A Solution Of The Conducig Plane Imag 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 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 haven’t found any of these anywhere in ˆ and the upper the literature, and I present them here. Let the charge be at dk surface of the conducting plane be the X-Y plane. On the plane, r ≡ (x2 +y 2 )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 + Eσ . ˆ 0 [2]. Now The electrostatic force per unit area on the surface is f = σ 2 (r)k/2 1

f(r) = σ(r)Eq (r) + fσ (r), where fσ (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 ˆ we get σ(r)/20 = Eq (r) · k ˆ = −qd/4π0 (r2 + d2 )3/2 (fig.1). So, σ(r) = with k −qd/2π(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/4π0 (r2 + d2 )3/2 , and we know (Eq )z is continuous. On the other hand (Eσ )z is discontinuous in the amount σ/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 (Eσ )z = −σ/20 . But below the plane, i.e. inside the conductor, the total field is zero. So, (Eq )z + (Eσ )z = 0. Hence, σ = −qd/2π(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. ˆ + sinαˆr) 3(a)) the field just above P due to the surface charges is −Eσ (cosαk (where ˆr is a unit vector in the plane from the origin O to the point P), then by ˆ − sinαˆr) (fig. 3(b)). Since, just above symmetry, just below P it is Eσ (cosαk P the net field is normal, we have Eσ sinα = Eq sinθ. And just below P the net field is zero, so Eσ cosα = Eq cosθ. Hence, Eσ = Eq and α = θ. Thus, we see ˆ Application of Gauss’s law to a that the net field just above P is −2Eq cosθk. pillbox at P that spans the surface (with zero contribution from the side of the box within the conductor) gives σ = −20 Eq cosθ = −qd/2π(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 q’s field in the conducting plane and then reversing its direction. This is exactly the field of an image charge -q ˆ but we notice that after the fact. placed at −dk,

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, Englewood 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. 3

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 ˆ from the foot of the perpendicular to the plane from the point charge is Eq · k or −Eq cosθ, i.e. −qd/4π0 (r2 + d2 )3/2 . ˆ above the X-Y plane. The region z¡0 is filled Fig. 2. A point charge q lies at dk 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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