This celebrated equation was first given by S.D. Poisson, although previously demonstrated by Laplace for the case when ρ = 0. It defines the condition which must be fulfilled by the potential at any and every point in an electric field, through which ρ is finite and the electric force continuous. It may be looked upon as an equation to determine ρ when V is given or vice versa. An exactly similar expression holds good in hydrokinetics, provided that for the electric potential we substitute velocity potential, and for the electric force the velocity of the liquid.
The Poisson equation cannot, however, be applied in the above form to a region which is partly within and partly without an electrified conductor, because then the electric force undergoes a sudden change in value from zero to a finite value, in passing outwards through the bounding surface of the conductor. We can, however, obtain another equation called the “surface characteristic equation” as follows:—Suppose a very small area dS described on a conductor having a surface density of electrification σ. Then let a small, very short cylinder be described of which dS is a section, and the generating lines are normal to the surface. Let V1 and V2 be the potentials at points just outside and inside the surface dS, and let n1 and n2 be the normals to the surface dS drawn outwards and inwards; then −dV1 / dn1 and −dV2 / dn2 are the normal components of the force over the ends of the imaginary small cylinder. But the force perpendicular to the curved surface of this cylinder is everywhere zero. Hence the total flux through the surface considered is −{(dV1 / dn1) + (dV2 / dn2)} dS, and this by a previous theorem must be equal to 4πσdS, or the total included electric quantity. Hence we have the surface characteristic equation,[17]
(dV1 / dn1) + (dV2 / dn2) + 4πσ = 0
(19).
Let us apply these theorems to a portion of a tube of electric force. Let the part selected not include any charged surface. Then since the generating lines of the tube are lines of force, the component of the electric force perpendicular to the curved surface of the tube is everywhere zero. But the electric force is normal to the ends of the tube. Hence if dS and dS′ are the areas of the ends, and +E and -E′ the oppositely directed electric forces at the ends of the tube, the surface integral of normal force on the flux over the tube is
EdS − E′dS′
(20),
and this by the theorem already given is equal to zero, since the tube includes no electricity. Hence the characteristic quality of a tube of electric force is that its section is everywhere inversely as the electric force at that point. A tube so chosen that EdS for one section has a value unity, is called a unit tube, since the product of force and section is then everywhere unity for the same tube.
In the next place apply the surface characteristic equation to any point on a charged conductor at which the surface density is σ. The electric force outward from that point is −dV/dn, where dn is a distance measured along the outwardly drawn normal, and the force within the surface is zero. Hence we have
−dV/dn = 4.0πσ or σ = −(¼π) dV/dn = E/4π.