A Treatise on Electricity and Magnetism/Part I/Chapter V

by James Clerk Maxwell

CHAPTER V.

MECHANICAL ACTION BETWEEN ELECTRIFIED BODIES.

103.] Let $V=C$ be any closed equipotential surface, $C$ being a particular value of a function $V$, the form of which we suppose known at every point of space. Let the value of $V$ on the outside of this surface be $V_{1}$, and on the inside $V_{2}$. Then, by Poisson’s equation

${\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}+{\frac {d^{2}V}{dz^{2}}}+4\pi \rho =0$

(1)

we can determine the density $\rho _{1}$ at every point on the outside, and the density $\rho _{2}$ at every point on the inside of the surface. We shall call the whole electrified system thus explored on the outside $E_{1}$, and that on the inside $E_{2}$. The actual value of $V$ arises from the combined action of both these systems.

Let $R$ be the total resultant force at any point arising from the action of $E_{1}$ and $E_{2}$, $R$ is everywhere normal to the equipotential surface passing through the point.

Now let us suppose that on the equipotential surface $V=C$ electricity is distributed so that at any point of the surface at which the resultant force due to $E_{1}$ and $E_{2}$ reckoned outwards is $R$, the surface-density is $\rho$, with the condition

$R=4\pi \sigma \ ;$

(2)

and let us call this superficial distribution the electrified surface $S$, then we can prove the following theorem relating to the action of this electrified surface.

If any equipotential surface belonging to a given electrified system be coated with electricity, so that at each point the surface-density $\sigma ={\frac {R}{4\pi }}$, where $R$ is the resultant force, due to the original electrical system, acting outwards from that point of the surface, then the potential due to the electrified surface at any point on the outside of that surface will be equal to the potential at the same point due to that part of the original system which was on the inside of the surface, and the potential due to the electrified surface at any point on the inside added to that due to the part of the original system on the outside will be equal to $C$, the potential of the surface.

For let us alter the original system as follows :

Let us leave everything the same on the outside of the surface, but on the inside let us make $V_{2}$ everywhere equal to $C$, and let us do away with the electrified system $E_{2}$ on the inside of the surface, and substitute for it a surface-density $\rho$ at every point of the surface $S$, such that

$R=4\pi \sigma .\,$

(3)

Then this new arrangement will satisfy the characteristics of $V$ at every point.

For on the outside of the surface both the distribution of electricity and the value of $V$ are unaltered, therefore, since $V$ originally satisfied Laplace’s equation, it will still satisfy it.

On the inside $V$ is constant and $\rho$ zero. These values of $V$ and $\rho$ also satisfy the characteristic equations.

At the surface itself, if $V_{1}$ is the potential at any point on the outside and $V_{2}$ that on the inside, then, if l, m, n are the direction-cosines of the normal to the surface reckoned outwards,

$l{\frac {dV_{1}}{dx}}+m{\frac {dV_{1}}{dy}}+n{\frac {dV_{1}}{dz}}=-R=-4\pi \sigma ;$

(4)

and on the inside the derivatives of $V$ vanish, so that the superficial characteristic

$l\left({\frac {dV_{1}}{dx}}-{\frac {dV_{2}}{dx}}\right)+m\left({\frac {dV_{1}}{dy}}-{\frac {dV_{2}}{dy}}\right)+n\left({\frac {dV_{1}}{dz}}-{\frac {dV_{2}}{dz}}\right)+4\pi \sigma =0$

(5)

is satisfied at every point of the surface.

Hence the new distribution of potential, in which it has the old value on the outside of the surface and a constant value on the inside, is consistent with the new distribution of electricity, in which the electricity in the space within the surface is removed and a distribution of electricity on the surface is substituted for it. Also, since the original value of $V_{1}$ vanishes at infinity, the new value, which is the same outside the surface, also fulfils this condition, and therefore the new value of $V$ is the sole and only value of $V$ belonging to the new arrangement of electricity.

On the Mechanical Action and Reaction of the Systems $E_{1}$ and $E_{2}$.

104.] If we now suppose the equipotential surface $V=C$ to become rigid and capable of sustaining the action of forces, we may prove the following theorem.

If on every element $dS$ of an equipotential surface a force ${\frac {1}{8\pi }}R^{2}dS$ be made to act in the direction of the normal reckoned outwards, where $R$ is the ‚electrical resultant force‘ along the normal, then the total statical effect of these forces on the surface considered as a rigid shell will be the same as the total statical effect of the electrical action of the electrified system $E_{1}$ outside the shell on the electrified system $E_{2}$ inside the shell, the parts of the interior system $E_{2}$ being supposed rigidly connected together.

We have seen that the action of the electrified surface in the last theorem on any external point was equal to that of the internal system $E_{2}$, and, since action and reaction are equal and opposite, the action of any external electrified body on the electrified surface, considered as a rigid system, is equal to that on the internal system $E_{2}$. Hence the statical action of the external system $E_{1}$ on the electrified surface is equal in all respects to the action of $E_{1}$ on the internal system $E_{2}$.

But at any point just outside the electrified surface the resultant force is $R$ in a direction normal to the surface, and reckoned positive when it acts outwards. The resultant inside the surface is zero, therefore, by Art. 79, the resultant force acting on the element $dS$ of the electrified surface is ${\frac {1}{2}}R\sigma dS$, where $\sigma$ is the surface- density.

Substituting the value of $\sigma$ in terms of $R$ from equation (2), and denoting by p dS the resultant force on the electricity spread over the element $dS$, we find

$p\ dS={\frac {1}{8\pi }}R^{2}dS$

This force always acts along the normal and outwards, whether $R$ be positive or negative, and may be considered as equal to a pressure $p={\frac {1}{8\pi }}R^{2}$ acting on the surface from within, or to a tension of the same numerical value acting from without.^[1]

Now $R$ is the resultant due to the combined action of the external system $E_{1}$ and the electrification of the surface $S$. Hence the effect of the pressure $p$ on each element of the inside of the surface considered as a rigid body is equivalent to this combined action.

But the actions of the different parts of the surface on each other form a system in equilibrium, therefore the effect of the pressure $p$ on the rigid shell is equivalent in all respects to the electric attraction of $E_{1}$ on the shell, and this, as we have before shewn, is equivalent to the electric attraction of $E_{1}$ on $E_{2}$ considered as a rigid system.

If we had supposed the pressure $p$ to act on the outside of the shell, the resultant effect would have been equal and opposite, that is, it would have been statically equivalent to the action of the internal system $E_{2}$ on the external system $E_{1}$.

Let us now take the case of two electrified systems $E_{1}$ and $E_{2}$, such that two equipotential surfaces $V=C_{1}$ and $V=C_{2}$, which we shall call $S_{1}$ and $S_{1}$ respectively, can be described so that $E_{1}$ is exterior to $S_{1}$, and $S_{1}$ surrounds $S_{2}$, and $E_{2}$ lies within $S_{2}$.

Then if $R_{1}$ and $R_{2}$ represent the resultant force at any point of $S_{1}$ and $S_{2}$ respectively, and if we make

$p_{1}={\frac {1}{8\pi }}R_{1}^{2}$ and $p_{2}={\frac {1}{8\pi }}R_{2}^{2}$

the mechanical action between $E_{1}$ and $E_{2}$ is equivalent to that between the shells $S_{1}$ and $S_{2}$, supposing every point of $S_{1}$ pressed inwards, that is, towards $S_{2}$ with a pressure $p_{1}$, and every point of $S_{2}$ pressed outwards, that is, towards $S_{1}$ with a pressure $p_{2}$.

105.] According to the theory of action at a distance the action between $E_{1}$ and $E_{2}$ is really made up of a system of forces acting in straight lines between the electricity in $E_{1}$ and that in $E_{2}$, and the actual mechanical effect is in complete accordance with this theory.

There is, however, another point of view from which we may examine the action between $E_{1}$ and $E_{2}$. When we see one body acting on another at a distance, before we assume that the one acts directly on the other we generally inquire whether there is any material connexion between the two bodies, and if we find strings, or rods, or framework of any kind, capable of accounting for the observed action between the bodies, we prefer to explain the action by means of the intermediate connexions, rather than admit the notion of direct action at a distance.

Thus when two particles are connected by a straight or curved rod, the action between the particles is always along the line joining them, but we account for this action by means of a system of internal forces in the substance of the rod. The existence of these internal forces is deduced entirely from observation of the effect of external forces on the rod, and the internal forces themselves are generally assumed to be the resultants of forces which act between particles of the rod. Thus the observed action between two distant particles is, in this instance, removed from the class of direct actions at a distance by referring it to the intervention of the rod ; the action of the rod is explained by the existence of internal forces in its substance ; and the internal forces are explained by means of forces assumed to act between the particles of which the rod is composed, that is, between bodies at distances which though small must be finite.

The observed action at a considerable distance is therefore explained by means of a great number of forces acting between bodies at very small distances, for which we are as little able to account as for the action at any distance however great.

Nevertheless, the consideration of the phenomenon, as explained in this way, leads us to investigate the properties of the rod, and to form a theory of elasticity which we should have overlooked if we had been satisfied with the explanation by action at a distance.

106.] Let us now examine the consequence of assuming that the action between electrified bodies can be explained by the intermediate action of the medium between them, and let us ascertain what properties of the medium will account for the observed action.

We have first to determine the internal forces in the medium, and afterwards to account for them if possible.

In order to determine the internal forces in any case we proceed as follows :

Let the system $M$ be in equilibrium under the action of the system of external forces $F$. Divide $M$ by an imaginary surface into two parts, $M_{1}$ and $M_{2}$, and let the systems of external forces acting on these parts respectively be $F_{1}$ and $F_{2}$. Also let the internal forces acting on $M_{1}$ in consequence of its connexion with $M_{2}$ be called the system $I$.

Then, since $M_{1}$ is in equilibrium under the action of $F_{1}$ and $I$, it follows that $I$ is statically equivalent to $F_{1}$ reversed.

In the case of the electrical action between two electrified systems $E_{1}$ and $E_{1}$, we described two closed equipotential surfaces entirely surrounding $E_{2}$ and cutting it off from $E_{1}$, and we found that the application of a certain normal pressure at every point of the inner side of the inner surface, and on the outer side of the outer surface, would, if these surfaces were each rigid, act on the outer surface with a resultant equal to that of the electrical forces on the outer system $E_{1}$, and on the inner surface with a resultant equal to that of the electrical forces on the inner system.

Let us now consider the space between the surfaces, and let us suppose that at every point of this space there is a tension in the direction of $R$ and equal to ${\frac {1}{8\pi }}R^{2}$ per unit of area. This tension will act on the two surfaces in the same way as the pressures on the other side of the surfaces, and will therefore account for the action between $E_{1}$ and $E_{2}$, so far as it depends on the internal force in the space between $S_{1}$ and $S_{2}$.

Let us next investigate the equilibrium of a portion of the shell bounded by these surfaces and separated from the rest by a surface everywhere perpendicular to the equipotential surfaces. We may suppose this surface generated by describing any closed curve on $S_{1}$, and drawing from every point of this curve lines of force till they meet $S_{2}$.

The figure we have to consider is therefore bounded by the two equipotential surfaces $S_{1}$ and $S_{2}$, and by a surface through which there is no induction, which we may call $S_{0}$.

Let us first suppose that the area of the closed curve on $S_{1}$ is very small, call it $dS_{1}$ and that $C_{2}=C_{1}+dV$.

The portion of space thus bounded may be regarded as an element of volume. If $\nu$ is the normal to the equipotential surface, and $dS$ the element of that surface, then the volume of this element is ultimately $dS\ d\nu$.

The induction through $dS_{1}$ is $R\ dS_{1}$, and since there is no induction through $S_{0}$, and no free electricity within the space considered, the induction through the opposite surface $dS_{2}$ will be equal and opposite, considered with reference to the space within the closed surface.

There will therefore be a quantity of electricity

$e_{1}=-{\frac {1}{4\pi }}R_{1}dS_{1}$

on the first equipotential surface, and a quantity

$e_{2}={\frac {1}{4\pi }}R_{2}dS_{2}$

on the second equipotential surface, with the condition

$e_{1}+e_{2}=0\,$

Let us next consider the resultant force due to the action of the electrified systems on these small electrified surfaces.

The potential within the surface $S_{1}$ is constant and equal to $C_{1}$, and without the surface $S_{2}$ it is constant and equal to $C_{2}$. In the shell between these surfaces it is continuous from $C_{1}$ to $C_{2}$.

Hence the resultant force is zero except within the shell.

The electrified surface of the shell itself will be acted on by forces which are the arithmetical means of the forces just within and just without the surface, that is, in this case, since the resultant force outside is zero, the force acting on the superficial electrification is one-half of the resultant force just within the surface.

Hence, if $X\ dS\ d\nu$ be the total moving force resolved parallel to $x$, due to the electrical action on both the electrified surfaces of the element $dS\ dv$,

$X\ dS\ d\nu =-{\frac {1}{2}}\left(e_{1}{\frac {dV_{1}}{dx}}+e_{2}{\frac {dV_{2}}{dx}}\right)$

where the suffixes denote that the derivatives of $\nu$ are to be taken at $dS_{1}$ and $dS_{2}$ respectively.

Let $l,m,n$ be the direction-cosines of $V$, the normal to the equipotential surface, then making

$dx=l\ d\nu ,\ dy=m\ d\nu ,\ \mathrm {and} \ dz=n\ d\nu ,$ $\left({\frac {dV}{dx}}\right)_{2}=\left({\frac {dV}{dx}}\right)_{1}+\left(l{\frac {d^{2}V}{dx^{2}}}+m{\frac {d^{2}V}{dx\ dy}}+n{\frac {d^{2}V}{dx\ dz}}\right)d\nu +\mathrm {etc} .;$

and since $e_{2}=-e_{1}$, we may write the value of $X$

$X\ dS\ d\nu ={\frac {1}{2}}e_{1}{\frac {d}{dx}}\left(l{\frac {dV}{dx}}+m{\frac {dV}{dy}}+n{\frac {dV}{dz}}\right)d\nu .$

But

$e_{1}=-{\frac {1}{4\pi }}R\ dS$ and $\left(l{\frac {dV}{dx}}+m{\frac {dV}{dy}}+n{\frac {dV}{dz}}\right)=-R;$

therefore

$X\ dS\ d\nu ={\frac {1}{8\pi }}R{\frac {dR}{dx}}dS\ d\nu ;$

or, if we write

$p={\frac {1}{8\pi }}R^{2}={\frac {1}{8\pi }}\left(\left({\frac {dV}{dx}}\right)^{2}+\left({\frac {dV}{dy}}\right)^{2}+\left({\frac {dV}{dz}}\right)^{2}\right),$

then

$X={\frac {1}{2}}{\frac {dp}{dx}},\ Y={\frac {1}{2}}{\frac {dp}{dy}},\ Z={\frac {1}{2}}{\frac {dp}{dz}};$

or the force in any direction on the element arising from the action of the electrified system on the two electrified surfaces of the element is equal to half the rate of increase of $p$ in that direction multiplied by the volume of the element.

This result is the same if we substitute for the forces acting on the electrified surfaces an imaginary force whose potential is $-{\tfrac {1}{2}}p$, acting on the whole volume of the element and soliciting it to move so as to increase ${\tfrac {1}{2}}p$.

If we now return to the case of a figure of finite size, bounded by the equipotential surfaces $S_{1}$ and $S_{2}$ and by the surface of no induction $S_{0}$, we may divide the whole space into elements by a series of equipotential surfaces and two series of surfaces of no induction. The charges of electricity on those faces of the elements which are in contact will be equal and opposite, so that the total effect will be that due to the electrical forces acting on the charges on the surfaces $S_{1}$ and $S_{2}$, and by what we have proved this will be the same as the action on the whole volume of the figure due to a system of forces whose potential is $-{\frac {1}{2}}p$.

But we have already shewn that these electrical forces are equivalent to a tension $p$ applied at all points of the surfaces $S_{1}$ and $S_{2}$. Hence the effect of this tension is to pull the figure in the direction in which $p$ increases. The figure therefore cannot be in equilibrium unless some other forces act on it.

Now we know that if a hydrostatic pressure $p$ is applied at every point of the surface of any closed figure, the effect is equal to that of a system of forces acting on the whole volume of the figure and having a potential $p$. In this case the figure is pushed in the direction in which $p$ diminishes.

We can now arrange matters so that the figure shall be in equilibrium.

At every point of the two equipotential surfaces $S_{1}$ and $S_{2}$, let a tension = $p$ be applied, and at every point of the surface of no induction $S_{0}$ let a pressure = $p$ be applied. These forces will keep the figure in equilibrium.

For the tension $p$ may be considered as a pressure $p$ combined with a tension 2$p$. We have then a hydrostatic pressure $p$ acting at every point of the surface, and a tension \frac{2p}{} acting on $S_{1}$ and $S_{2}$ only.

The effect of the tension $2p$ at every point of $S_{1}$ and $S_{2}$ is double of that which we have just calculated, that is, it is equal to that of forces whose potential is $-p$ acting on the whole volume of the figure. The effect of the pressure $p$ acting on the whole surface is by hydrostatics equal and opposite to that of this system of forces, and will keep the figure in equilibrium.

107.] We have now determined a system of internal forces in the medium which is consistent with the phenomena so far as we have examined them. We have found that in order to account for the electric attraction between distant bodies without admitting direct action, we must assume the existence of a tension $p$ at every point of the medium in the direction of the resultant force $R$ at that point. In order to account for the equilibrium of the medium itself we must further suppose that in every direction perpendicular to $R$ there is a pressure $p$^[2].

By establishing the necessity of assuming these internal forces in the theory of an electric medium, we have advanced a step in that theory which will not be lost though we should fail in accounting for these internal forces, or in explaining the mechanism by which they can be maintained in air, glass, and other dielectric media.

We have seen that the internal stresses in solid bodies can be ascertained with precision, though the theories which account for these stresses by means of molecular forces may still be doubtful. In the same way we may estimate these internal electrical forces before we are able to account for them.

In order, however, that it may not appear as if we had no explanation of these internal forces, we shall shew that on the ordinary theory they must exist in a shell bounded by two equipotential surfaces, and that the attractions and repulsions of the electricity on the surfaces of the shell are sufficient to account for them.

Let the first surface $S_{1}$ be electrified so that the surface-density is

$\sigma _{1}=-{\frac {1}{4\pi }}R_{1},$

and the second surface $S_{2}$ so that the surface-density is

$\sigma _{2}={\frac {1}{4\pi }}R_{2};$

then, if we suppose that the value of $V$ is $C_{1}$ at every point within $S_{1}$ and $C_{2}$ at every point outside of $S_{2}$, the value of $V$ between these surfaces remaining as before, the characteristic equation of $V$ will be satisfied everywhere, and $V$ is therefore the true value of the potential.

We have already shewn that the outer and inner surfaces of the shell will be pulled towards each other with a force the value of which referred to unit of surface is $p$, or in other words, there is a tension $p$ in the substance of the shell in the direction of the lines of force.

If we now conceive the shell divided into two segments by a surface of no induction, the two parts will experience electrical forces the resultants of which will tend to separate the parts with a force equivalent to the resultant force due to a pressure $p$ acting on every part of the surface of no induction which divides them.

This illustration is to be taken merely as an explanation of what is meant by the tension and pressure, not as a physical theory to account for them.

108.] We have next to consider whether these internal forces are capable of accounting for the observed electrical forces in every case, as well as in the case where a closed equipotential surface can be drawn surrounding one of the electrified systems.

The statical theory of internal forces has been investigated by writers on the theory of elasticity. At present we shall require only to investigate the effect of an oblique tension or pressure on an element of surface.

Let $p$ be the value of a tension referred to unit of a surface to which it is normal, and let there be no tension or pressure in any direction normal to $p$. Let the direction-cosines of p be l, m, n. Let dy dz be an element of surface normal to the axis of x, and let the effect of the internal force be to urge the parts on the positive side of this element with a force whose components are

$p_{xx}dy\ dz$	in the direction of	x,
$p_{xy}dy\ dz$		y, and
$p_{xz}dy\ dz$		z.

From every point of the boundary of the element dy dz let lines be drawn parallel to the direction of the tension $p$, forming a prism whose axis is in the line of tension, and let this prism be cut by a plane normal to its axis.

The area of this section will be l dy dx, and the whole tension upon it will be pl dy dz and since there is no action on the sides of the prism, which are normal to p, the force on the base dy dz must be equivalent to the force pl dy dx acting in the direction (l, m, n). Hence the component in the direction of x,

$p_{xx}dy\ dz=pl^{2}dy\ dz\,$ ; or

${\begin{array}{lll}&&p_{xx}=pl^{2}.\\\mathrm {Similarly} &&p_{xy}=plm,\\&&p_{xz}=pln.\end{array}}$

(1)

If we now combine with this tension two tensions $p'$ and $p''$ in directions $(l',m',n')$ and $(l'',m'',n'')$ respectively, we shall have

${\begin{array}{llll}p_{xx}&=pl^{2}&+p'l'^{2}&+p''l''^{2}\\p_{xy}&=plm&+p'l'm'&+p''l''m''\\p_{xz}&=pln&+p'l'n'&+p''l''n''\end{array}}$

(2)

In the case of the electrical tension and pressure the pressures are numerically equal to the tension at every point, and are in directions at right angles to the tension and to each other. Hence, putting

$p'=p''=-p,\,$

(3)

$l^{2}+l'^{2}+l''^{2}=1,\ lm+l'm'+l''m''=0,\ ln+l'n'+l''n''=0,$

(4)

we find

${\begin{array}{l}p_{xx}=\left(2l^{2}-1\right)p,\\p_{xy}=2lmp,\\p_{xz}=2lnp.\end{array}}$

(5)

for the action of the combined tension and pressures.

Also, since $p={\frac {1}{8\pi }}R^{2}$, where $R$ denotes the resultant force, and since $Rl=X,\ Rm=Y,\ Rn=Z,$,

${\begin{array}{ll}p_{xx}&={\frac {1}{8\pi }}\left(X^{2}-Y^{2}-Z^{2}\right),\\\\p_{xy}&={\frac {1}{8\pi }}2XY=p_{yx},\\\\p_{xz}&={\frac {1}{8\pi }}2XZ=p_{zx};\end{array}}$

(6)

where X, Y, Z are the components of $R$, the resultant electromotive force.

The expressions for the component internal forces on surfaces normal to $y$ and $z$ may be written down from symmetry.

To determine the conditions of equilibrium of the element dx dy dz.

This element is bounded by the six planes perpendicular to the axes of coordinates passing through the points (x, y, z) and ($x+dx,\ y+dy,\ z+dz$).

The force in the direction of $x$ which acts on the first face dy dz is $-p_{xx}dy\ dz$, tending to draw the element towards the negative side. On the second face dy dz, for which $x$ has the value $x+dx$, the tension $p_{xx}$ has the value

$p_{xx}dy\ dz+\left({\frac {d}{dx}}p_{xx}\right)dx\ dy\ dz$

and this tension tends to draw the element in the positive direction. If we next consider the two faces dz dx with respect to the tangential forces urging them in the direction of $x$, we find the force on the first face $-p_{yx}dz\ dx$, and that on the second

$p_{yx}dz\ dx+\left({\frac {d}{dy}}p_{yz}\right)dz\ dx\ dy$

Similarly for the faces dx dy, we find that a force $-p_{zx}dx\ dy$ acts on the first face, and

$p_{zx}dx\ dy+\left({\frac {d}{dz}}p_{zx}\right)dx\ dy\ dz$

on the second in the direction of $x$.

If $\xi \ dx\ dy\ dz$ denotes the total effect of all these internal forces acting parallel to the axis of $x$ on the six faces of the element, we find

$\xi \ dx\ dy\ dz=\left({\frac {d}{dx}}p_{xx}+{\frac {d}{dy}}p_{yx}+{\frac {d}{dz}}p_{zx}\right)dx\ dy\ dz;$

or, denoting by $\xi$ the internal force, referred to unit of volume, and resolved parallel to the axis of $x$,

$\xi ={\frac {d}{dx}}p_{xx}+{\frac {d}{dy}}p_{yz}+{\frac {d}{dz}}p_{zx},$

(7)

with similar expressions for $\eta$ and $\zeta$, the component forces in the other directions^[3].

Differentiating the values of $p_{xx},\ p_{yz}$, and $p_{zx}$ given in equations (6), we find

$\xi ={\frac {1}{4\pi }}X\left({\frac {dX}{dx}}+{\frac {dY}{dy}}+{\frac {dZ}{dz}}\right).$

(8)

But by Art. 77

$\left({\frac {dX}{dx}}+{\frac {dY}{dy}}+{\frac {dZ}{dz}}\right)=4\pi \rho .$

(9)

${\begin{array}{lll}\mathrm {Hence} &&\xi =\rho X.\\\mathrm {Similarly} &&\eta =\rho Y,\\&&\zeta =\rho Z.\end{array}}$

(10)

Thus, the resultant of the tensions and pressures which we have supposed to act upon the surface of the element is a force whose components are the same as those of the force, which, in the ordinary theory, is ascribed to the action of electrified bodies on the electricity within the element.

If, therefore, we admit that there is a medium in which there is maintained at every point a tension $p$ in the direction of the resultant electromotive force $R$, and such that $R^{2}=8\pi p$, combined with an equal pressure $p$ in every direction at right angles to the resultant $R$, then the mechanical effect of these tensions and pressures on any portion of the medium, however bounded, will be identical with the mechanical effect of the electrical forces according to the ordinary theory of direct action at a distance.

109.] This distribution of stress is precisely that to which Faraday was led in his investigation of induction through dielectrics. He sums up in the following words :—

‚(1297) The direct inductive force, which may be conceived to be exerted in lines between the two limiting and charged conducting surfaces, is accompanied by a lateral or transverse force equivalent to a dilatation or repulsion of these representative lines (1224.); or the attracting force which exists amongst the particles of the dielectric in the direction of the induction is accompanied by a repulsive or a diverging force in the transverse direction.

‚(1298) Induction appears to consist in a certain polarized state of the particles, into which they are thrown by the electrified body sustaining the action, the particles assuming positive and negative points or parts, which are symmetrically arranged with respect to each other and the inducting surfaces or particles. The state must be a forced one, for it is originated and sustained only by force, and sinks to the normal or quiescent state when that force is removed. It can be continued only in insulators by the same portion of electricity, because they only can retain this state of the particles.

This is an exact account of the conclusions to which we have been conducted by our mathematical investigation. At every point of the medium there is a state of stress such that there is tension along the lines of force and pressure in all directions at right angles to these lines, the numerical magnitude of the pressure being equal to that of the tension, and both varying as the square of the resultant force at the point.

The expression ‚electric tension‘ has been used in various senses by different writers. I shall always use it to denote the tension along the lines of force, which, as we have seen, varies from point to point, and is always proportional to the square of the resultant force at the point.

110.] The hypothesis that a state of stress of this kind exists in a fluid dielectric, such as air or turpentine, may at first sight appear at variance with the established principle that at any point in a fluid the pressures in all directions are equal. But in the deduction of this principle from a consideration of the mobility and equilibrium of the parts of the fluid it is taken for granted that no action such as that which we here suppose to take place along the lines of force exists in the fluid. The state of stress which we have been studying is perfectly consistent with the mobility and equilibrium of the fluid, for we have seen that, if any portion of the fluid is devoid of electric charge, it experiences no resultant force from the stresses on its surface, however intense these may be. It is only when a portion of the fluid becomes charged, that its equilibrium is disturbed by the stresses on its surface, and we know that in this case it actually tends to move. Hence the supposed state of stress is not inconsistent with the equilibrium of a fluid dielectric.

The quantity $Q$, which was investigated in Thomson’s theorem, Art. 98, may be interpreted as the energy in the medium due to the distribution of stress. It appears from that theorem that the distribution of stress which satisfies the ordinary conditions also makes $Q$ an absolute minimum. Now when the energy is a minimum for any configuration, that configuration is one of equilibrium, and the equilibrium is stable. Hence the dielectric, when subjected to the inductive action of electrified bodies, will of itself take up a state of stress distributed in the way we have described.

It must be carefully borne in mind that we have made only one step in the theory of the action of the medium. We have supposed it to be in a state of stress, but we have not in any way accounted for this stress, or explained how it is maintained. This step, however, seems to me to be an important one, as it explains, by the action of the consecutive parts of the medium, phenomena which were formerly supposed to be explicable only by direct action at a distance.

111.] I have not been able to make the next step, namely, to account by mechanical considerations for these stresses in the dielectric. I therefore leave the theory at this point, merely stating what are the other parts of the phenomenon of induction in dielectrics.

I. Electric Displacement. When induction takes place in a dielectric a phenomenon takes place which is equivalent to a displacement of electricity in the direction of the induction. For instance, in a Leyden jar, of which the inner coating is charged positively and the outer coating negatively, the displacement in the substance of the glass is from within outwards.

Any increase of this displacement is equivalent, during the time of increase, to a current of positive electricity from within outwards, and any diminution of the displacement is equivalent to a current in the opposite direction.

The whole quantity of electricity displaced through any area of a surface fixed in the dielectric is measured by the quantity which we have already investigated (Art. 75) as the surface-integral of induction through that area, multiplied by ${\frac {1}{4\pi }}K$, where $K$ is the specific inductive capacity of the dielectric.

II. Superficial Electrification of the Particles of the Dielectric. Conceive any portion of the dielectric, large or small, to be separated (in imagination) from the rest by a closed surface, then we must suppose that on every elementary portion of this surface there is an electrification measured by the total displacement of electricity through that element of surface reckoned inwards.

In the case of the Leyden jar of which the inner coating is charged positively, any portion of the glass will have its inner side charged positively and its outer side negatively. If this portion be entirely in the interior of the glass, its superficial electrification will be neutralized by the opposite electrification of the parts in contact with it, but if it be in contact with a conducting body which is incapable of maintaining in itself the inductive state, the superficial electrification will not be neutralized, but will constitute that apparent electrification which is commonly called the Electrification of the Conductor.

The electrification therefore at the bounding surface of a conductor and the surrounding dielectric, which on the old theory was called the electrification of the conductor, must be called in the theory of induction the superficial electrification of the surrounding dielectric.

According to this theory, all electrification is the residual effect of the polarization of the dielectric. This polarization exists throughout the interior of the substance, but it is there neutralized by the juxtaposition of oppositely electrified parts, so that it is only at the surface of the dielectric that the effects of the electrification become apparent.

The theory completely accounts for the theorem of Art. 77, that the total induction through a closed surface is equal to the total quantity of electricity within the surface multiplied by 4$\pi$. For what we have called the induction through the surface is simply the electric displacement multiplied by 4$\pi$, and the total displacement outwards is necessarily equal to the total electrification within the surface.

The theory also accounts for the impossibility of communicating an absolute charge to matter. For every particle of the dielectric is electrified with equal and opposite charges on its opposite sides, if it would not be more correct to say that these electrifications are only the manifestations of a single phenomenon, which we may call Electric Polarization.

A dielectric medium, when thus polarized, is the seat of electrical energy, and the energy in unit of volume of the medium is numerically equal to the electric tension on unit of area, both quantities being equal to half the product of the displacement and the resultant electromotive force, or

$p={\frac {1}{2}}{\mathfrak {DE}}={\frac {1}{8\pi }}K{\mathfrak {E}}^{2}={\frac {2\pi }{K}}{\mathfrak {D}}^{2},$

where $p$ is the electric tension, ${\mathfrak {D}}$ the displacement, ${\mathfrak {E}}$ the electro motive force, and $K$ the specific inductive capacity.

If the medium is not a perfect insulator, the state of constraint, which we call electric polarization, is continually giving way. The medium yields to the electromotive force, the electric stress is relaxed, and the potential energy of the state of constraint is converted into heat. The rate at which this decay of the state of polarization takes place depends on the nature of the medium. In some kinds of glass, days or years may elapse before the polarization sinks to half its original value. In copper, this change may occupy less than the billionth of a second.

We have supposed the medium after being polarized to be simply left to itself. In the phenomenon called the electric current the constant passage of electricity through the medium tends to restore the state of polarization as fast as the conductivity of the medium allows it to decay. Thus the external agency which maintains the current is always doing work in restoring the polarization of the medium, which is continually becoming relaxed, and the potential energy of this polarization is continually becoming transformed into heat, so that the final result of the energy expended in maintaining the current is to raise the temperature of the conductor.

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1. See Sir W. Thomson ‚On the Attractions of Conducting and Non-conducting Electrified Bodies‘, Cambridge Mathematical Journal, May 1843, and Reprint, Art. VII, § 147.
2. See Faraday, Exp. Res. (1224) and (1297).
3. This investigation may be compared with that of the ‚equation of continuity in hydrodynamics‘, and with others in which the effect on an element of volume is deduced from the values of certain quantities at its bounding surface.