A Solution Of The Conducting Plane Image Problem Without Using The Method Of Images

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

∫∫ E ⋅ da = Q

enc

/ε 0 . It is explicitly stated sometimes1, for good, that the boundary

S

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 E i of a charge q i over a boundary S to a connected region R enc is

⎧ ⎪ 0 ⎪ ⎪q Ω ∫∫S Ei ⋅ da = ⎨ 4πi ε0i ⎪ ⎪ qi ⎪ ε ⎩ 0

if q i lies outside S if q i lies on S at a point where inside solid angle is Ω i

(1)

if q i is enclosed by S

We denote by ri the position of q i . Let’s first consider the case A when the charge lies outside the boundary S. In this case E i 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 ri lies on S. Let the inside solid angle formed on S at the point ri be Ω i in measure. We chose an arbitrary non-zero radius δ sufficiently small, say

less than a critical radius δ c , such that the part Sδ of the spherical surface of radius δ centred at ri which lies not outside the region R enc enclosed by S

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i) Divides R enc into two parts R 1δ and R 2δ such that R 2δ , the one not ′ adjacent to ri , is enclosed by a closed surface S2δ which has the inner

surfaces of S as its inner surfaces in case R enc has cavities. ii) Has no missing patches, i.e. Sδ is bounded by a single closed curve Pδ and not by a group of closed curves. Hence, the closed curve Pδ divides the outer surface of S into two parts, S1δ and S2δ , where S2δ and Sδ together with the inner surfaces of S form the closed ′ surface S2δ which encloses R 2δ . Also S1δ and Sδ together form the closed ′ surface S1δ which encloses the compact region R 1δ .

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

∫∫ E ⋅ da = ∫∫ E ⋅ da + ∫∫ E ⋅ da i

S

S1δ

′

i

S2δ

′

i

Because when we add the flux q i area of Sδ 4πε 0 δ2

Φ=

Of E i through Sδ 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 E i through S can now be easily obtained qiΩi

∫∫ E ⋅ da = lim ∫∫ E ⋅ da = lim ∫∫ E ⋅ da = lim Φ = 4πε i

S

δ →0

i

S

δ →0

S1δ

′

i

δ→0

0

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Finally we consider the case C when ri lies enclosed by S. We consider any plane passing through ri and denote by S0 the part of the plane that is enclosed by S. We denote again by R 1 and R 2 the regions into which S0 divides R enc , and by

′ ′ S1 and S2 the closed surfaces that enclose these regions. So we have again

∫∫ E ⋅ da = ∫∫ E ⋅ da + ∫∫ E ⋅ da i

′ S1

S

i

S2

′

i

Only this time we are already with an identity, for this time the flux Φ of E i through S0 is zero. Now according to the result of case B, both the terms on the right of equality are q i /2ε 0 , so that the flux is qi

∫∫ E ⋅ da = ε i

S

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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

∫∫ (∑ E ) ⋅ da , i

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 flux

∫∫ E ⋅ da i

∑ (∫∫ E ⋅ da) . Since the i

of any charge q i enclosed by S is q i /ε o , the sum of the isolated

fluxes of all charges enclosed by S is Q enc /ε 0 , where Q enc 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 Q con /2ε 0 , where Q con 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

Φd =

Ωiqi discon 4πε o

∑

(2)

For we can’t say anything about the inner solid angles Ω i 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

∫∫ E ⋅ da = S

Q enc 1 Q con + + Φd ε0 2 ε0

(3)

Where Q enc equals the net charge enclosed by S, Q con equals the net charge residing on the continuities of S and Φ d 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 Q con 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 Φ d 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

∑q

i

= 0 . Now since − q i ≤ q i ≤ q i , we have

discon

−

∑

discon

q i Ωi 4πε 0

≤

q Ω qiΩi ≤ ∑ i i. discon 4πε 0 discon 4πε 0

∑

The central term in the inequalities is Φ d . So, Φd ≤

∑

discon

qi Ωi 4πε0

=

q i Ωi

∑

discon

4πε0

And since 0 ≤ Ω i ≤ 4π , 0 ≤ q i Ω i ≤ q i 4π . Hence 0≤

∑

discon

q i Ωi 4π

≤

∑q

i

.

discon

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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 Φ d 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

∫∫ E ⋅ da = S

Q enc 1 Q con + ε0 2 ε0

(4)

Where Q enc is the net charge enclosed by S and Q con is the net charge contained by S.” Superposing the fluxes we get as a corollary that for any kind of charge configuration

∫∫ E ⋅ da = S

Q enc 1 Q sur + + Φ pl ε0 2 ε0

(5)

Where Q enc is the net charge enclosed by S, Q sur is the net surface charge residing on S and Φ pl 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 Φ pl = Q p /2ε 0 + Q l /2ε 0 , where Q p equals the net point charge and Q l 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 E1 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:

E1 (r) − E(r) =

σ(r ) nˆ 1 2ε 0

(6)

Where nˆ 1 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:

E1 − E =

σ σ σ σ nˆ 1 ; E 2 − E = nˆ 2 = − nˆ 1 ; E1 − E 2 = nˆ 1 2ε 0 2ε 0 2ε 0 ε0

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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 0 ≥ z ≥ −h , 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 0 ≥ z ≥ − h , the electric field of a point charge q placed at dzˆ . We denote by σ(r ) an arbitrary such charge density and consider the net field E of the point charge q at dˆz 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

E 1 (r ) − E(r ) = −

σ(r ) ˆz 2ε 0

As E is zero everywhere in 0 ≥ z ≥ − h , the limit E1 (r ) is zero. So, we have σ(r ) = E(r ) ⋅ ˆz 2ε 0

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Now, from the principal of superposition E(r ) = Eq (r ) + ES (r ) , where Eq equals the electric field of the point charge q and ES 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 Eq (r ) =

q d(− zˆ ) + r 4πε 0 (d 2 + r 2 ) 3/2

So that we have Eq (r ) ⋅ zˆ =

q −d 2 4πε 0 (d + r 2 )3/2

Because r and zˆ are orthogonal. Utilizing this advancement we get σ(r ) qd =− + ES (r ) ⋅ ˆz . 2ε 0 4πε 0 (d 2 + r 2 ) 3/2 Now comes the special character of this simple image problem: it follows from the radial character of Coulomb’s law that ES (r ) is orthogonal to zˆ . So σ(r ) =

qd −1 2 2π (r + d 2 )3/2

(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 1

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

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Admittedly, 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 3

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

2006), pp. 106-107 4

Reference 3, p.117

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