Furthermore it can be shown by the application of the calculus of variations that the condition for a minimum value of the function W, is that ∇V = 0. Hence that distribution of potential which is necessary to satisfy Laplace’s equation is also one which makes the potential energy a minimum and therefore the energy stable. Thus the actual distribution of electricity on the conductor in the field is not merely a stable distribution, it is the only possible stable distribution.
| Fig. 5. |
Method of Electrical Images.—A very powerful method of attacking problems in electrical distribution was first made known by Lord Kelvin in 1845 and is described as the method of electrical images.[20] By older mathematical methods it had only been possible to predict in a few simple cases the distribution of electricity at rest on conductors of various forms. The notion of an electrical image may be easily grasped by the following illustration: Let there be at A (see fig. 5) a point-charge of positive electricity +q and an infinite conducting plate PO, shown in section, connected to earth and therefore at zero potential. Then the charge at A together with the induced surface charge on the plate makes a certain field of electric force on the left of the plate PO, which is a zero equipotential surface. If we remove the plate, and yet by any means can keep the identical surface occupied by it a plane of zero potential, the boundary conditions will remain the same, and therefore the field of force to the left of PO will remain unaltered. This can be done by placing at B an equal negative point-charge −q in the place which would be occupied by the optical image of A if PO were a mirror, that is, let −q be placed at B, so that the distance BO is equal to the distance AO, whilst AOB is at right angles to PO. Then the potential at any point P in this ideal plane PO is equal to q/AP − q/BP = O, whilst the resultant force at P due to the two point charges is 2qAO/AP³, and is parallel to AB or normal to PO. Hence if we remove the charge −q at B and distribute electricity over the surface PO with a surface density σ, according to the Coulomb-Poisson law, σ = qAO / 2πAP³, the field of force to the left of PD will fulfil the required boundary conditions, and hence will be the law of distribution of the induced electricity in the case of the actual plate. The point-charge −q at B is called the “electrical image” of the point-charge +q at A.
We find a precisely analogous effect in optics which justifies the term “electrical image.” Suppose a room lit by a single candle. There is everywhere a certain illumination due to it. Place across the room a plane mirror. All the space behind the mirror will become dark, and all the space in front of the mirror will acquire an exalted illumination. Whatever this increased illumination may be, it can be precisely imitated by removing the mirror and placing a second lighted candle at the place occupied by the optical image of the first candle in the mirror, that is, as far behind the plane as the first candle was in front. So the potential distribution in the space due to the electric point-charge +q as A together with −q at B is the same as that due to +q at A and the negative induced charge erected on the infinite plane (earthed) metal sheet placed half-way between A and B.
| Fig. 6. |
The same reasoning can be applied to determine the electrical image of a point-charge of positive electricity in a spherical surface, and therefore the distribution of induced electricity over a metal sphere connected to earth produced by a point-charge near it. Let +q be any positive point-charge placed at a point A outside a sphere (fig. 6) of radius r, and centre at C, and let P be any point on it. Let CA = d. Take a point B in CA such that CB·CA = r², or CB = r²/d. It is easy then to show that PA : PB = d : r. If then we put a negative point-charge −qr/d at B, it follows that the spherical surface will be a zero potential surface, for
| q | − | rq | · | 1 | = 0 |
| PA | d | PB |
(24).
Another equipotential surface is evidently a very small sphere described round A. The resultant force due to these two point-charges must then be in the direction CP, and its value E is the vector sum of the two forces along AP and BP due to the two point-charges. It is not difficult to show that
E = − (d² − r²) q / rAP³