PARTICULAR CASES OF ELECTRIFICATION

Excerpts

• PARTICULAR CASES OF ELECTRIFICATION

  *Definition*.---The electric or electromotive force at a point is the force which would be experienced by a small body charged with the unit of positive electricity and placed at that point, the electrification of the system being supposed to remain undisturbed by the presence of this unit of electricity.

• PARTICULAR CASES OF ELECTRIFICATION

  Hence the idea of an electrified point is a mere mathematical fiction which can never be realised in nature.

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  But in the vulgar language of the time when dynamical science was unknown, all the words relating to exertion, such as force, energy, power, &c., were confounded with each other, though some of the schoolmen endeavoured to introduce a greater precision into their language.

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  Now the electric capacity of a body in a given field is measured by the charge which raises its potential to unity. Hence the electric capacity of a conducting sphere placed in air at a considerable distance from any other conductor is numerically equal to the radius of the sphere.

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  or the electric force close to the surface of an electrified sphere is at right angles to the surface and is equal to the surface-density multiplied by 4 .

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  This circle is an example of a line of equilibrium, for the resultant force vanishes at every point of this line.

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  The smaller the distance between the surfaces and the greater the area of the surfaces, the greater the capacity of the jar.

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  Now the quantity of electricity in a body is measured, according to Faraday’s ideas, by the *number* of lines of force, or rather of induction, which proceed from it. These lines of force must all terminate somewhere, either on bodies in the neighbourhood, or on the walls and roof of the room, or on the earth, or on the heavenly bodies, and wherever they terminate there is a quantity of electricity exactly equal and opposite to that on the part of the body from which they proceeded.

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  We must bear in mind, however, that it is physically impossible to charge the small sphere with more than a certain quantity of electricity on each unit of area of its surface. If the surface-density exceed this limit, electricity will fly off in the form of the brush discharge. Hence the idea of an electrified point is a mere mathematical fiction which can never be realised in nature.

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  If we suppose the radius of the inner sphere to become very small till at last the sphere cannot be distinguished from a point, we may imagine the whole charge concentrated at this point, and we may then express our result by saying that the electric action of a uniformly electrified sphere at any point outside the sphere is the same as that of the whole charge of the sphere would be if concentrated at the centre of the sphere.

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  The electric potential at any point is the work which must be expended in order to bring a body charged with unit of electricity from an infinite distance to that point.

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  This example may serve to illustrate the principle of the Leyden jar, which consists of two metallic surfaces separated by insulating material. The smaller the distance between the surfaces and the greater the area of the surfaces, the greater the capacity of the jar.

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  Very few, however, even of scientific men, are careful to observe these distinctions; hence we often hear of the force of a cannon-ball when either its energy or its momentum is meant, and of the force of an electrified body when the quantity of its electrification is meant.

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  In other cases we cannot afford to despise the humbler method of actually drawing tentative figures on paper, and selecting that which appears least unlike the figure we require.

Equations

• PARTICULAR CASES OF ELECTRIFICATION

  E = -e\text{.}

  The charge on the internal surface of the outer sphere is equal and opposite to the charge on the inner sphere.

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  s = 4 \pi r^2

  The surface of a sphere of radius r is 4 pi r squared.

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  S = 4 \pi R^2

  The surface of a sphere of radius R is 4 pi R squared.

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  e = s \sigma

  The whole charge on a surface is its area multiplied by its surface density.

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  E = S \Sigma

  The whole charge on the internal surface of the outer sphere is its area multiplied by its surface density.

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  \sigma = \frac{e}{4 \pi r^2}

  The surface density on the inner sphere equals its charge divided by its area.

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  \Sigma = \frac{E}{4 \pi R^2}

  The surface density on the internal surface of the outer sphere equals its charge divided by its area.

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  \Sigma = \frac{-e}{4 \pi R^2}

  Substituting E = -e, the surface density on the internal surface of the outer sphere is minus e divided by its area.

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  f = \frac{ee'}{r^2}

  The force between two small charged bodies is the product of their charges divided by the square of their distance; it is a repulsion for like charges and an attraction for unlike charges.

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  \mathfrak{E} = \frac{e}{r'^2}

  At a point outside a uniformly electrified sphere at distance r' from its centre, the electric force is the sphere's charge divided by r' squared, directed from the centre.

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  \mathfrak{E} = \frac{e}{r^2} = 4 \pi \sigma

  Close to the surface of an electrified sphere, the electric force equals the surface density multiplied by 4 pi.

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  \mathfrak{E} = 4 \pi \sigma

  The electric force at a conductor's surface is at right angles to the surface and equals 4 pi times the surface density; this is Coulomb's law in complete form.

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  e = 4 \pi r^2 \sigma

  The total charge of a uniformly electrified sphere is 4 pi r squared times its surface density.

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  \psi' - \psi = \overline{BA} \ldot \mathfrak{E}

  If the electric force is constant and equal to E along BA, the work done per unit charge from B to A equals the distance BA times E.

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  \left(\frac{1}{b} - \frac{1}{a}\right)\frac{a}{b} < B - A < \left(\frac{1}{b} - \frac{1}{a}\right) \frac{a}{b}

  Bounds on the potential difference B - A between points at distances a and b from a unit charge. FLAG: the printed lower bound reads a/b, but the derivation just above gives (1/b - 1/a) times b/a as the lower bound; the printed left-hand bound looks like a printing slip in the book.

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  \left(\frac{1}{z} - \frac{1}{a}\right) \frac{1}{p} < Z - A < \left(\frac{1}{z} - \frac{1}{y}\right) p

  Summed bounds on Z - A obtained by adding the stepwise inequalities, with p the greatest ratio of successive distances. FLAG: the upper bound is printed as (1/z - 1/y) p; adding the steps gives (1/z - 1/a) p, so the y appears to be a printing slip in the book.

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  Z - A = \frac{1}{z} - \frac{1}{a}

  In the limit of an infinitely fine subdivision, the potential difference between two points is the difference of the reciprocals of their distances from a unit charge.

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  Z = \frac{1}{z}

  With the potential at infinity taken as zero, the potential at a point due to unit charge at distance z is the reciprocal of z.

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  \psi = \frac{e}{r}

  For points outside a uniformly electrified spherical surface, the potential equals the total charge divided by the distance from the centre.

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  \psi_a = \frac{e}{a}

  The potential at the surface of a sphere of radius a due to its own charge is the charge divided by the radius.

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  e = a

  When the potential of a sphere is unity its charge is numerically equal to its radius, which gives the capacity of an isolated sphere.

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  e\xp\left(\dfrac{1}{r} - \dfrac{1}{b} \right)

  The potential at a point between the two concentric spherical surfaces, at distance r from the centre, is e times (1/r minus 1/b).

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  e\xp\left(\dfrac{1}{a} - \dfrac{1}{b} \right)

  The potential of the inner sphere, which is uniform throughout it, is e times (1/a minus 1/b).

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  e = \frac{1}{\dfrac{1}{a} - \dfrac{1}{b}} = \frac{ab}{b - a}

  The capacity of the inner sphere, the charge needed to raise its potential to unity, equals ab divided by (b - a).

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  \sigma = \frac{A - B}{4 \pi c}

  For two parallel plane electrodes at distance c with potentials A and B, the surface density on the upper plane is the potential difference divided by 4 pi c.

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  \xp\dfrac{A-B}{c}

  The electric force between two parallel planes, away from their edges, has magnitude (A - B) divided by c and acts from A to B.

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  e = \frac{A - B}{4 \pi c} S

  The charge on an area S cut from the upper plane is the surface density times S.

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  Q = \tfrac{1}{2}\{Ae + B(-e)\} = \tfrac{1}{2} (A - B)e

  The electrical energy of the two equal and opposite charges on the parallel planes is half the potential difference times the charge.

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  Q = \frac{2 \pi}{S} e^2c

  The electrical energy of the parallel-plane pair, written in terms of the charge e and the separation c.

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  Q' = \frac{2 \pi}{S} e^2c'

  With the charges kept fixed and the separation increased to c', the electrical energy becomes this value.

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  Q' - Q = \frac{2 \pi}{S} e^2(c' - c)

  The increase in electrical energy when the planes are pulled apart is 2 pi e squared over S times the increase in separation.

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  F(c' - c) = \frac{2 \pi}{S} e^2(c' - c)

  The work done by external agency in pulling the planes apart equals the attraction F times the increase in separation, which equals the increase in electrical energy.

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  F = \frac{2 \pi}{S} e^2

  The electric attraction between two parallel plane areas S carrying equal and opposite charge e is 2 pi e squared divided by S.

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  e = \sqrt\frac{FS}{2 \pi}

  The charge on an area S can be measured from the dynamically measured attraction F between the planes.

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  A - B = 4 \pi c \frac{e}{S} = c \sqrt{\frac{8 \pi F}{S}}

  The potential difference between the two planes equals 4 pi c times the surface density, and also c times the square root of 8 pi F over S.

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  V = \xp\dfrac{E}{r}

  The potential at distance r from a single small electrified body with charge E.

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  r = \xp\dfrac{E}{V}

  The radius of the equipotential sphere of potential V around a single charge E.

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  V_1 + V_2 = V

  The potential due to two centres of force is the sum of the potentials due to each.

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  2 \pi E(1 - cos \theta)

  The surface-integral of induction through the part of a surface cut off by a cone of half-angle theta from a centre of charge E is 2 pi E times (1 minus cos theta).

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  E(1 - cos \theta) = 2 \Psi

  Defines Psi as half the induction through the bounded surface, so that the induction equals E times (1 minus cos theta) over 2.

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  \theta = cos^{-1}\left(1 - 2 \frac{\Psi}{E}\right)

  The angle of a line of force with the axis is the inverse cosine of 1 minus 2 Psi over E.

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  1 - 2 \xp\dfrac{\Psi}{E}

  The cosine of the angle between the asymptote of a line of force and the axis of a finite system, where E is the total electrification.

Problems

No exercises in this chapter.