A Treatise on Electricity and Magnetism/Volume 1/Part 2/Chapter 10

by James Clerk Maxwell

CHAPTER X.

CONDUCTION IN DIELECTRICS.

325.] We have seen that when electromotive force acts on a dielectric medium it produces in it a state which we have called electric polarization, and which we have described as consisting of electric displacement within the medium in a direction which, in isotropic media,, coincides with that of the electromotive force, combined with a superficial charge on every element of volume into which we may suppose the dielectric divided, which is negative on the side towards which the force acts, and positive on the side from which it acts.

When electromotive force acts on a conducting medium it also produces what is called an electric current.

Now dielectric media, with very few, if any, exceptions, are also more or less imperfect conductors, and many media which are not good insulators exhibit phenomena of dielectric induction. Hence we are led to study the state of a medium in which induction and conduction are going on at the same time.

For simplicity we shall suppose the medium isotropic at every point, but not necessarily homogeneous at different points. In this case, the equation of Poisson becomes, by Art. 83,

${\frac {d}{dx}}\left(K{\frac {dV}{dx}}\right)+{\frac {d}{dy}}\left(K{\frac {dV}{dy}}\right)+{\frac {d}{dz}}\left(K{\frac {dV}{dz}}\right)+4\pi \rho =0$

(1)

where $K$ is the 'specific inductive capacity.'

The 'equation of continuity' of electric currents becomes

${\frac {d}{dx}}\left({\frac {1}{r}}{\frac {dV}{dx}}\right)+{\frac {d}{dy}}\left({\frac {1}{r}}{\frac {dV}{dy}}\right)+{\frac {d}{dz}}\left({\frac {1}{r}}{\frac {dV}{dz}}\right)-{\frac {d\rho }{dt}}=0$

(2)

where $r$ is the specific resistance referred to unit of volume.

When $K$ or $r$ is discontinuous, these equations must be transformed into those appropriate to surfaces of discontinuity.

In a strictly homogeneous medium $r$ and $K$ are both constant, so that we find

${\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}+{\frac {d^{2}V}{dz^{2}}}=-4\pi {\frac {\rho }{K}}=r{\frac {d\rho }{dt}},$

(3)

whence

$\rho =Ce^{-{\frac {4\pi }{Kr}}};$

(4)

or, if we put

$T={\frac {Kr}{4\pi }},\quad \quad \rho =Ce^{-{\frac {t}{T}}}$

(5)

This result shews that under the action of any external electric forces on a homogeneous medium, the interior of which is originally charged in any manner with electricity, the internal charges will die away at a rate which does not depend on the external forces, so that at length there will be no charge of electricity within the medium, after which no external forces can either produce or maintain a charge in any internal portion of the medium, provided the relation between electromotive force, electric polarization and conduction remains the same. When disruptive discharge occurs these relations cease to be true, and internal charge may be produced.

On Conduction through a Condenser.

326.] Let $C$ be the capacity of a condenser, $R$ its resistance, and $E$ the electromotive force which acts on it, that is, the difference of potentials of the surfaces of the metallic electrodes.

Then the quantity of electricity on the side from which the electromotive force acts will be $CE,$ and the current through the substance of the condenser in the direction of the electromotive force will be ${\frac {E}{R}}.$

If the electrification is supposed to be produced by an electromotive force $E$ acting in a circuit of which the condenser forms part, and if ${\frac {dQ}{dt}}$represents the current in that circuit, then

${\frac {dQ}{dt}}={\frac {E}{R}}+C{\frac {dE}{dt}}.$

(6)

Let a battery of electromotive force $E_{0}$ and resistance $r$ be introduced into this circuit, then

${\frac {dQ}{dt}}={\frac {E_{0}-E}{r}}={\frac {E}{R}}+C{\frac {dE}{dt}}.$

(7)

Hence, at any time $t_{1},$

$E(=E_{1})=E_{0}{\frac {R}{R+r_{1}}}\left(1-e^{-{\frac {t_{1}}{T_{1}}}}\right)\quad {\mbox{where}}\quad T_{1}={\frac {CRr_{1}}{R+r_{1}}}.$

(8)

Next, let the circuit $r_{1}$ be broken for a time $t_{2},$

$E(=E_{2})=E_{1}e^{-{\frac {t_{2}}{T_{2}}}}\quad {\mbox{where}}\quad T_{2}=CR.$

(9)

Finally, let the surfaces of the condenser be connected by means of a wire whose resistance is $r_{3}$ for a time $t_{3},$

$E(=E_{3})=E_{2}e^{-{\frac {t_{3}}{T_{3}}}}\quad {\mbox{where}}\quad T_{3}={\frac {CRr_{3}}{R+r_{3}}}.$

(10)

If $Q_{3}$ is the total discharge through this wire in the time $t_{3},$

$Q_{3}=E_{0}{\frac {CR^{2}}{(R+r_{1})(R+r_{3})}}\left(1-e^{-{\frac {t_{1}}{T_{1}}}}\right)e^{-{\frac {t_{2}}{T_{2}}}}\left(1-e^{-{\frac {t_{3}}{T_{3}}}}\right).$

(11)

In this way we may find the discharge through a wire which is made to connect the surfaces of a condenser after being charged for a time $t_{l},$ and then insulated for a time $t_{2}.$ If the time of charging is sufficient, as it generally is, to develope the whole charge, and if the time of discharge is sufficient for a complete discharge, the discharge is

$Q_{3}=E_{0}{\frac {CR^{2}}{(R+r_{1})(R+r_{3})}}e^{-{\frac {t_{2}}{CR}}}.$

(12)

327.] In a condenser of this kind, first charged in any way, next discharged through a wire of small resistance, and then insulated, no new electrification will appear. In most actual condensers, however, we find that after discharge and insulation a new charge is gradually developed, of the same kind as the original charge, but inferior in intensity. This is called the residual charge. To account for it we must admit that the constitution of the dielectric medium is different from that which we have just described. We shall find, however, that a medium formed of a conglomeration of small pieces of different simple media would possess this property.

Theory of a Composite Dielectric.

328.] We shall suppose, for the sake of simplicity, that the dielectric consists of a number of plane strata of different materials and of area unity, and that the electric forces act in the direction of the normal to the strata.

Let $a_{1},\,a_{2},$ &c. be the thicknesses of the different strata.

$X_{l},\,X_{2},$ &c. the resultant electrical force within each stratum.

$p_{1},\,p_{2},$ &c. the current due to conduction through each stratum.

$f_{1},\,f_{2},$ &c. the electric displacement.

$u_{1},u_{2},$ &c. the total current, due partly to conduction and partly to variation of displacement.

$r_{1},\,r_{2},$ &c. the specific resistance referred to unit of volume.

$K_{1},\,K_{2},$ &c. the specific inductive capacity.

$k_{1},\,k_{2},$ &c. the reciprocal of the specific inductive capacity.

$E$ the electromotive force due to a voltaic battery, placed in the part of the circuit leading from the last stratum towards the first, which we shall suppose good conductors.

$Q$ the total quantity of electricity which has passed through this part of the circuit up to the time $t$.

$R$ the resistance of the battery with its connecting wires.

$\sigma _{12}$ the surface-density of electricity on the surface which separates the first and second strata.

Then in the first stratum we have, by Ohm's Law,

$X_{1}=r_{1}p_{1}$

(1)

By the theory of electrical displacement,

$X_{1}=4\pi k_{1}f_{1}.$

(2)

By the definition of the total current,

$u_{1}=p_{1}+{\frac {df_{1}}{dt}},$

(3)

with similar equations for the other strata, in each of which the quantities have the suffix belonging to that stratum.

To determine the surface-density on any stratum, we have an equation of the form

$\sigma _{12}=f_{2}-f_{1},$

(4)

and to determine its variation we have

${\frac {\sigma _{12}}{dt}}=p_{1}-p_{2}.$

(5)

By differentiating (4) with respect to $t,$ and equating the result to (5), we obtain

$p_{1}+{\frac {df_{1}}{dt}}=p_{2}+{\frac {df_{2}}{dt}}=u,\quad {\mbox{say,}}$

(6)

or, by taking account of (3),

$u_{1}=u_{2}=\mathrm {\&c.} =u.$

(7)

That is, the total current u is the same in all the strata, and is equal to the current through the wire and battery.

We have also, in virtue of equations (1) and (2),

$u={\frac {1}{r_{1}}}X_{1}+{\frac {1}{4\pi k_{1}}}{\frac {dX_{1}}{dt}},$

(8)

from which we may find $X_{l}$ by the inverse operation on $u,$

$X_{1}=\left({\frac {1}{r_{1}}}+{\frac {1}{4\pi k}}{\frac {d}{dt}}\right)^{-1}u.$

(9)

The total electromotive force $E$ is

$E=a_{1}X_{1}+a_{2}X_{2}+\mathrm {\&c.} ,$

(10)

or

$E=\left\{a_{1}\left({\frac {1}{r_{1}}}+{\frac {1}{4\pi k_{1}}}{\frac {d}{dt}}\right)^{-1}+a_{2}\left({\frac {1}{r_{2}}}+{\frac {1}{4\pi k_{2}}}{\frac {d}{dt}}\right)^{-1}+\mathrm {\&c.} \right\}u,$

(11)

an equation between $E,$ the external electromotive force, and $u,$ the external current.

If the ratio of $r$ to $k$ is the same in all the strata, the equation reduces itself to

$E+{\frac {r}{4\pi k}}{\frac {dE}{dt}}=(a_{1}r_{1}+a_{2}r_{2}+\mathrm {\&c.} )u,$

(12)

which is the case we have already examined, and in which, as we found, no phenomenon of residual charge can take place.

If there are $n$ substances having different ratios of $r$ to $k,$ the general equation (11), when cleared of inverse operations, will be a linear differential equation, of the nth order with respect to $E$ and of the $(n-1)$th order with respect to $u,\,t$ being the independent variable.

From the form of the equation it is evident that the order of the different strata is indifferent, so that if there are several strata of the same substance we may suppose them united into one without altering the phenomena.

329.] Let us now suppose that at first $f_{1},\,f_{2},$ &c. are all zero, and that an electromotive force $E$ is suddenly made to act, and let us find its instantaneous effect.

Integrating (8) with respect to $t,$ we find

$Q=\int u\,dt={\frac {1}{r_{1}}}\int X_{1}\,dt+{\frac {1}{4\pi k_{1}}}X_{1}+\mathrm {const.}$

(13)

Now, since $X_{1}$ is always in this case finite, $\int X_{1}\,dt,$ must be insensible when $t$ is insensible, and therefore, since $X_{1}$ is originally zero, the instantaneous effect will be

$X_{1}=4\pi k_{1}Q.$

(14)

Hence, by equation (10),

$E+4\pi (k_{1}a_{1}+k_{2}a_{2}+\mathrm {\&c.} )Q,$

(15)

and if $C$ be the electric capacity of the system as measured in this instantaneous way,

$C={\frac {Q}{E}}={\frac {1}{4\pi (k_{1}a_{1}+k_{2}a_{2}+\mathrm {\&c.} )}}.$

(16)

This is the same result that we should have obtained if we had neglected the conductivity of the strata.

Let us next suppose that the electromotive force $E$ is continued uniform for an indefinitely long time, or till a uniform current of conduction equal top is established through the system.

We have then $X_{1}=r_{1}p,$ and therefore

$E=(r_{1}+a_{1}+r_{2}a_{2}+\mathrm {\&c.} )p.$

(17)

If $R$ be the total resistance of the system,

$R={\frac {E}{P}}=r_{1}a_{1}+r_{2}a_{2}+\mathrm {\&c.}$

(18)

In this state we have by (2),

$f_{1}={\frac {r_{1}}{4\pi k_{1}}}p,$

so that

$\sigma _{12}=\left({\frac {r_{2}}{4\pi k_{2}}}-{\frac {r_{1}}{4\pi k_{1}}}\right)p.$

(19)

If we now suddenly connect the extreme strata by means of a conductor of small resistance, $E$ will be suddenly changed from its original value $E_{0}$ to zero, and a quantity $Q$ of electricity will pass through the conductor.

To determine $Q$ we observe that if $X_{1}'$ be the new value of $X_{l},$ then by (13),

$X_{1}'=X_{1}+4\pi k_{1}Q.$

(20)

Hence, by (10), putting $E=0,$

$0=a_{1}X_{1}+\mathrm {\&c.} +4\pi (a_{1}k_{1}+a_{2}k_{2}+\mathrm {\&c.} )Q,$

(21)

or

$0=E_{0}+{\frac {1}{C}}Q.$

(22)

Hence $Q=-CE_{0}$ where $C$ is the capacity, as given by equation (16). The instantaneous discharge is therefore equal to the instantaneous charge.

Let us next suppose the connexion broken immediately after this discharge. We shall then have $u=0,$ so that by equation (8),

$X_{1}=X'e^{-{\frac {4\pi k_{1}}{r_{1}}}t},$

(23)

where $X'$ is the initial value after the discharge.

Hence, at any time $t,$

$X_{1}=E_{0}\left\{{\frac {r_{1}}{R}}-4\pi k_{1}C\right\}e^{-{\frac {4\pi k_{1}}{r_{1}}}}.$

The value of $E$ at any time is therefore

$E=E_{0}\left\{\left({\frac {a_{1}r_{1}}{R}}-4\pi a_{1}k_{1}C\right)e^{-{\frac {4\pi k_{1}}{r_{1}}}t}+\left({\frac {a_{2}r_{2}}{R}}-4\pi a_{2}k_{2}C\right)e^{-{\frac {4\pi k_{2}}{r_{2}}}t}+\mathrm {\&c.} \right\},$

(24)

and the instantaneous discharge after any time $t$ is $EC.$ This is called the residual discharge.

If the ratio of $r$ to $k$ is the same for all the strata, the value of $E$ will be reduced to zero. If, however, this ratio is not the same, let the terms be arranged according to the values of this ratio in descending order of magnitude.

The sum of all the coefficients is evidently zero, so that when $t=0,\,E=0.$ The coefficients are also in descending order of magnitude, and so are the exponential terms when $t$ is positive. Hence, when $t$ is positive, $E$ will be positive, so that the residual discharge is always of the same sign as the primary discharge.

When $t$ is indefinitely great all the terms disappear unless any of the strata are perfect insulators, in which case $r_{1}$ is infinite for that stratum/, and $R$ is infinite for the whole system, and the final value of $E$ is not zero but

$E=E_{0}(1-4\pi a_{1}k_{1}C)$

(25)

Hence, when some, but not all, of the strata are perfect insulators, a residual discharge may be permanently preserved in the system.

330.] We shall next determine the total discharge through a wire of resistance $R_{0}$ kept permanently in connexion with the extreme strata of the system, supposing the system first charged by means of a long-continued application of the electromotive force $E.$

At any instant we have

$E=a_{1}r_{1}p_{1}+a_{2}r_{2}p_{2}+\mathrm {\&c.} +R_{0}u=0,$

(26)

and also, by (3),

$u=p_{1}+{\frac {df_{1}}{dt}}.$

(27)

Hence

$(R+R_{0})u=a_{1}r_{1}{\frac {df_{1}}{dt}}+a_{2}r_{2}{\frac {df_{2}}{dt}}+\mathrm {\&c.}$

(28)

Integrating with respect to $t$ in order to find $Q,$ we get

$(R+R_{0})Q=a_{1}r_{1}(f_{1}'-f_{1})+a_{2}r_{2}(f_{2}'-f_{2})+\mathrm {\&c.} ,$

(29)

where $f_{1}$ is the initial, and $f_{1}'$ the final value of $f_{1}.$

In this case $f_{1}'=0,$ and $f_{1}=E\left({\frac {r_{1}}{4\pi k_{1}R}}-C\right).$

Hence

$(R+R_{0})Q={\frac {E_{0}}{4\pi R}}\left({\frac {a_{1}r_{1}^{2}}{k_{1}}}+{\frac {a_{2}r_{2}^{2}}{k_{2}}}+\mathrm {\&c.} \right)-E_{0}CR,$

(30)

$=-{\frac {CE_{0}}{R}}\Sigma \Sigma \left[a_{1}a_{2}k_{1}k_{2}\left({\frac {r_{1}}{k_{1}}}-{\frac {r_{2}}{k_{2}}}\right)^{2}\right],$

(31)

where the summation is extended to all quantities of this form belonging to every pair of strata.

It appears from this that $Q$ is always negative, that is to say, in the opposite direction to that of the current employed in charging the system.

This investigation shews that a dielectric composed of strata of different kinds may exhibit the phenomena known as electric absorption and residual discharge, although none of the substances of which it is made exhibit these phenomena when alone. An investigation of the cases in which the materials are arranged otherwise than in strata would lead to similar results, though the calculations would be more complicated, so that we may conclude that the phenomena of electric absorption may be expected in the case of substances composed of parts of different kinds, even though these individual parts should be microscopically small.

It by no means follows that every substance which exhibits this phenomenon is so composed, for it may indicate a new kind of electric polarization of which a homogeneous substance may be capable, and this in some cases may perhaps resemble electro-chemical polarization much more than dielectric polarization.

The object of the investigation is merely to point out the true mathematical character of the so-called electric absorption, and to shew how fundamentally it differs from the phenomena of heat which seem at first sight analogous.

331.] If we take a thick plate of any substance and heat it on one side, so as to produce a flow of heat through it, and if we then suddenly cool the heated side to the same temperature as the other, and leave the plate to itself, the heated side of the plate will again become hotter than the other by conduction from within.

Now an electrical phenomenon exactly analogous to this can be produced, and actually occurs in telegraph cables, but its mathematical laws, though exactly agreeing with those of heat, differ entirely from those of the stratified condenser.

In the case of heat there is true absorption of the heat into the substance with the result of making it hot. To produce a truly analogous phenomenon in electricity is impossible, but we may imitate it in the following way in the form of a lecture-room experiment.

Let $A_{1},\,A_{2},$ &c. be the inner conducting surfaces of a series of condensers, of which $B_{0},\,B_{1},\,B_{2},$ &c. are the outer surfaces.

Let $A_{1},\,A_{2},$ &c. be connected in series by connexions of resistance $R,$ and let a current be passed along this series from left to right.

Let us first suppose the plates $B_{0},\,B_{1},\,B_{2},$ each insulated and free from charge. Then the total quantity of electricity on each of the plates $B$ must remain zero, and since the electricity on the plates $A$ is in each case equal and opposite to that of the opposed

Fig. 25.

surface they will not be electrified, and no alteration of the current will be observed.

But let the plates $B$ be all connected together, or let each be connected with the earth. Then, since the potential of $A_{1}$ is positive, while that of the plates $B$ is zero, $A_{1}$ will be positively electrified and $B_{1}$ negatively.

If $P_{1},\,P_{2},$ &c. are the potentials of the plates $A_{1},\,A_{2},$ &c., and $C$ the capacity of each, and if we suppose that a quantity of electricity equal to $Q_{0}$ passes through the wire on the left, $Q_{l}$ through the connexion $R_{1},$ and so on, then the quantity which exists on the plate $A_{1}$ is $Q_{0}-Q_{1},$ and we have

$Q_{0}-Q_{1}=C_{1}P_{1}.$

Similarly

$Q_{1}-Q_{2}=C_{2}P_{2},$

and so on.

But by Ohm's Law we have

${\begin{aligned}P_{1}-P_{2}&=R_{1}{\frac {dQ_{1}}{dt}},\\P_{2}-P_{3}&=R_{2}{\frac {dQ_{2}}{dt}}.\end{aligned}}$

If we suppose the values of $C$ the same for each plate, and those of $R$ the same for each wire, we shall have a series of equations of the form

${\begin{aligned}Q_{0}-2Q_{1}+Q_{2}&=RC{\frac {dQ_{1}}{dt}},\\Q_{1}-2Q_{2}+Q_{3}&=RC{\frac {dQ_{2}}{dt}}.\end{aligned}}$

If there are n quantities of electricity to be determined, and if either the total electromotive force, or some other equivalent conditions be given, the differential equation for determining any one of them will be linear and of the nth order.

By an apparatus arranged in this way, Mr. Varley succeeded in imitating the electrical action of a cable 12,000 miles long.

When an electromotive force is made to act along the wire on the left hand, the electricity which flows into the system is at first principally occupied in charging the different condensers beginning with $A_{1},$ and only a very small fraction of the current appears at the right hand till a considerable time has elapsed. If galvanometers be placed in circuit at $R_{1},\,R_{2},$ &c. they will be affected by the current one after another, the interval between the times of equal indications being greater as we proceed to the right.

332.] In the case of a telegraph cable the conducting wire is separated from conductors outside by a cylindrical sheath of guttapercha, or other insulating material. Each portion of the cable thus becomes a condenser, the outer surface of which is always at potential zero. Hence, in a given portion of the cable, the quantity of free electricity at the surface of the conducting wire is equal to the product of the potential into the capacity of the portion of the cable considered as a condenser.

If $a_{1},a_{2}$ are the outer and inner radii of the insulating sheath, and if $K$ is its specific dielectric capacity, the capacity of unit of length of the cable is, by Art. 126,

$c={\frac {K}{2\log {\frac {a_{1}}{a_{2}}}}}.$

(1)

Let $v$ be the potential at any point of the wire, which we may consider as the same at every part of the same section.

Let $Q$ be the total quantity of electricity which has passed through that section since the beginning of the current. Then the quantity which at the time $t$ exists between sections at $x$ and at $x+\delta x$, is

$Q-\left(Q+{\frac {dQ}{dx}}\delta x\right),\qquad {\mbox{or}}\qquad -{\frac {dQ}{dx}}\delta x,$

and this is, by what we have said, equal to $cv\delta x.$

Hence

$cv=-{\frac {dQ}{dx}}.$

(2)

Again, the electromotive force at any section is $-{\frac {dv}{dx}},$ and by Ohm's Law,

$-{\frac {dv}{dx}}=k{\frac {dQ}{dt}},$

(3)

where $k$ is the resistance of unit of length of the conductor, and ${\frac {dQ}{dt}}$is the strength of the current. Eliminating $Q$ between (2) and (3), we find

$ck{\frac {dv}{dt}}={\frac {d^{2}v}{dx^{2}}}.$

(4)

This is the partial differential equation which must be solved in order to obtain the potential at any instant at any point of the cable. It is identical with that which Fourier gives to determine the temperature at any point of a stratum through which heat is flowing in a direction normal to the stratum. In the case of heat $c$ represents the capacity of unit of volume, or what Fourier calls $CD,$ and $k$ represents the reciprocal of the conductivity.

If the sheath is not a perfect insulator, and if $k_{1}$ is the resistance of unit of length of the sheath to conduction through it in a radial direction, then if $\rho _{1}$ is the specific resistance of the insulating material,

$k_{1}=2\rho _{1}\log _{e}{\frac {r_{1}}{r_{2}}}.$

(5)

The equation (2) will no longer be true, for the electricity is expended not only in charging the wire to the extent represented by $cv,$ but in escaping at a rate represented by ${\frac {v}{k_{1}}}.$ Hence the rate of expenditure of electricity will be

$-{\frac {d^{2}}{dx}}{\frac {Q}{dt}}=c{\frac {dv}{dt}}+{\frac {1}{k_{1}}}v,$

(6)

whence, by comparison with (3), we get

$ck{\frac {dv}{dt}}={\frac {d^{2}v}{dx^{2}}}-{\frac {k}{k_{1}}}v,$

(7)

and this is the equation of conduction of heat in a rod or ring as given by Fourier^[1].

333.] If we had supposed that a body when raised to a high potential becomes electrified throughout its substance as if electricity were compressed into it, we should have arrived at equations of this very form. It is remarkable that Ohm himself, misled by the analogy between electricity and heat, entertained an opinion of this kind, and was thus, by means of an erroneous opinion, led to employ the equations of Fourier to express the true laws of conduction of electricity through a long wire, long before the real reason of the appropriateness of these equations had been suspected.

Mechanical Illustration of the Properties of a Dielectric.

Fig. 26.

334.] Five tubes of equal sectional area $A,\,B,\,C,\,D$ and $P$ are arranged in circuit as in the figure. $A,\,B,\,C$ and $D$ are vertical and equal, and $P$ is horizontal.

The lower halves of $A,\,B,\,C,\,D$ are filled with mercury, their upper halves and the horizontal tube $P$ are filled with water.

A tube with a stopcock $Q$ connects the lower part of $A$ and $B$ with that of $C$ and $D,$ and a piston $P$ is made to slide in the horizontal tube.

Let us begin by supposing that the level of the mercury in the four tubes is the same, and that it is indicated by $A_{0},\,B_{0},\,C_{0},\,D_{0},$ that the piston is at $P_{0},$ and that the stopcock $Q$ is shut.

Now let the piston be moved from $P_{0}$ to $P_{1},$ a distance $a.$ Then, since the sections of all the tubes are equal, the level of the mercury in $A$ and $C$ will rise a distance $a,$ or to $A_{1}$ and $C_{1},$ and the mercury in $B$ and $D$ will sink an equal distance $a,$ or to $B_{1}$ and $D_{1}.$

The difference of pressure on the two sides of the piston will be represented by $4a.$

This arrangement may serve to represent the state of a dielectric acted on by an electromotive force $4a.$

The excess of water in the tube $D$ may be taken to represent a positive charge of electricity on one side of the dielectric, and the excess of mercury in the tube $A$ may represent the negative charge on the other side. The excess of pressure in the tube $P$ on the side of the piston next $D$ will then represent the excess of potential on the positive side of the dielectric.

If the piston is free to move it will move back to $P_{0}$ and be in equilibrium there. This represents the complete discharge of the dielectric.

During the discharge there is a reversed motion of the liquids throughout the whole tube, and this represents that change of electric displacement which we have supposed to take place in a dielectric.

I have supposed every part of the system of tubes filled with incompressible liquids, in order to represent the property of all electric displacement that there is no real accumulation of electricity at any place.

Let us now consider the effect of opening the stopcock $Q$ while the piston $P$ is at $P_{1}.$

The level of $A_{1}$ and $D_{1}$ will remain unchanged, but that of $B$ and $C$ will become the same, and will coincide with $B_{0}$ and $C_{0}.$

The opening of the stopcock $Q$ corresponds to the existence of a part of the dielectric which has a slight conducting power, but which does not extend through the whole dielectric so as to form an open channel.

The charges on the opposite sides of the dielectric remain insulated, but their difference of potential diminishes.

In fact, the difference of pressure on the two sides of the piston sinks from $4a$ to $2a$ during the passage of the fluid through $Q.$

If we now shut the stopcock $Q$ and allow the piston $P$ to move freely, it will come to equilibrium at a point $P_{2},$ and the discharge will be apparently only half of the charge.

The level of the mercury in $A$ and $B$ will be ${\frac {1}{2}}a$ above its original level, and the level in the tubes $C$ and $D$ will be ${\frac {1}{2}}a$ below its original level. This is indicated by the levels $A_{2},\,B_{2},\,C_{2},\,D_{2}.$

If the piston is now fixed and the stopcock opened, mercury will flow from $B$ to $C$ till the level in the two tubes is again at $B_{0}$ and $C_{0}.$ There will then be a difference of pressure ${}=a$ on the two sides of the piston $P.$ If the stopcock is then closed and the piston $P$ left free to move, it will again come to equilibrium at a point $P_{3},$ half way between $P_{2}$ and $P_{0}.$ This corresponds to the residual charge which is observed when a charged dielectric is first discharged and then left to itself. It gradually recovers part of its charge, and if this is again discharged a third charge is formed, the successive charges diminishing in quantity. In the case of the illustrative experiment each charge is half of the preceding, and the discharges, which are ½, ¼, &c. of the original charge, form a series whose sum is equal to the original charge.

If, instead of opening and closing the stopcock, we had allowed it to remain nearly, but not quite, closed during the whole experiment, we should have had a case resembling that of the electrification of a dielectric which is a perfect insulator and yet exhibits the phenomenon called 'electric absorption.'

To represent the case in which there is true conduction through the dielectric we must either make the piston leaky, or we must establish a communication between the top of the tube $A$ and the top of the tube $D.$

In this way we may construct a mechanical illustration of the properties of a dielectric of any kind, in which the two electricities are represented by two real fluids, and the electric potential is represented by fluid pressure. Charge and discharge are represented by the motion of the piston $P,$ and electromotive force by the resultant force on the piston.

1. Théorie de la Chaleur, art. 105.