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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

December 1, 2009

A point charge is placed at a distance d from an infinite conducting plane;

what is the charge density on the plane at a distance r from the foot of the

perpendicular to the plane from the point charge? This is the simplest problem

for which the method of images is invoked [1]. Solutions without using the

uniqueness theorems are possible. I havent found any of these anywhere in

the literature, and I present them here. Let the charge be at dk and the upper

surface of the conducting plane be the X-Y plane. On the plane, r (x2+y2)1/2

is the distance from the origin. At any point in space the total field is due in

part to q and in part to the surface charges induced on the plane: E = Eq + E.

The electrostatic force per unit area on the surface is f = 2(r)k/20 [2]. Now

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f(r) = (r)Eq(r) + f(r), where f(r) is the electrostatic force per unit area on

the X-Y plane due to the charges on the plane. By symmetry, it can have no

Z-component. Equating the above two values of f and taking the dot product

with k we get (r)/20 = Eq(r) k = qd/40(r2 + d2)3/2 (fig.1). So, (r) =

qd/2(r2 + d2)3/2.

In this solution I have talked of f and not E as the field on the surface is

discontinuous and it has no well defined value on the surface. Let me present

another solution.

At a general point P on the plane whose distance from the origin is r we have

(Eq)z = qd/40(r2 + d2)3/2, and we know (Eq)z is continuous. On the other

hand (E)z is discontinuous in the amount /0 [3], and by symmetry is the

same above and below in magnitude and its direction on both sides is either

towards the plane or away from the plane. Thus immediately below the plane

(E)z = /20. But below the plane, i.e. inside the conductor, the total field

is zero. So, (Eq)z + (E)z = 0. Hence, = qd/2(r2 + d2)3/2.

Let me present yet another solution. I learnt it from Dr. J.D.Jackson in the

reply to a mail conveying my solution.

Consider any point P on the plane (fig.2). On the conducting plane, the electric

field caused by the point charge and the distribution of surface charges must be

normal to the plane. Otherwise, the charge, free to move, will readjust itself.

Now, since the tangential component of electric field is continuous at a surface

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charge, the net field is normal to the plane both above and below P. If (as in fig.

3(a)) the field just above P due to the surface charges is E(cosk + sinr)

(where r is a unit vector in the plane from the origin O to the point P), then by

symmetry, just below P it is E(cosk sinr) (fig. 3(b)). Since, just above

P the net field is normal, we have Esin = Eqsin. And just below P the net

field is zero, so Ecos = Eqcos. Hence, E = Eq and = . Thus, we see

that the net field just above P is 2Eqcosk. Application of Gausss law to a

pillbox at P that spans the surface (with zero contribution from the side of the

box within the conductor) gives = 20Eqcos = qd/2(r2 + d2)3/2 (fig.1).

By arguing without using an image charge we, in our solution, showed that near

any point on the conducting plane the field due to the induced surface charges

on the plane is constructed by first reflecting qs field in the conducting plane

and then reversing its direction. This is exactly the field of an image charge -q

placed at dk, but we notice that after the fact.

References

[1] Edward.M.Purcell, Electricity and Magnetism, (McGraw-Hill, New York,

1963), 1st ed., pp. 92-93.

[2] David.J.Griffiths, Introduction to Electrodynamics, (Prentice Hall, Engle-

wood Cliffs, NJ, 1989), 2nd ed., eq-3.10, p.121.

[3] J.D.Jackson, Classical Electrodynamics, (John Wiley, New York, 1999), 3rd

ed., eq-1.22, p.31.

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FIGURE CAPTIONS

Fig. 1. A point charge q is held at a distance d from an infinite conducting

plane. The normal component of the field of q at a point on the plane distant r

from the foot of the perpendicular to the plane from the point charge is Eq k

or Eqcos, i.e. qd/40(r2 + d2)3/2.

Fig. 2. A point charge q lies at dk above the X-Y plane. The region z0 is filled

with grounded conducting material.

Fig.3. The electric fields of the point charge and the surface charge (a) just

above and (b) just below the X-Y plane. The vertical line shown is a line

through P parallel to the z axis.

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