If we define the positive direction along a tube of electric force as the direction in which a small body charged with positive electricity would tend to move, we can summarize the above facts in a simple form by saying that, if we have any closed surface described in any manner in an electric field, the excess of the number of unit tubes which leave the surface over those which enter it is equal to 4π-times the algebraic sum of all the electricity included within the surface.
Every tube of electric force must therefore begin and end on electrified surfaces of opposite sign, and the quantities of positive and negative electricity on its two ends are equal, since the force E just outside an electrified surface is normal to it and equal to σ/4π, where σ is the surface density; and since we have just proved that for the ends of a tube of force EdS = E′dS′, it follows that σdS = σ′dS′, or Q = Q′, where Q and Q′ are the quantities of electricity on the ends of the tube of force. Accordingly, since every tube sent out from a charged conductor must end somewhere on another charge of opposite sign, it follows that the two electricities always exist in equal quantity, and that it is impossible to create any quantity of one kind without creating an equal quantity of the opposite sign.
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| Fig. 4. |
We have next to consider the energy storage which takes place when electric charge is created, i.e. when the dielectric is strained or polarized. Since the potential of a conductor is defined to be the work required to move a unit of positive electricity from the surface of the earth or from an infinite distance from all electricity to the surface of the conductor, it follows that the work done in putting a small charge dq into a conductor at a potential v is v dq. Let us then suppose that a conductor originally at zero potential has its potential raised by administering to it small successive doses of electricity dq. The first raises its potential to v, the second to v′ and so on, and the nth to V. Take any horizontal line and divide it into small elements of length each representing dq, and draw vertical lines representing the potentials v, v′, &c., and after each dose. Since the potential rises proportionately to the quantity in the conductor, the ends of these ordinates will lie on a straight line and define a triangle whose base line is a length equal to the total quantity Q and height a length equal to the final potential V. The element of work done in introducing the quantity of electricity dq at a potential v is represented by the element of area of this triangle (see fig. 4), and hence the work done in charging the conductor with quantity Q to final potential V is 12QV, or since Q = CV, where C is its capacity, the work done is represented by 12CV2 or by 12Q2 / C.
If σ is the surface density and dS an element of surface, then ∫σdS is the whole charge, and hence 12 ∫ VσdS is the expression for the energy of charge of a conductor.
We can deduce a remarkable expression for the energy stored up in an electric field containing electrified bodies as follows:[1] Let V denote the potential at any point in the field. Consider the integral
| W = | 1 | ∭{( | dV | ) | 2 | + ( | dV | ) | 2 | + ( | dV | ) | 2 | } dx dy dz. |
| 8π | dx | dy | dz |
where the integration extends throughout the whole space unoccupied by conductors. We have by partial integration
| ∭( | dV | ) | 2 | dx dy dz = ∬ V | dV | dy dz − ∭ V | d 2V | dx dy dz, |
| dx | dx | dx2 |
and two similar equations in y and z. Hence
| 1 | ∭ {( | dV | ) | 2 | + ( | dV | ) | 2 | + ( | dV | ) | 2 | } dx dy dz = |
| 8π | dx | dy | dz |
| 1 | ∬ V | dV | dS − | 1 | ∭ V∇V dx dy dz |
| 8π | dn | 8π |
where dV/dn means differentiation along the normal, and ∇ stands for the operator d 2/dx2 + d 2/dy2 + d 2/dz2. Let E be the resultant electric force at any point in the field. Then bearing in mind that σ = (1/4π) dV/dn, and ρ = −(1/4π) ∇V, we have finally
| 1 | ∭ E2dV = | 1 | ∬ Vσ dS + | 1 | ∭ Vρ dV. |
| 8π | 2 | 2 |
The first term on the right hand side expresses the energy of the surface electrification of the conductors in the field, and the second the energy of volume density (if any). Accordingly the term on the left hand side gives us the whole energy in the field.
Suppose that the dielectric has a constant K, then we must multiply both sides by K and the expression for the energy per unit of volume of the field is equivalent to 12DE where D is the displacement or polarization in the dielectric.
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.
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| 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.[2] 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.
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| 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 = r 2, or CB = r 2/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 |
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
in other words, the force at P is inversely as the cube of the distance from A. Suppose then we remove the negative point-charge, and let the sphere be supposed to become conductive and be connected to earth. If we make a distribution of negative electricity over it, which has a density σ varying according to the law
that distribution, together with the point-charge +q at A, will make a distribution of electric force at all points outside the sphere
