images and green's functions

8/13/2019 Images and Green's Functions

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Chapter 5

IMAGES AND GREENS

FUNCTIONS

5.1 Introduction

The potential due to a point charge qplaced above a grounded conducting plane can be found as

a sum of two potentials, one due to the charge itself, and the other due to an image charge, q,placed at the mirror point with respect to the plane. This simple example indicates a possibility

of greatly simplifying boundary value problems by introducing appropriate images for a given

boundary shape.

In this Chapter, image charges for simple boundary shapes will be discussed. Finding an

appropriate image enables us to derive a formula for the potential due to a prescribed boundary

potential on the surface without necessarily going through the methods developed in Chapter 3. Aswill be shown, solving a boundary value problem is essentially identical to nding an appropriate

scalar function (Greens function) for a given boundary shape.

5.2 Image for Flat Surfaces

Consider the conguration shown in Figure 5.1. The grounded, wide, conducting plane can be

replaced by an imageqplaced at the mirror point of the real charge +qas long as the potentialand electric eld in the space above the plane are concerned. The potential above the plane is thus

given by

= ++ = 140

qr+

qr

(5.1)

wherer+ and rare the distances from the observing point to the respective charges. In Eq. (5.1),

the rst term in RHS is the contribution from the charge q. The potential + satises the singular

Poissons equation

r2+ = q0

(r r0) (5.2)

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image charge

conductor

q

-q

d

d

r

r

+

-

Figure 5-1: Image charge for a at conducting plate.

where r0= dez is the location of the charge q. The image charge qis below the plane. Therefore,

in the space above the plane, the potential due to the image charge should satisfy the Laplaceequaion without any singularities,

r2= 0 (z >0) (5.3)

In other words, the potential can be regarded as a general solution to the singular Poissons

equation in Eq. (5.2) which is added in order that the total potential satisfy the boundary condition,

= 0 on the plane (5.4)

Obviously, the solution+ is the particular solution to the Poissons equation. Note that we have a

freedom to add general solutions satisfying the Laplace equation to a particular solution satisfying

the Poissons equation.

Images for a grounded wedge can be similarly found. For example, if the wedge angle is 90,

three images appear as shown in Figure 5.2. The potential is given by the sum of four contributions

from each charge. The positive image charge is the image of the negative image charges.

The potential due to a charge qplaced above a dielectric body can also be worked out by images.

In this case, we may assume the potential in the air

> = 1

40

q

r1 q

0

r2

(5.5)

and the potential in the dielectric< =

1

40

q00

r3(5.6)

where the distances r1; r2 andr3 are indicated in Fig. 5.3, q0 is the image charge in the dielectric,

andq00 is another image located at the same position as the real charge q. Note that the potential

in the dielectric remains regular (no singularity). To determine q0 and q00, we rst impose the

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q

q

-q

-q

90 degree wedge

Figure 5-2: Images for 90 degree conductor wedge.

continuity of the potential at the boundary, r1= r2= r3. This yields

1

0(q q0) =1

q00 (5.7)

The second boundary condition is the continuity in the displacement vector, Dz (normal compo-

nent). Since

r1=p

x2 + (z d)2 (5.8)

we have@

@z1

r1=

z d[x2 + (z d)]3=2

(5.9)

Similarly,@

@z

1

r2

= z+ d

[x2 + (z+ d)2]3=2

@

@z

1

r3

z d

[x2 + (d z)2]3=2 (z


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q

q'

q''

r

r

r

1

2

3

0

(z < 0)

(z > 0)

Figure 5-3: Images q0 andq00 for a at dielectric boundary.

Solving Eqs. (5.7) and (5.13) for q0 andq00, we nd

q0 = 0 + 0

(5.14)

q00 = 2

+ 0q (5.15)

5.3 Image for Cylindrical Surfaces (Two Dimensional)

The potential due to a long line charge (C/m) placed parallel to a grounded conducting cylinder

can be found by the method of image. In Chapter 2, we saw that two opposite line charges of equal

magnitudes,+ and create a family of equipotential cylindrical surfaces,

(r+; r) =

20ln

rr+

(5.16)

where r+ and r are the distances to the respective line charges, + and. Consider a line

charge at a distance 0

from the axis of a cylinder having a radius a. A negative line chargeat a distance 00 =a2=1 and the positive line charge make the cylinder surface an equipotential

surface

s =

20ln

a

0

(5.17)

(Check this by considering the potentials at A and B in Figure 5.4.) In order to make the cylinder

surface at zero potential, we have to subtract this potential from the potential given in Eq. (5.16)

which is based on the choice = 0 on the midplane, r+ = r. Therefore, the solution to the

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potential becomes

(r) =

20

ln

rr+

ln

a

1

(5.18)

But

r+= 2 + 02 20 cos 1=2 (5.19)

r=

"2 +

a2

0

2 2 a

2

0 cos

#1=2(5.20)

Substituting these into Eq. (5.18), we nally obtain

(; ) =

20ln

26664

202

a2 + a2 20 cos

1=2(2 + 02 20 cos )1=2

37775 (5.21)

When= a, this indeed vanishes.The surface charge density induced on the cylinder surface is given by

= 0 @@

=a

= 2

21a a

02 + a2 2a0 cos (C/m2) (5.22)

Therefore, the total charge (per unit length) on the cylinder surface is

ql = 2

02 a2 Z 20

d02 + a2 2a0 cos (5.23)

The integral reduces to2

02 a2 (5.24)

when 1> a. Therefore, the charge per unit length of the cylinder is simply, as expected.

5.4 Image for a Sphere

We consider a charge qat a distance r0 from the center of a grounded conducting sphere of radius

a. The potential is symmetric about the line connecting the center and the charge. Therefore, an

image charge should be on the same line. We assume a negative charge q0 placed at a distance xfrom the center. The potential at the point A is

A = 1

40

q

r0 a q0

a x

(5.25)

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a

'''

r+

r-

Figure 5-4: A line charge at a distance 0 from the center of a conducting cylinder of radius aand its image line charge at 00 = a2=0 make the cylinder surface an equipotential surface ats =

2"0

ln (a=0) :

This should vanish if the image is to replace the sphere without aecting the potential outside the

sphere. Therefore,q

r0 a = q0

a x (5.26)

Similarly, the potential at B is given by

B = 1

40

q

D+ a q

0

a + x

= 0 (5.27)

q

r0

+ a

= q0

a + x

(5.28)

Solving Eqs. (5.26) and (5.28) for x and q0, we nd

q0 = a

r0q (5.29)

x=a2

r0 (5.30)

Therefore, a chargeaq=r0 placed at a distance a2=r0 from the center can be used to replace thesphere as far as the potential and electric eld outside the sphere are concerned. Outside the sphere,

the potential is given by

= q40

1r+

a=r0

r

(5.31)

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a

r

r''

r'

q-q

r+

r-

AB

Figure 5-5: Image for a sphere. Charge qat distance r 0 from the center of a sphere of radius a andits imageqatr 00 =a2=r0 make the potential vanish on the sphere surface r = a:

where r1 and r2 are the distances between the observing point and respective charges, and given,

in terms of the coordinate (r; ) at the observing point, by

r+ =

r2 + r02 2rr0 cos 1=2r =

"r2 +

a2

r0

2 2 a

2

r0r cos

#1=2(5.32)

The surface charge density on the sphere surface can be evaluated from

() = 0 @@r

r=a

= 1

4

a r

02

a

q

(a2 + r02 2ar0 cos )3=2 (5.33)

Therefore, the total charge residing on the outer surface of the sphere is given by the integral,

qs =

Z 0

()2a2 sin d

=

a r 02

a

a2 Z 1

1

1

(a2 + r02 2ar0)3=2 d (= cos )= a

r0q= q0

This result is expected from the Gauss law because the q0 is the only charge enclosed by thesphere surface.

Example

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A charge qis placed at a distance D from the center of an isolated (oating) conducting sphere of

a radius a. When the sphere carries no charges, what will the sphere potential be?

We rst make the sphere potential zero by placing a chargeq0 =aq=D at r0 =a2=D as inthe preceding discussion. Since the sphere carries no net charge, if a charge +q0 = +aq=D is placed

at the center of the sphere, the sphere potential will be raised (when q >0) to

s = 1

40

aq=D

a

= 1

40

q

D (5.34)

Note that this is independent of the sphere radius a. When the sphere carries a net charge Q, the

sphere potential becomes

s = 1

40

q

D+

Q

a

(5.35)

5.5 Greens Theorem and Greens Function

The method of image charges has a more important application in potential boundary value prob-

lems. When a potential is prescribed on a closed surface, it uniquely determines the potential in the

space surrounding the surface and also in the space surrounded by the surface. The most general

formulation for the potential is given in terms of a scalar function called Greens function, which

is closely related to the potential due to a charge and its image for a given surface shape.

For arbitrary scalar functions and, the following identity immediately follows from Gauss

theorem, Z r (r)dV = I r dS (5.36)

The LHS is equal to Zr r+ r2 dV (5.37)

Therefore, Zr r+ r2 dV = I r dS (5.38)

Exchanging and , we obtain

Zr r + r2 dV = I r dS (5.39)

Substracting Eq. (5.39) from Eq. (5.38), yields

Zr2 r2 dV = I (r r) dS (5.40)

This is known as the Greens theorem. Its usefulness in potential problems can be appreciated as

follows.

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In Eq. (5.40), we identify as the desired scalar potential (r), and as a solution to the

following singular Poissons equation

r2G= r r0 (5.41)Since the potential(r) satises Poissons equation,

r2 = 10

(5.42)

and Z (r0)(r r0)dV0 = (r) (5.43)

Eq. (5.40) yields,

(r) =1

0 Z G(r; r0)(r0)dV0 Is

(sr0G Gr0s) dS0 (5.44)

where s(r0) is the surface potential prescribed on S, and r0 indicates derivative with respect

to the surface coordinates r0. When there are no surfaces, and the potential is entirely due to a

prescribed charge density distribution , the Greens function is

G(r; r0) = 1

4

1

jr r0j (5.45)

and the potential is given by the familiar form,

(r) = 1

40Z (r0)jr r0jdV

0 (5.46)

In the absence of charges (= 0), the potential is determined by the surface potential s and

its derivative, rs, which is of course related to the electric eld on the surface.

The Greens function in Eq. (5.45) is essentially the potential due to a charge located at r0.

However, in addition to this particular solution, any general solutions satisfying

r2Gg = 0 (5.47)

can be added to the Greens function so that it vanishes on a given surface. In this case (G= 0on

S), the potential due to a prescribed surface potential distribution becomes,

(r) = I

s(r0)

@G

@n0dS0 (5.48)

where n0 is the normal vector on the surface, which is directed away from the region in which we

wish to evaluate the potential. (Fig. 5.6) This is known as the Dirichlets formulation for potential

boundary value problems.

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Solving a boundary value potential problem is now reduced to nding a suitable Greens function

for a given surface shape, on which G= 0. This is where the method of image will be very useful.

Greens Function for a Sphere

Consider a spherical surface having radius a. We place a charge q at r0. An image charge

q0 =

aq=r0 at

r00 =

a2

r02r0 (5.49)

and the original charge qat r0 make the surface potential vanish. Therefore, the Greens function

for a sphere is immediately written as

G(r; r0) = 1

4

1

jr r0j a=r0

jr r00j

(5.50)

Noting,

r r0

=

r2 + r02 2rr0 cos

1=2

(5.51)

r r00="

r2 +

a2

r0

2

2 a2

r0r cos

#1=2

(5.52)

Fig. 5.6 Geometry for the Dirichlet boundary value problems,

Fig. 5.7 Greeens function for a spherical boundary is essentially equivalent to the potential due

to a charge near a grounded conducting sphere of the same radius.

we nd

G(r; r0) = 1

4

0

BBB@1

(r2

+ r02

2rr0

cos )1=2

ar0

r2 +a2r02 2a2r0r cos 1

=2

1

CCCA

= 1

4

0B@ 1

(r2 + r02 2rr0 cos )1=2 1

r2r02

a2 + a2 2rr0 cos

1=21CA

When we wish to evaluate the potential outside the sphere, the normal derivative is given by

@G

@n0 = @G

@r 0

r0=a

= 14

0BB@ a r cos (r2 + a2 2ra cos )3=2r2

a r cos (r2 + a2 2ra cos )3=2

1CCA=

1

4

a r

2

a

1

(r2 + a2 2ra cos )3=2 (r > a) (5.53)

For interior problems, the negative of the above equation is to be employed. Recall thatcos in

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Eq. (5.53) is given by

cos = cos cos 0 + sin sin 0 cos( 0) (5.54)

in terms of the angular locations of the vectors r = (r;;) and r0 = (r0; 0; 0). Substituting Eq.

(5.53) into Eq. (5.48), we nally obtain the desired exterior potential,

(r) = a2

4

r2

a a

Z 0

sin 0d0Z 2

0

d0 s(

0; 0)

(r2 + a2 2ra cos )3=2 (5.55)

Let us apply this formula to the problem of oppositely charged hemisphere which we solved in

Chapter 3. The surface potential s()is described by,

s() =

8>>>:

+V 0

images and green's functions

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