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A Solution of the Conducting Plane Image Problem

Without Using The Method Of Images

Ishnath Pathak

B.Tech student, Department of Civil Engineering, Indian Institute of

Technology, North Guwahati, Guwahati 781039

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ABSTRACT

The problem of evaluation of the induced surface charges on an infinite

conducting plane in response to the placement of a point charge near it is solved

without using the method of images or any uniqueness theorem. For so doing a

generalization of gauss’s theorem is discussed which allows charges to lie on the

Gaussian surface, and some common special cases are considered.

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I. INTRODUCTION

With the possibility of electrons being point charges, we don’t know if a finite

amount of charge can reside completely in a region of zero volume. But still such

idealizations in the theory of electrodynamics such as point, line and surface

charges cannot be avoided at any rate. One of the most important theorems of

electrostatics is the Gauss’s theorem. The well known theorem states

0encS

/εQ=⋅∫∫ daE . It is explicitly stated sometimes1, for good, that the boundary

of the region enclosed must not contain any point, line or surface charges. But

what happens if the surface contains such charges in any idealized problem? To

my knowledge, the generalization2 of Gauss’s theorem has not been discussed

anywhere till now, but anyway it is and will remain to be among the trivia. To

include this consideration in an introductory course, a simple exposition such as

presented in the next section is required. This pedagogy may be employed if one

aims at presenting in an introductory course that elegant solution to the

conducting plane image problem which has been discussed in the following

section.

II. THE GENERALIZATION OF GAUSS’S THEOREM

In this section we digress to discuss what missing terms appear in the gauss’s

theorem if we permit presence of charges on the surface of integration. The

solution to the conducting plane image problem using results of this section

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instead of any special trick of electrostatics is discussed in the next section. This

digression will be pretty long, but the results we’ll arrive at will be worth the

length.

A. FLUX OF THE FIELD OF A POINT CHARGE THROUGH A

SURFACE

We prove in this section that the flux of the electric field iE of a charge iq over a

boundary S to a connected region encR is

Sbyenclosedisq if

ΩisanglesolidinsidewherepointaatSonliesqif

Soutsideliesqif

εq

4πΩq

0

i

i

i

i

0

i

0

ii

S

⎪⎪⎪

⎩

⎪⎪⎪

⎨

⎧

ε=⋅∫∫ daiE (1)

We denote by ir the position of iq . Let’s first consider the case A when the

charge lies outside the boundary S. In this case iE is differentiable properly over

an open connected region containing the surface S along with its enclosure.

Application of divergence theorem will yield the result. Let’s now consider the

case B when ir lies on S. Let the inside solid angle formed on S at the point ir be

iΩ in measure. We chose an arbitrary non-zero radius δ sufficiently small, say

less than a critical radius cδ , such that the part δS of the spherical surface of

radius δ centred at ir which lies not outside the region encR enclosed by S

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i) Divides encR into two parts 1δR and 2δR such that 2δR , the one not

adjacent to ir , is enclosed by a closed surface ′2δS which has the inner

surfaces of S as its inner surfaces in case encR has cavities.

ii) Has no missing patches, i.e. δS is bounded by a single closed curve δP

and not by a group of closed curves.

Hence, the closed curve δP divides the outer surface of S into two parts, 1δS and

2δS , where 2δS and δS together with the inner surfaces of S form the closed

surface ′2δS which encloses 2δR . Also 1δS and δS together form the closed

surface ′1δS which encloses the compact region 1δR .

Now, we can write the flux of electric field through S as

dadada ⋅+⋅=⋅ ∫∫∫∫∫∫′′

2δ1δ SSSiii EEE

Because when we add the flux

2δ

0

i

δSofarea

4πqΦε

=

Of iE through δS to the second integral on the right of equality and subtract the

same from the first one, we get identity. Now according to the result of case A the

second integral must vanish. The flux of iE through S can now be easily obtained

0

ii

0δS

0S

0S 4π

ΩqΦlimlimlim1

ε==⋅=⋅=⋅

→′

→δ→δ ∫∫∫∫∫∫δ

dadada iii EEE

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Finally we consider the case C when ir lies enclosed by S. We consider any plane

passing through ir and denote by 0S the part of the plane that is enclosed by S.

We denote again by 1R and 2R the regions into which 0S divides encR , and by

′1S and ′

2S the closed surfaces that enclose these regions. So we have again

∫∫∫∫∫∫′′

⋅+⋅=⋅21 SSS

dadada iii EEE

Only this time we are already with an identity, for this time the flux Φ of iE

through 0S is zero. Now according to the result of case B, both the terms on the

right of equality are 0i /2εq , so that the flux is

0

i

S εq

=⋅∫∫ daiE

B. THE GENERALIZED GAUSS’S THEOREM

The principal of superposition for electrostatic field allows us to insist that the

flux of the electric field over a closed surface S is da⋅∫∫ ∑ )( iE , where the

summation is done over all charges in the charge configuration. The linearity of

the operation of flux enables us to insist that the flux is ∑ ∫∫ ⋅ )( daiE . Since the

flux ∫∫ ⋅daiE of any charge iq enclosed by S is oi /εq , the sum of the isolated

fluxes of all charges enclosed by S is 0enc/εQ , where encQ is the net charge

enclosed by S. Similarly we get that the sum of the isolated fluxes of all charges

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lying outside S is zero, and that of all charges residing on the continuities of S is

0con /2εQ , where conQ is the net charge residing on the continuities of S. By

continuities of S, we mean a point on S where the inside solid angle formed is

equal to 2π . The sum of the isolated fluxes of the remaining charges (all of which

lie at the discontinuities of S) is left as a summation

∑ ε=

discon o

iid 4π

qΩΦ (2)

For we can’t say anything about the inner solid angles iΩ without a particular

knowledge of the geometry of the closed surface and the locations of the charges

lying at the discontinuities. Hence, we have the generalized Gauss’s theorem:

“ The flux of the electrostatic field E over any closed surface S is

d0

con

0

enc

S

Φε

Q21

εQ

++=⋅∫∫ daE (3)

Where encQ equals the net charge enclosed by S, conQ equals the net charge

residing on the continuities of S and dΦ as described by (2) equals the flux of the

electric field of the charges lying at the discontinuities of S”

C. THEOREM GUISE UNDER EXPLICIT CONFINEMENT

STIPULATIONS

We in this section consider those cases in which the generalized Gauss’s theorem

takes a beautiful form which we call as the active guise of the theorem. We begin

by confining ourselves to an electric field caused by a configuration of charges

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not containing any point or line charges. In this special case conQ of equation (3)

can be elegantly interpreted as the net charge contained by S, for the net charge

lying on the discontinuities of S is zero anyway, because S now contains no point

or line charges- which were the only varieties which could accumulate to a finite

amount by assembling only at discontinuities. Also dΦ vanishes in case of an

electric field caused by such a configuration. This needs some explanation. Let’s

imagine a different charge configuration- the one in which each charge is replaced

by a corresponding positive one of an equal magnitude. The fact that the net

charge lying on the discontinuities of S is zero whenever the charge configuration

does not contain any point or line charges implies that for the original charge

configuration we’ll have 0qi =∑discon

. Now since iii qqq ≤≤− , we have

∑∑∑ ε≤

ε≤

ε−

discondiscondiscon 0

ii

0

ii

0

ii

4πΩq

4πΩq

4πΩq

.

The central term in the inequalities is dΦ . So,

∑∑ πε=

πε≤

discondiscon 0

ii

0

iid 4

Ωq4Ωq

Φ

And since 4πΩ0 i ≤≤ , 4πqΩq0 iii ≤≤ . Hence

∑∑ ≤≤discondiscon

iii q

4πΩq

0 .

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As the rightmost term in the inequalities is zero, as argued earlier, we have the

middle term equal to zero. From here we conclude that dΦ vanishes. Therefore if

the source charge configuration is free from point and line charges, then

“ The flux of electrostatics field over any closed surface is

0

con

0

enc

S εQ

21

εQ

+=⋅∫∫ daE (4)

Where encQ is the net charge enclosed by S and conQ is the net charge contained

by S.”

Superposing the fluxes we get as a corollary that for any kind of charge

configuration

pl0

sur

0

enc

S

Φε

Q21

εQ

++=⋅∫∫ daE (5)

Where encQ is the net charge enclosed by S, surQ is the net surface charge

residing on S and plΦ is the flux of the electric field of the point and line charges

residing on S.

Till now (4) was referred to as applicable only in case of charge

configurations that didn’t contain point and line charges. From (5) it can be seen

that (4) holds ‘whenever no point or line charge lies anywhere on the

discontinuities of S’, for in that case we see (in light of equation (3)) that

0l0ppl /2εQ/2εQΦ += , where pQ equals the net point charge and lQ equals the

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net line charge residing on S. And then, since the recent most restriction holds if S

has no discontinuity at all, (4) holds ‘whenever S is throughout continuous’. We

shall call equation (4) as the ‘active guise of generalized Gauss’s theorem’ and

see that it is almost always applicable.

D. THE ABSOLUTE ELECTROSTATIC BOUNDARY EQUATION

Let S be a surface on which, near a point r , only surface charges lie. Let’s denote

the charge density by )σ(r . If one side of the surface, say side 1, contains no

point or line charges near r , then the application of the generalized Gauss’s

theorem in the usual manner3 provides us with the following boundary condition

concerning the difference between the limit 1E of electric field as we approach r

from side 1, and the value E of the electric field at the point r on the surface:

11 nrE(r)(r)E ˆ02ε)σ(

=− (6)

Where 1n is the normal unit vector to S at r towards side 1. This absolute

boundary condition will be used in the next section to solve the conducting plane

image problem. Before that I’d like to call your attention to the fact that the

absolute boundary equation can be used to arrive at the usual boundary equation

in this manner:

12112211 nEEnnEEnEE ˆˆˆˆ0000 εσ;

2εσ

2εσ;

2εσ

=−−==−=−

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III. A SOLUTION OF THE CONDUCTING PLANE IMAGE

PROBLEM WITHOUT USING THE METHOD OF IMAGES

A conductor in electrostatics arranges its charge on the surface in such a

manner that it cancels the external electric field everywhere inside the volume of

the conductor4. In our problem we have a conductor whose one side is the X-Y

plane. The other side can be of any geometry; all we are certain about is that some

slice of space, say hz0 −≥≥ , is occupied by the volume of the conductor. Now,

we’ve to show that there exists one and only one surface charge density )σ(r

defined on the X-Y plane such that the electrostatic field of the surface charge

distribution on the X-Y plane cancels everywhere in the volume hz0 −≥≥ , the

electric field of a point charge q placed at zd . We denote by )σ(r an arbitrary

such charge density and consider the net field E of the point charge q at zd and

the surface charges on the X-Y plane. Applying the absolute electrostatic

boundary equation on the X-Y plane at a point r with side 1 as the interior of the

conductor we get

zrrErE1 ˆ02ε)σ()()( −=−

As E is zero everywhere in hz0 −≥≥ , the limit )(rE1 is zero. So, we have

zrEr ˆ⋅= )(2ε

)σ(

0

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Now, from the principal of superposition )()()( rErErE Sq += , where qE equals

the electric field of the point charge q and SE equals the electric field of the

surface charges on the X-Y plane. It follows straight from the Coulomb’s law that

for any point r on the X-Y plane

3/2220 )r(d

)d(4π

q)(+

+−ε

=rzrEq

ˆ

So that we have

3/2220 )r(d

d4π

q)(+−

ε=⋅zrEq ˆ

Because r and z are orthogonal. Utilizing this advancement we get

zrErS ˆ⋅+

+ε−= )(

)r(d4πqd

2ε)σ(

3/22200

.

Now comes the special character of this simple image problem: it follows from

the radial character of Coulomb’s law that )(rES is orthogonal to z . So

3/222 )d(rqd

2π1)σ(

+−

=r (7)

Since we started with )σ(r as an arbitrary suitable charge density, we get that the

charge density is unique. Direct integration can be used to verify that this charge

density cancels the electric field of the point charge.

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ACKNOWLEDGEMENTS

I owe gratitude to IIT Guwahati for all the library facility and access to science

direct that I have enjoyed. I also thank my father, Mr. Shyam Rudra Pathak, for

providing some very nice suggestions after proof reading.

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FOOTNOTES AND REFERENCES

1E.M.Purcell, Electricity & Magnetism, 1st edition, (McGraw Hill, 1965), p.22

2Admittedly, there is another generalization of Gauss’s theorem which is in fact

the four dimensional analogue to the Gauss’s theorem, but that’s certainly not

what we are talking about. See L.D.Landau & E.M.Lifshitz, The Classical Theory

of Fields, 4th revised English edition, (Butterworth Heinemann, 1975), p.20

3D.J.Griffiths, Introduction to Electrodynamics, 3rd edition, (Dorling Kindersley,

2006), pp. 106-107

4Reference 3, p.117