒±𝑖𝜙, 𝑌10 = √

3

4𝜋cos 𝜃, ….

Spherical harmonics form an orthonormal basis for 𝜃 and 𝜙:

∫𝑌𝑙𝑚∗ (𝜃, 𝜙) 𝑌𝑙′𝑚′(𝜃, 𝜙) 𝑑Ω = 𝛿𝑙𝑙′𝛿𝑚𝑚′

The corresponding completeness relation is written as

∑ ∑ 𝑌𝑙𝑚∗ (𝜃′, 𝜙′) 𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

= 𝛿(𝜙 − 𝜙′)𝛿(cos𝜃 − cos 𝜃′)

General Solution

The general solution for a boundary-value problem in spherical coordinates can be written as

Φ(𝑟, 𝜃, 𝜙) =∑ ∑ [𝐴𝑙𝑚𝑟𝑙 + 𝐵𝑙𝑚𝑟

−(𝑙+1)] 𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

3.2 Boundary-Value Problems with Azimuthal Symmetry

We consider physical situations with complete rotational symmetry about the z-axis (azimuthal

symmetry or axial symmetry). This means that the general solution is independent of 𝜙, i.e.,

𝑚 = 0 in Eq. 3.25:

Φ(𝑟, 𝜃) =∑[𝐴𝑙𝑟𝑙 + 𝐵𝑙𝑟

−(𝑙+1)] 𝑃𝑙(cos𝜃)

∞

𝑙=0

The coefficients 𝐴𝑙 and 𝐵𝑙 can be determined by the boundary conditions.

Spherical Shell

Suppose that the potential 𝑉(𝜃) is specified on the surface of a spherical shell of radius 𝑎.

Inside the shell, 𝐵𝑙 = 0 for all 𝑙 because the potential at origin must be finite. The boundary

condition at 𝑟 = 𝑎 leads to

𝑉(𝜃) =∑𝐴𝑙𝑎𝑙 𝑃𝑙(cos𝜃)

∞

𝑙=0

Using the orthogonality relation Eq. 3.17, we can evaluate the coefficients 𝐴𝑙,

𝐴𝑙 =2𝑙 + 1

2𝑎𝑙∫ 𝑉(𝜃)𝑃𝑙(cos 𝜃)𝜋

0

sin 𝜃 𝑑𝜃

(3.23)

(3.25)

(3.26)

(3.27)

(3.28)

(3.22)

(3.24)

5

On the other hand, outside the shell, 𝐴𝑙 = 0 for all 𝑙 because the potential for 𝑟 → ∞ must be

finite and the boundary condition gives rise to

𝐵𝑙 =2𝑙 + 1

2𝑎𝑙+1∫ 𝑉(𝜃)𝑃𝑙(cos𝜃)

𝜋

0

sin 𝜃 𝑑𝜃

Two hemispheres at equal and opposite potentials

Consider a conducting sphere composed of two hemispheres at equal and opposite potentials

as shown in Fig. 3.1. Then, inside the sphere, the coefficients 𝐴𝑙 are

𝐴𝑙 =2𝑙+1

2𝑎𝑙𝑉 [∫ 𝑃𝑙(𝑥)

1

0𝑑𝑥 − ∫ 𝑃𝑙(𝑥)

0

−1𝑑𝑥]

=2𝑙+1

2𝑎𝑙𝑉 [∫ 𝑃𝑙(𝑥)

1

0𝑑𝑥 − ∫ 𝑃𝑙(−𝑥)

1

0𝑑𝑥]

=2𝑙+1

2𝑎𝑙𝑉[1 − (−1)𝑙] ∫ 𝑃𝑙(𝑥)

1

0𝑑𝑥

Thus, for even 𝑙, 𝐴𝑙 = 0, and for odd 𝑙, using Rodrigues formula, we obtain

𝐴𝑙 =2𝑙 + 1

𝑎𝑙𝑉∫ 𝑃𝑙(𝑥)

1

0

𝑑𝑥 = (−1

2)

𝑙−1

2 (2𝑙 + 1)(𝑙 − 2)‼

2 (𝑙 + 1

2) !

𝑉

𝑎𝑙

Similarly, outside the sphere, for even 𝑙, 𝐵𝑙 = 0, and for odd 𝑙

𝐵𝑙 = 𝑎2𝑙+1𝐴𝑙

Using Eq. 3.30 and 3.31, we obtain the potential in entire space:

Φ(𝑟, 𝜃) = 𝑉 {3

2(𝑟

𝑎)𝑃1(cos𝜃)−

7

8(𝑟

𝑎)3𝑃3(cos𝜃)+

11

16(𝑟

𝑎)5𝑃5(cos𝜃)⋯, for 𝑟<𝑎

3

2(𝑎

𝑟)2𝑃1(cos𝜃)−

7

8(𝑎

𝑟)4𝑃3(cos 𝜃)+

11

16(𝑎

𝑟)6𝑃5(cos𝜃)⋯, for 𝑟>𝑎

z

x

y

+V

-V

Fig 3.1. Conducting sphere of radius a

made up of two hemispherical shells

separated by a thin insulating ring. They

are kept at different potentials, +V and –V.

(3.29)

(3.30)

(3.31)

(3.32)

6

Metal sphere in a uniform electric field

An uncharged metal sphere of radius 𝑎 is placed in an

otherwise uniform electric field 𝐄 = 𝐸0𝐞𝑧 as shown in Fig. 3.2.

The potential is given by

Φ(𝑟, 𝜃) ≅ −𝐸0𝑧 = −𝐸0𝑟 cos𝜃

at large distances from the ball, where Φ = 0 in the

equatorial plane at 𝑧 = 0. Accordingly, the boundary

condition at the surface is Φ(𝑎, 𝜃) = 0. Referring to the

general solution Eq. 3.26, we can immediately set all 𝐴𝑙

except for 𝐴1(= −𝐸0) equal to zero. At 𝑟 = 𝑎, we have

Φ(𝑎, 𝜃) = 0 = −𝐸0𝑎 cos 𝜃 +∑𝐵𝑙𝑎𝑙+1

𝑃𝑙(cos 𝜃)

∞

𝑙=0

We can determine 𝐵𝑙 from this equation: 𝐵1 = 𝐸0𝑎3 and 𝐵𝑙 = 0 for 𝑙 ≠ 1. We finally have

Φ(𝑟, 𝜃) = −𝐸0 (𝑟 −𝑎3

𝑟2) cos 𝜃

The (normal) electric field at the surface is

𝐸𝑟 = −(𝜕Φ

𝜕𝑟)𝑟=𝑎

= 3𝐸0 cos 𝜃

The surface charge density is

𝜎(𝜃) = 휀0𝐸𝑟 = 3휀0𝐸0 cos𝜃

Point charge on the z-axis

The potential at x due to a unit point charge at x’ can be expressed as an important expansion:

1

|𝐱−𝐱′|=∑

𝑟<𝑙

𝑟>𝑙+1 𝑃𝑙(cos 𝛾)

∞

𝑙=0

Where 𝑟< and 𝑟> are the smaller and larger of 𝑟 and 𝑟′, respectively, and 𝛾 is the angle between

x and x’. We can prove this equation by aligning x’ along the z-axis as shown in Fig. 3.3. This

system has azimuthal symmetry, thus we can expand the potential as

1

|𝐱−𝐱′|= ∑[𝐴𝑙𝑟

𝑙 + 𝐵𝑙𝑟−(𝑙+1)] 𝑃𝑙(cos𝜃)

∞

𝑙=0

(3.33)

Fig 3.2.

(3.34)

(3.35)

(3.36)

(3.37)

(3.38)

(3.39)

z

x

y

+ +++ ++

- -- --

a

7

Fig. 3.3

If the point x is on the z-axis,

1

|𝐱−𝐱′|=

1

|𝑟−𝑟′|=1

𝑟>∑(

𝑟<𝑟>)𝑙

=∑[𝐴𝑙𝑟𝑙 + 𝐵𝑙𝑟

−(𝑙+1)]

∞

𝑙=0

∞

𝑙=0

Therefore, we can write

1

|𝐱−𝐱′|=∑

𝑟<𝑙

𝑟>𝑙+1 𝑃𝑙(cos 𝜃)

∞

𝑙=0

In fact, this equation is independent of coordinate system because 𝜃 is the angle between x and

x’, i.e., 𝜃 = 𝛾. This proves the general result of Eq. 3.38.

Charged circular ring

The potential on the z-axis due to a charged ring shown in Fig. 3.4 is

z

x

y

x’ xq

f

),,( fqrP

)',0','(' fq rP

x-x’

z

x

y

),0,( fq rP

bc

z=r

aq

dla

qdq

2

dl

(3.40)

(3.41)

Fig 3.4. Charged ring of radius 𝑎

and total charge of 𝑞 located on

the z-axis with center at 𝑧 = 𝑏.

8

Φ(𝑧 = 𝑟) =1

4𝜋휀0∫

𝑑𝑞

(𝑟2 + 𝑐2 − 2𝑐𝑟 cos 𝛼)1/2=

𝑞

4𝜋휀0

1

(𝑟2 + 𝑐2 − 2𝑐𝑟 cos 𝛼)1/2

(3.42)

where 𝑐2 = 𝑎2 + 𝑏2 and tan 𝛼 =𝑎

𝑏. We can expand Eq. 3.42 using Eq. 3.38.

Φ(𝑧 = 𝑟) =𝑞

4𝜋휀0{∑

𝑐𝑙

𝑟𝑙+1𝑃𝑙(cos 𝛼)

∞𝑙=0 , for 𝑟 > 𝑐

∑𝑟𝑙

𝑐𝑙+1𝑃𝑙(cos 𝛼)

∞𝑙=0 , for 𝑟 < 𝑐

Thus the potential at any point in space is

Φ(𝑟, 𝜃) =𝑞

4𝜋휀0{∑

𝑐𝑙

𝑟𝑙+1𝑃𝑙(cos 𝛼)𝑃𝑙(cos 𝜃)

∞𝑙=0 , for 𝑟 > 𝑐

∑𝑟𝑙

𝑐𝑙+1𝑃𝑙(cos 𝛼)

∞𝑙=0 𝑃𝑙(cos 𝜃), for 𝑟 < 𝑐

3.3 Electric Fields Near a Sharp Point of Conductor

We discuss how electric fields behave near a sharp point of conductor. We consider a conical

conducting tip which possesses azimuthal symmetry as shown in Fig. 3.5.

The basic solution to the Laplace boundary-value problem of Fig. 3.5 is 𝐴𝑟𝜈𝑃𝜈(cos 𝜃), where the

order parameter 𝜈 is determined by the opening angle 𝛽. Since the potential must vanish at

𝜃 = 𝛽,

𝑃𝜈(cos𝛽) = 0

The general solution for the potential is expressed as

Φ(𝑟, 𝜃) = ∑ 𝐴𝑘𝑟𝜈𝑘𝑃𝜈𝑘(cos𝜃)

∞

𝑘=0

z

),,( fqrP

x

q

(3.43)

(3.44)

Fig 3.5. Conical conducting tip of

opening angle 𝛽 located at the

origin

(3.45)

(3.46)

9

The potential near the tip is approximately

Φ(𝑟, 𝜃) ≅ 𝐴0𝑟𝜈0𝑃𝜈0(cos𝜃)

where 𝜈0 is the smallest root of Eq. 3.45. The components of electric field and the surface-

charge density near the tip are

{

𝐸𝑟 = −

𝜕Φ

𝜕𝑟≅ −𝜈0𝐴0𝑟

𝜈0−1𝑃𝜈0(cos 𝜃)

𝐸𝜃 = −1

𝑟

𝜕Φ

𝜕𝜃≅ 𝐴0𝑟

𝜈0−1 sin 𝜃 𝑃𝜈0′ (cos𝜃)

𝜎(𝑟) = −휀0𝐸𝜃|𝜃=𝛽 ≅ −휀0𝐴0𝑟𝜈0−1 sin 𝛽 𝑃𝜈0

′ (cos𝛽)

The field and charge density all vary as 𝑟𝜈0−1 as 𝑟 → 0, therefore they are singular at 𝑟 = 0 for

𝜈0 < 1. When the tip is sharp, i.e., 𝛽 → 𝜋, an approximate expression for 𝜈0 as a function 𝛽 is

𝜈0 ≅ [2 ln (2

𝜋 − 𝛽)]−1

Figure 3.6 show 𝜈0as a function of 𝛽 for 150∘ < 𝛽 < 180∘. This indicates that the fields near a

sharp point of conductor vary as 𝑟−1+𝜖 where 𝜖 ≪ 1.

3.4 Laplace Equation in Cylindrical Coordinates

In cylindrical coordinates (𝜌, 𝜙, 𝑧), the Laplace equation takes the form:

1

𝜌

𝜕

𝜕𝜌(𝜌𝜕Φ

𝜕𝜌) +

1

𝜌2𝜕2Φ

𝜕𝜙2+𝜕2Φ

𝜕𝑧2= 0

Separating the variables by making the substitution

Φ(𝜌, 𝜙, 𝑧) = 𝑅(𝜌)𝑄(𝜙)𝑍(𝑧)

155 160 165 170 175 180

0.05

0.10

0.15

0.20

0.25

0.30

0.35

(degrees)

0

(3.47)

(3.48)

Fig 3.6. The order parameter 𝜈0

as a function of the opening

angle 𝛽 for 𝛽 → 𝜋

(3.49)

(3.51)

(3.50)

10

Then we obtain the three ordinary differential equations:

𝑑2𝑍

𝑑𝑧2− 𝑘2𝑍 = 0

𝑑2𝑄

𝑑𝜙2+ 𝜈2𝑄 = 0

𝑑2𝑅

𝑑𝜌2+

1

𝜌

𝑑𝑅

𝑑𝜌+ (𝑘2 −

𝑣2

𝜌2)𝑅 = 0

The solutions of the first two equations are easily obtained:

𝑍(𝑧) = 𝑒±𝑘𝑧 and 𝑄(𝜙) = 𝑒±𝑖𝑚𝜙

When the full azimuthal angle is allowed, i.e., 0 ≤ 𝜙 ≤ 2𝜋, 𝑚 must be an integer. We rewrite

the radial equation (Eq. 3.54) by chaning the variable 𝑥 = 𝑘𝜌.

𝑑𝑅

𝑑𝑥2+1

𝑥

𝑑𝑅

𝑑𝑥+ (1 −

𝑚2

𝑥2)𝑅 = 0

The solutioins to this equation are best rexpressed as a power series in 𝑥. There are two

independet solutions, 𝐽𝑚(𝑥) and 𝑁𝑚(𝑥), called Bessel functions of the first kind and Neumann

functions, respectively. The Bessel function is defined as

𝐽𝑚(𝑥) = (𝑥

2)𝑚

∑(−1)𝑗

𝑗! (𝑚 + 𝑗)!(𝑥

2)2𝑗

∞

𝑗=0

The limiting forms of 𝐽𝑚(𝑥) and 𝑁𝑚(𝑥) for small and large 𝑥 are usuful to analyze the physical

properties of the given bounary-value problem.

For 𝑥 ≪ 1 𝐽𝑚(𝑥) ≅1

𝑚!(𝑥

2)𝑚

𝑁𝑚(𝑥) ≅ {

2

𝜋[ln (

𝑥

2) + 0.5772⋯ ] , 𝑚 = 0

−(𝑚−1)!

𝜋(2

𝑥)𝑚, 𝑚 ≠ 0

For 𝑥 ≫ 1 𝐽𝑚(𝑥) ≅ √2

𝜋𝑥cos (𝑥 −

𝑚𝜋

2−𝜋

4)

𝑁𝑚(𝑥) ≅ √2

𝜋𝑥sin (𝑥 −

𝑚𝜋

2−𝜋

4)

(3.52)

(3.53)

(3.54)

(3.55)

)

(3.56)

(3.57)

(3.58)

(3.59)

(3.60)

(3.61)

11

Fig. 3.7. First few Bessel functions and Neumann functions

The roots of Bessel functions are important for many boundary-value problems:

𝐽𝑚(𝑥𝑚𝑛) = 0, 𝑛 = 1,2,3,…

𝑥𝑚𝑛 is the 𝑛th root of 𝐽𝑚(𝑥):

𝑚 = 0, 𝑥0𝑛 = 2.405, 5.520, 8.654,… 𝑚 = 1, 𝑥1𝑛 = 3.832, 7.016, 10.173,…𝑚 = 2, 𝑥2𝑛 = 5.136, 8.417, 11.620,…

Orthogonal complete set of functions

√𝜌 𝐽𝑚 (𝑥𝑚𝑛𝜌

𝑎) for fixed 𝑚 and 𝑛 = 1,2,3, … form an orthogonal set on the interval 0 ≤ 𝜌 ≤ 𝑎.

The normalization integral is

∫ 𝜌𝐽𝑚 (𝑥𝑚𝑛′𝜌

𝑎) 𝐽𝑚 (𝑥𝑚𝑛

𝜌

𝑎)

𝑎

0

𝑑𝜌 =𝑎2

2[𝐽𝑚+1(𝑥𝑚𝑛)]

2𝛿𝑛𝑛′

Cylindrical Cavity

We consider a cylinder of radius 𝑎 and height 𝐿 as shown

in Fig. 3.8. The potential on the side and the bottom is

zero, while the top has a potential Φ = 𝑉(𝜌, 𝜙). We want

to find the potential at any point inside the cylinder.

Boundary conditions

Φ = 0 at 𝑧 = 0 leads to 𝑍(𝑧) = sinh 𝑘𝑧.

Φ = 0 at 𝜌 = 𝑎, i.e., 𝐽𝑚(𝑘𝑚𝑛𝑎) = 0,

therefore, 𝑘𝑚𝑛 =𝑥𝑚𝑛

𝑎 (see, Eq. 3.62)

2 4 6 8 10

0.4

0.2

0.2

0.4

0.6

0.8

1.0

2 4 6 8 10

1.0

0.5

0.5)(0 xJ

)(1 xJ)(2 xJ

)(0 xN)(1 xN )(2 xN

(a) Bessel functions (a) Neumann functions

(3.62)

Fig 3.8.

(3.63)

z

x

y

La

0

0

),( fV

12

Then, the general solution can be expressed as

Φ(𝜌, 𝜙, 𝑧) = ∑ ∑𝐽𝑚(𝑘𝑚𝑛𝜌) sinh(𝑘𝑚𝑛𝑧) (𝐴𝑚𝑛 sin𝑚𝜙 + 𝐵𝑚𝑛 cos𝑚𝜙)

∞

𝑛=1

∞

𝑚=0

(3.64)

At 𝑧 = 𝐿, we have Φ = 𝑉(𝜌, 𝜙), thus

V(𝜌, 𝜙) = ∑ ∑sinh(𝑘𝑚𝑛𝐿) 𝐽𝑚(𝑘𝑚𝑛𝜌)(𝐴𝑚𝑛 sin𝑚𝜙 + 𝐵𝑚𝑛 cos𝑚𝜙)

∞

𝑛=1

∞

𝑚=0

(3.65)

We can determine the coefficients 𝐴𝑚𝑛 and 𝐵𝑚𝑛 using the orthogol relations of the sinusoidal

and Bessel functions (see Eq. 3.63).

{𝐴𝑚𝑛𝐵𝑚𝑛

} =2 csch(𝑘𝑚𝑛𝐿)

𝜋𝑎2𝐽𝑚+12 (𝑘𝑚𝑛𝑎)

∫ 𝑑𝜙2𝜋

0

∫ 𝜌𝑑𝜌𝑎

0

𝑉(𝜌, 𝜙)𝐽𝑚(𝑘𝑚𝑛𝜌) {sin𝑚𝜙

cos𝑚𝜙}

for 𝑚 = 0, we use 1

2𝐵0𝑛 in the series. (3.66)

3.5 Poisson Equation and Green Functions in Spherical Coordinates

Addition thorem for spherical harmonics

Fig 3.9.

The potential at x (x’) due to a unit point charge at x’ (x) is an exceedingly important physical

quantity in electrostatics. When the two coordinate vectors x and x’ have an angle 𝛾 between

them, it can be expressed as an important expansion (see Eq. 3.38. We proved this equation in

Section 3.2.):

z

x

q

'q

x’ ),,( fqrP

)',','(' fqrP

x

yf

'f

x-x’

13

1

|𝐱−𝐱′|= ∑

𝑟<𝑙

𝑟>𝑙+1 𝑃𝑙(cos 𝛾)

∞𝑙=0

where 𝑟< and 𝑟> are the smaller and larger of 𝑟 and 𝑟′, respectively. It is of great interest and

use to express this equation in spherical coordiantes. The addition theorem of spherical

harmonics is an useful mathematical result for this purpose.

The addition theorem expresses a Legendre polynomial of order 𝑙 in the angle 𝛾 In spherical

coordiates as shown in Fig. 3.9:

𝑃𝑙(cos 𝛾) =4𝜋

2𝑙 + 1∑ 𝑌𝑙𝑚

∗ (𝜃′, 𝜙′)𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

Combining Eq. 3.32 and 3.67, we obtain a completely factorized form in the coordinates x and

x’.

1

|𝐱−𝐱′|= ∑

4𝜋

2𝑙 + 1

𝑟<𝑙

𝑟>𝑙+1 ∑ 𝑌𝑙𝑚

∗ (𝜃′, 𝜙′)𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

We immediately see that Eq. 3.68 is the expansion of the Green function in spherical

coordinates for the case of no boundary surfaces, except at infinity.

Green function with a spherical boundary

The Green function appropriate for Dirichlet boundary conditions on the sphere of radius a

satisfies the equation (see Eq. 1.27)

∇2𝐺(𝐱, 𝐱′) = −4𝜋𝛿(𝐱 − 𝐱′) and is expressed as (see Eq. 2.13 and 2.14)

𝐺(𝐱, 𝐱′) = 𝐺(𝐱′, 𝐱) =1

|𝐱−𝐱′|+ 𝐹(𝐱, 𝐱′) (2.13)

where ∇′2𝐹(𝐱, 𝐱′) = 0 for 𝑟′ > 𝑎 and 𝐺(𝐱, 𝐱′) = 0 for 𝑟 = 𝑟′ = 𝑎. The discussion of the

conducting sphere with the method of images indicates that the Green function can take the

form

𝐺(𝐱, 𝐱′) =1

|𝐱− 𝐱′|−

𝑎

𝑟′ |𝐱 −𝑎2

𝑟′2𝐱′|

Using Eq. 3.68 we rewrite Eq. 2.14 as

𝐺(𝐱, 𝐱′) =∑4𝜋

2𝑙 + 1[𝑟<𝑙

𝑟>𝑙+1−1𝑎(𝑎2

𝑟𝑟′)

𝑙+1

] ∑ 𝑌𝑙𝑚∗(𝜃′, 𝜙′)𝑌𝑙𝑚(𝜃,𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

(3.38)

(3.67)

1)

(3.68)

(2.14)

(3.69)

(1.27)

14

The radial factors inside and outside the sphere can be separately expressed as

𝑟<𝑙

𝑟>𝑙+1 −

1

𝑎(𝑎2

𝑟𝑟′)

𝑙+1

=

{

1

𝑟′𝑙+1(𝑟𝑙 −

𝑎2𝑙+1

𝑟𝑙+1) , 𝑟 < 𝑟′

1

𝑟𝑙+1(𝑟′𝑙 −

𝑎2𝑙+1

𝑟′𝑙+1) , 𝑟 > 𝑟′

General structure of Green function in spherical coordinates

The Green function appropriate for Dirichlet boundary conditions satisfies the equation (see Eq.

1.27)

∇2𝐺(𝐱, 𝐱′) = −4𝜋𝛿(𝐱 − 𝐱′)

In spherical coordinates the delta function can be written

𝛿(𝐱 − 𝐱′) =1

𝑟2𝛿(𝑟 − 𝑟′)𝛿(𝜙 − 𝜙′)𝛿(cos 𝜃 − cos 𝜃′)

Using the completeness relation for spherical harmonics (Eq. 3.24), we obtain

𝛿(𝐱 − 𝐱′) =1

𝑟2𝛿(𝑟 − 𝑟′)∑ ∑ 𝑌𝑙𝑚

∗ (𝜃′, 𝜙′) 𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

Equation 1.27 and 3.72 leads to the expansion of the Green function

𝐺(𝐱, 𝐱′) =∑ ∑ 𝑔𝑙(𝑟, 𝑟′)𝑌𝑙𝑚

∗ (𝜃′, 𝜙′) 𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

and the equation for the radial Green function

1

𝑟

𝑑2

𝑑𝑟2[𝑟𝑔𝑙(𝑟, 𝑟

′)] −𝑙(𝑙 + 1)

𝑟2𝑔𝑙(𝑟, 𝑟

′) = −4𝜋

𝑟2𝛿(𝑟 − 𝑟′)

The general solution of this equation for 𝑟 ≠ 𝑟′ (see Eq. 3.5) can be written as

𝑔𝑙(𝑟, 𝑟′) = {

𝐴(𝑟′)𝑟𝑙 +𝐵(𝑟′)

𝑟𝑙+1 for 𝑟 < 𝑟′

𝐴′(𝑟′)𝑟𝑙 +𝐵′(𝑟′)

𝑟𝑙+1 for 𝑟 > 𝑟′

The coefficients A, B, A’, B’ are functions of r’ to be determined by the boundary conditions, the

discontinuity at 𝑟 = 𝑟′, and the symmetry of 𝑔𝑙(𝑟, 𝑟′) = 𝑔𝑙(𝑟

′, 𝑟).

(3.70)

(3.71)

(3.72)

(1.27)

(3.73)

(3.75)

(3.74)

15

Green function with noboundary

For the case of no boundary, 𝑔𝑙 must be finite for 𝑟 → 0 and ∞, therefore 𝐵 and 𝐴′ are zero.

𝑔𝑙(𝑟, 𝑟′) = {

𝐴(𝑟′)𝑟𝑙 for 𝑟 < 𝑟′𝐵′(𝑟′)

𝑟𝑙+1 for 𝑟 > 𝑟′

Then, the symmetry of of 𝑔𝑙(𝑟, 𝑟′) = 𝑔𝑙(𝑟

′, 𝑟) leads to

𝑔𝑙(𝑟, 𝑟′) = 𝑔𝑙(𝑟

′, 𝑟) = 𝐶𝑟<𝑙

𝑟>𝑙+1

where 𝑟< and 𝑟> are the smaller and larger of 𝑟 and 𝑟′, respectively. We determine the constant

𝐶 using the discontinuity at 𝑟 = 𝑟′. Integrating the radial equation 3.74 over the infinitesimally

narrow interval from 𝑟 = 𝑟′ − 𝜖 to 𝑟 = 𝑟′ + 𝜖 with very small 𝜖, we obtain

𝑑

𝑑𝑟[𝑟𝑔𝑙(𝑟, 𝑟

′)]|𝑟′−𝜖

𝑟′+𝜖

= −4𝜋

𝑟′

Substituning Eq. 3.77 into Eq. 3.78, we find

−𝐶(𝑙 + 1)

𝑟′−𝐶𝑙

𝑟′= −

4𝜋

𝑟′ → 𝐶 =

4𝜋2𝑙 + 1

→ 𝑔𝑙(𝑟, 𝑟′) =

4𝜋2𝑙 + 1

𝑟<𝑙

𝑟>𝑙+1

This reduces to Eq. 3.68

𝐺(𝐱, 𝐱′) =1

|𝐱−𝐱′|= ∑

4𝜋

2𝑙 + 1

𝑟<𝑙

𝑟>𝑙+1 ∑ 𝑌𝑙𝑚

∗ (𝜃′, 𝜙′)𝑌𝑙𝑚(𝜃, 𝜙)

𝑙

𝑚=−𝑙

∞

𝑙=0

Green function with two concentric spheres

Suppose that the boundary surfaces are concentric spheres at 𝑟 = 𝑎 and 𝑟 = 𝑏. 𝐺(𝐱, 𝐱′) = 0 on

the surfaces gives rise to

𝑔𝑙(𝑟, 𝑟′) =

{

𝐴(𝑟𝑙 −

𝑎2𝑙+1

𝑟𝑙+1) for 𝑟 < 𝑟′

𝐵′ (1

𝑟𝑙+1−

𝑟𝑙

𝑏2𝑙+1) for 𝑟 > 𝑟′

The symmetry of of 𝑔𝑙(𝑟, 𝑟′) = 𝑔𝑙(𝑟

′, 𝑟) requires

𝑔𝑙(𝑟, 𝑟′) = 𝐶 (𝑟<

𝑙 −𝑎2𝑙+1

𝑟<𝑙+1 )(

1

𝑟>𝑙+1 −

𝑟>𝑙

𝑏2𝑙+1)

(3.76)

(3.77)

(3.78)

(3.79)

(3.68)

(3.80)

(3.81)

16

Using the discontinuity at 𝑟 = 𝑟′ (Eq. 3.78), we obtain the constant 𝐶:

𝐶 =4𝜋

(2𝑙+1)[1−(𝑎

𝑏)2𝑙+1

]

Combining Eq. 3.73, 3.81, and 3.82, we find the expansion of the Green function

𝐺(𝐱, 𝐱′) =∑ ∑4𝜋𝑌𝑙𝑚

∗ (𝜃′, 𝜙′)𝑌𝑙𝑚(𝜃, 𝜙)

(2𝑙 + 1) [1 − (𝑎

𝑏)2𝑙+1

](𝑟<

𝑙 −𝑎2𝑙+1

𝑟<𝑙+1 )(

1

𝑟>𝑙+1 −

𝑟>𝑙

𝑏2𝑙+1)

𝑙

𝑚=−𝑙

∞

𝑙=0

(3.83)

Charged ring inside a grounded sphere

We consider a spherical cavity of radius 𝑏 with a concnetric righ of charge of radius 𝑎 and total

charge 𝑄 as shown in Fig. 3.10. The charge density can be written as

𝜌(𝐱′) =𝑄

2𝜋𝑎2𝛿(𝑟′ − 𝑎)𝛿(cos𝜃′)

The general solution of the Poisson equation (Eq. 1.31) is given as

Φ(𝐱) =1

4𝜋휀0∫𝜌(𝐱′)𝑉

𝐺(𝐱, 𝐱′)𝑑3𝑥′ −1

4𝜋∫ [Φ(𝐱′)

𝜕𝐺(𝐱, 𝐱′)

𝜕𝑛′]

𝑆

𝑑𝑎′

Since Φ = 0 on the sphere

Φ(𝐱) =1

4𝜋휀0∫𝜌(𝐱′)𝑉

𝐺(𝐱, 𝐱′)𝑑3𝑥′

Using Eq. 3.83 with 𝑎 → 0,

Φ(𝐱) =𝑄

4𝜋휀0∑ 𝑟<

𝑙 (1

𝑟>𝑙+1 −

𝑟>𝑙

𝑏2𝑙+1)𝑃𝑙(0)𝑃𝑙(cos𝜃)

∞

𝑙=0

z

x

y

0

Q

a

b

(3.82)

Fig 3.10. Charged ring of radius 𝑎 and

total charge Q inside a grounded,

conducting sphere of radius b.

(1.31)

(3.84)

(3.84)

17

=𝑄

4𝜋휀0∑

(−1)𝑛(2𝑛 − 1)‼

2𝑛𝑛! 𝑟<2𝑛 (

1

𝑟>2𝑛+1 −

𝑟>2𝑛

𝑏4𝑛+1)𝑃2𝑛(cos 𝜃)

∞

𝑛=0

where 𝑟< and 𝑟> are the smaller and larger of 𝑟 and 𝑎, respectively.

Uniform line charge inside a grounded sphere

We consider a spherical cavity of radius 𝑏 with a uniform line charge of length 2𝑏 and total

charge 𝑄 as shown in Fig. 3.11. The charge density can be written as

𝜌(𝐱′) =𝑄

2𝑏

1

2𝜋𝑟′2 [𝛿(cos 𝜃′ − 1) + 𝛿(cos𝜃′ + 1)]

Using Eq. 3.84 and 𝑃𝑙(−1) = (−1)𝑙, we obtain the solution

Φ(𝐱) =𝑄

8𝜋휀0𝑏∑[1 + (−1)𝑙]𝑃𝑙(cos𝜃)∫ 𝑟<

𝑙 (1

𝑟>𝑙+1 −

𝑟>𝑙

𝑏2𝑙+1)

𝑏

0

𝑑𝑟′∞

𝑙=0

(3.87)

The integral must be broken up into two intervals

∫ 𝑟<𝑙 (

1

𝑟>𝑙+1 −

𝑟>𝑙

𝑏2𝑙+1)

𝑏

0𝑑𝑟′ = (

1

𝑟𝑙+1−

𝑟𝑙

𝑏2𝑙+1) ∫ 𝑟′𝑙

𝑟

0𝑑𝑟′ + 𝑟𝑙 ∫ (

1

𝑟′𝑙+1−

𝑟′𝑙

𝑏2𝑙+1)

𝑏

𝑟𝑑𝑟′

=(2𝑙+1)

𝑙(𝑙+1)[1 − (

𝑟

𝑏)𝑙

] for 𝑙 ≥ 1 (3.88)

For 𝑙 = 0, this result is indeterminate, and hence we use its limiting behavior

∫ (1

𝑟>−

1

𝑏)

𝑏

0𝑑𝑟′ = ∫ (

1

𝑟−

1

𝑏)

𝑟

0𝑑𝑟′ + ∫ (

1

𝑟′−

1

𝑏)

𝑏

𝑟𝑑𝑟′

= 1 −𝑟

𝑏+ ln (

𝑏

𝑟) − (1 −

𝑟

𝑏) = ln (

𝑏

𝑟) for 𝑙 = 0 (3.89)

z

x

y

0

Qb

(3.85)

Fig 3.11. Uniform line charge of

length 2𝑏 total charge Q inside a

grounded, conducting sphere of

radius b.

(3.86)

18

Substituting Eq. 3.88 and 3.98 into Eq. 3.87, we find

Φ(𝐱) =𝑄

4𝜋𝜀0𝑏{ln (

𝑏

𝑟) + ∑

(4𝑛+1)

2𝑛(2𝑛+1)[1 − (

𝑟

𝑏)2𝑛

] 𝑃2𝑛(cos 𝜃)∞𝑛=1 } (3.90)

The surface charge density on the grounded sphere is

σ(θ) = ε0∂Φ

𝜕𝑟|𝑟=𝑏

= −𝑄

4𝜋𝑏2[1 +∑

(4𝑛 + 1)

(2𝑛 + 1)𝑃2𝑛(cos 𝜃)

∞

𝑛=1

]

3.6 Conducting Plane with a Circular Hole

We consider a grounded conducting plane in which a circular hole is punctured. Figure 3.12

illustrates the geometry. The electric field far from hole has only a z component: 𝐸𝑧 = −𝐸0 for

𝑧 > 0 and 𝐸𝑧 = 0 for 𝑧 < 0. From the limiting behavior of the field, we can deduce that the

potential takes the form

Φ = {𝐸0𝑧 + Φ

(1), 𝑧 > 0

Φ(1), 𝑧 < 0

where Φ(1) is the potential due to a rearrangement of surface charge near the hole, 𝜎(1).

Because of the azimuthal symmetry, Φ(1) is independent of 𝜙, and it can be expressed as

Φ(1)(𝜌, 𝑧) =1

4𝜋휀0∬

𝜎(1)(𝜌′)𝜌′𝑑𝜌′𝑑𝜙′

√𝜌2 + 𝜌′2 − 2𝜌𝜌′ cos𝜙′ + 𝑧2

z

x

y

0

a

0E),,( zP f

)0,','(' fP

x

x’

x-x’

)'(

(3.91)

Fig 3.12. A circular hole of radius 𝑎 is

punctured in a grounded conducting

plane. The limiting behavior of the

electric field far from the hole is

𝐄 = {−𝐸0𝐞𝑧, 𝑧 > 0

0, 𝑧 < 0

(3.92)

(3.93)

19

Mixed boundary conditions

Φ(1) is even in z, and hence 𝐸𝑧(1)

is odd. Since the total z component of electric field must be

continuous across 𝑧 = 0 in the hole, we must have (for 𝜌 < 𝑎)

−𝐸0 + 𝐸𝑧(1)|𝑧=0+

= 𝐸𝑧(1)|𝑧=0−

Since 𝐸𝑧(1)

is odd,

𝐸𝑧(1)|𝑧=0+

= −𝐸𝑧(1)|𝑧=0−

=1

2𝐸0

This relation and the ground potential of the conducting surface complete the boundary

conditions for the entire xy plane. We therefore have mixed boundary conditions

{

𝜕Φ(1)

∂z|𝑧=0+

= −1

2𝐸0, 0 ≤ 𝜌 < 𝑎

Φ(1)|𝑧=0

= 0 , 𝑎 ≤ 𝜌 < ∞

Solution of Laplace equation in cylindrical coordinates

The general solution of the Laplace equation in cylindrical coordiates (see Eq. 3.51, 3.55 and

3.57) is

Φ = 𝑅(𝜌)𝑄(𝜙)𝑍(𝑧) = 𝐽𝑚(𝑘𝜌)𝑒±𝑖𝑚𝜙𝑒±𝑘𝑧

Because of the azimuthal symmetry, Φ(1) is independent of 𝜙, and hence 𝑚 = 0. Therefore,

Φ(1) can be written in terms of cylindrical coordinates

Φ(1)(𝜌, 𝑧) = ∫ 𝐴(𝑘)𝑒−𝑘|𝑧|𝐽0(𝑘𝜌)∞

0

𝑑𝑘

Boundary conditions and integral equatoins

The boundary conditions (Eq. 3.96) for the general solution (Eq. 3.98) give rise to the integral

equatoins of the coefficient 𝐴(𝑘):

{

∫ 𝑘𝐴(𝑘)𝐽0(𝑘𝜌)

∞

0

𝑑𝑘 =1

2𝐸0, 0 ≤ 𝜌 < 𝑎

∫ 𝐴(𝑘)𝐽0(𝑘𝜌)∞

0

𝑑𝑘 = 0 , 𝑎 ≤ 𝜌 < ∞

There exists an analytic solution of these interal equations.

𝐴(𝑘) =𝐸0𝑎

2

𝜋𝑗1(𝑘𝑎) =

𝐸0𝑎2

𝜋(sin 𝑘𝑎

𝑘2𝑎2−cos 𝑘𝑎

𝑘𝑎)

where 𝑗1(𝑥) is the spherical Bessel function of order 1.

(3.94)

(3.95)

(3.96)

(3.97)

(3.98)

(3.99)

(3.100)

20

Multipole expansion in the far-field region

In the far-feld region, i.e., in the region for |𝑧| and/or 𝜌 ≫ 𝑎, the integral in Eq. 3.98 is mainly

determined by the contributions around 𝑘 = 0, more precisely, for 𝑘 ≪1

𝑎. The expansion of

𝐴(𝑘) for small 𝑘𝑎 takes the form

𝐴(𝑘) ≅𝐸0𝑎

2

3𝜋[𝑘𝑎 −

(𝑘𝑎)3

10+⋯]

The leading term gives rise to the asymptotic potential

Φ(1)(𝜌, 𝑧) ≅𝐸0𝑎

3

3𝜋∫ 𝑘𝑒−𝑘|𝑧|𝐽0(𝑘𝜌)∞

0𝑑𝑘

= −𝐸0𝑎

3

3𝜋

𝑑

𝑑|𝑧|∫ 𝑒−𝑘|𝑧|𝐽0(𝑘𝜌)∞

0𝑑𝑘

= −𝐸0𝑎

3

3𝜋

𝑑

𝑑|𝑧|(

1

√𝜌2+𝑧2) =

𝐸0𝑎3

3𝜋

|𝑧|

(𝜌2+𝑧2)32

Here we have the asymptotic potential

Φ(1)(𝜌, 𝑧) ≅𝐸0𝑎

3

3𝜋

|𝑧|

𝑟3

falling off with distance as 𝑟−2 and having an effective electric dipole moment,

𝐩 = ∓4

3휀0𝐄0𝑎

3

where – for 𝑧 > 0 and + for 𝑧 < 0.

Potential in the near-field region

The potential in the neighborhood of the hole must be calculated from the exact expression

Φ(1)(𝜌, 𝑧) =𝐸0𝑎

2

𝜋∫ 𝑗1(𝑘𝑎)𝑒

−𝑘|𝑧|𝐽0(𝑘𝜌)∞

0

𝑑𝑘

An integration by parts and Laplace transforms results in

Φ(1)(𝜌, 𝑧) =𝐸0𝑎

𝜋[√𝑅 − 𝜆

2−|𝑧|

𝑎tan−1 (√

2

𝑅 + 𝜆)]

where 𝜆 =1

𝑎2(𝑧2 + 𝜌2 − 𝑎2), 𝑅 = √𝜆2 +

4𝑧2

𝑎2 (3.106)

(3.101)

(3.102)

(3.103)

(3.104)

(3.105)

21

For 𝜌 = 0,

Φ(1)(0, 𝑧) =𝐸0𝑎

𝜋[1 −

|𝑧|

𝑎tan−1 (

𝑎

|𝑧|)]

For |𝑧| ≫ 𝑎,

Φ(1)(0, 𝑧) ≅𝐸0𝑎

𝜋{1 −

|𝑧|

𝑎[𝑎

|𝑧|−1

3(𝑎

|𝑧|)3

+⋯]} =𝐸0𝑎

3

3𝜋

1

|𝑧|2

This is consistent with the dipole approximation in Eq. 3.102.

In the plane of the opening (𝑧 = 0),

Φ(1)(𝜌, 0) =𝐸0𝜋√𝑎2 − 𝜌2 for 0 ≤ 𝜌 < 𝑎

The tangential electric field in the opening is a radial field,

𝐄∥(𝜌, 0) =𝐸0𝜋

𝛒

√𝑎2 − 𝜌2

The boundary condition in Eq. 3.96 indicates the normal component of electric field in

the opening,

𝐸𝑧(𝜌, 0) = −1

2𝐸0

az /

2 1 1 2

0.05

0.10

0.15

0.20

0.25

0.30

aE

z

0

)1(),0(

az /

a

z

a

z

ax /

aE

z

0

)1(),(

aE

z

0

),(

(3.107)

(3.108)

(3.109)

(3.110)

(3.111)

Fig 3.13. Equipotential

contours of the additional

potential Φ(1) and the total

potential Φ in the vicinity of

the hole. Φ(1)(0, 𝑧) as a

function of z is also shown.