A Treatise on Electricity and Magnetism/Volume 1/Part 2/Chapter 9

by James Clerk Maxwell

CHAPTER IX.

CONDUCTION THROUGH HETEROGENEOUS MEDIA.

On the Conditions to be Fulfilled at the Surface of Separation between Two Conducting Media.

310.] There are two conditions which the distribution of currents must fulfil in general, the condition that the potential must be continuous, and the condition of 'continuity' of the electric currents.

At the surface of separation between two media the first of these conditions requires that the potentials at two points on opposite sides of the surface, but infinitely near each other, shall be equal. The potentials are here understood to be measured by an electrometer put in connexion with the given point by means of an electrode of a given metal. If the potentials are measured by the method described in Arts. 222, 246, where the electrode terminates in a cavity of the conductor filled with air, then the potentials at contiguous points of different metals measured in this way will differ by a quantity depending on the temperature and on the nature of the two metals.

The other condition at the surface is that the current through any element of the surface is the same when measured in either medium.

Thus, if $V_{1}$ and $V_{2}$ are the potentials in the two media, then at any point in the surface of separation

$V_{1}=V_{2},$

(1)

and if $u_{1},v_{1},w_{1}$ and $u_{2},v_{2},w_{2}$ are the components of currents in the two media, and $l,m,n$ the direction-cosines of the normal to the surface of separation,

$u_{1}l+v_{1}m+w_{1}n=u_{2}l+v_{2}m+w_{2}n.$

(2)

In the most general case the components $u,v,w$ are linear functions of the derivatives of $V,$ the forms of which are given in the equations

$\left.{\begin{aligned}u&=r_{1}X+p_{3}Y+q_{2}Z,\\v&=q_{3}X+r_{2}Y+p_{1}Z,\\w&=p_{2}X+q_{1}Y+r_{3}Z,\end{aligned}}\right\}$

(3)

where $X,Y,Z$ are the derivatives of $V$ with respect to $x,y,z$ respectively.

Let us take the case of the surface which separates a medium having these coefficients of conduction from an isotropic medium having a coefficient of conduction equal to $r.$

Let $X',Y',Z'$ be the values of $X,Y,Z$ in the isotropic medium, then we have at the surface

$V=V',$

(4)

or

$X\,dx+Y\,dy+Z\,dz=X'\,dx+Y'\,dy+Z'\,dz,$

(5)

when

$l\,dx+m\,dy+n\,dz=0.$

(6)

This condition gives

$X'=X+4\pi \sigma l,\quad \quad Y'=Y+4\pi \sigma m,\quad \quad Z'=Z+4\pi \sigma n,$

(7)

where $\sigma$ is the surface-density.

We have also in the isotropic medium

$u'=rX',\quad \quad v'=rY',\quad \quad w'=rZ',$

(8)

and at the boundary the condition of flow is

$u'l+v'm+w'n=ul+vm+wn,$

(9)

or

${\begin{aligned}&\qquad r(lX+mY+nZ+4\pi \sigma )\\&l(r_{1}X+p_{3}Y+q_{2}Z)+m(q_{3}X+r_{2}Y+p_{1}Z)+n(p_{2}X+q_{1}Y+r_{3}Z),\end{aligned}}$

(10)

whence

${\begin{aligned}4\pi \sigma r=(l(r_{1}-r)+mq_{3}+np_{2})X+(lp_{3}+m(r_{2}-r)+nq_{1})Y\qquad \\+(lq_{2}+mp_{1}+n(r_{3}-r))Z.\end{aligned}}$

(11)

The quantity $\sigma$ represents the surface-density of the charge on the surface of separation. In crystallized and organized substances it depends on the direction of the surface as well as on the force perpendicular to it. In isotropic substances the coefficients $p$ and $q$ are zero, and the coefficients $r$ are all equal, so that

$4\pi \sigma =({\frac {r_{1}}{r}}-1)(lX+mY+nZ),$

(12)

where $r_{1}$ is the conductivity of the substance, $r$ that of the external medium, and $l,m,n$ the direction-cosines of the normal drawn towards the medium whose conductivity is $r$.

When both media are isotropic the conditions may be greatly simplified, for if $k$ is the specific resistance per unit of volume, then

$u=-{\frac {1}{k}}{\frac {dV}{dx}},\quad \quad v=-{\frac {1}{k}},\quad \quad w=-{\frac {1}{k}}{\frac {dV}{dz}},$

(13)

and if $\nu$ is the normal drawn at any point of the surface of separation from the first medium towards the second, the conduction of continuity is

${\frac {1}{k_{1}}}{\frac {dV_{1}}{d\nu }}={\frac {1}{k_{2}}}{\frac {dV_{2}}{d\nu }}.$

(14)

If $\theta _{1}$ and $\theta _{2}$ are the angles which the lines of flow in the first and second media respectively make with the normal to the surface of separation, then the tangents to these lines of flow are in the same plane with the normal and on opposite sides of it, and

$k_{1}\tan \theta _{1}=k_{2}\tan \theta _{2}.$

(15)

This may be called the law of refraction of lines of flow.

311.] As an example of the conditions which must be fulfilled when electricity crosses the surface of separation of two media, let us suppose the surface spherical and of radius $a$, the specific resistance being $k_{1}$ within and $k_{2}$ without the surface.

Let the potential, both within and without the surface, be expanded in solid harmonics, and let the part which depends on the surface harmonic $S_{i}$ be

$V_{1}=\left(A_{1}r^{i}+B_{1}r^{-(i+1)}\right),$

(1)

$V_{2}=\left(A_{2}r^{i}+B_{2}r^{-(i+1)}\right)S_{i}$

(2)

within and without the sphere respectively.

At the surface of separation where $r=a$ we must have

$V_{1}=V_{2}\quad \quad {\mbox{and}}\quad \quad {\frac {1}{k_{1}}}{\frac {dV_{1}}{dr}}={\frac {1}{k_{2}}}{\frac {dV_{2}}{dr}}.$

(3)

From these conditions we get the equations

$\left.{\begin{array}{l}(A_{1}-A_{2})a^{2i+1}+B_{1}-B_{2}=0,\\\left({\frac {1}{k_{1}}}A_{1}-{\frac {1}{k_{2}}}A_{2}\right)ia^{2i+1}-\left({\frac {1}{k_{1}}}B_{1}-{\frac {1}{k_{2}}}B_{2}\right)(i+1)=0.\end{array}}\right\}$

(4)

These equations are sufficient, when we know two of the four quantities $A_{1},\,A_{2},\,B_{1},\,B_{2},$ to deduce the other two.

Let us suppose $A_{1}$ and $B_{1}$ known, then we find the following expressions for $A_{2}$ and $B_{2}$,

$\left.{\begin{array}{l}A_{2}={\frac {(k_{1}(i+1)+k_{2}i)A_{1}+(k_{1}-k_{2})(i+1)B_{1}a^{-(2i+1)}}{k_{1}(2i+1)}},\\\\B_{2}={\frac {(k_{1}-k_{2})iA_{1}a^{2i+1}+(k_{1}i+k_{2}(i+1))B_{1}}{k_{1}(2i+1)}}.\end{array}}\right\}$

(5)

In this way we can find the conditions which each term of the harmonic expansion of the potential must satisfy for any number of strata bounded by concentric spherical surfaces.

312.] Let us suppose the radius of the first spherical surface to be $a_{1},$ and let there be a second spherical surface of radius $a_{2}$ greater than $a_{1}$ beyond which the specific resistance is $k_{3}.$ If there are no sources or sinks of electricity within these spheres there will be no infinite values of $V,$ and we shall have $B_{1}=0.$

We then find for $A_{3}$ and $B_{3},$ the coefficients for the outer medium,

$\left.{\begin{aligned}A_{3}k_{1}k_{2}(2i+1)^{2}=&{\bigg [}\{k_{1}(i+1)+k_{2}i\}\{k_{2}(i+1)+k_{3}i\}\\&\;\;+i(i+1)(k_{1}-k_{2})(k_{2}-k_{3})\left({\frac {a_{1}}{a_{2}}}\right)^{2i+1}{\bigg ]}A_{1},\\B_{3}k_{1}k_{2}(2i+1)^{2}=&[i\{k_{1}(i+1)+k_{2}i\}(k_{2}-k_{3})a_{2}^{2i+1}\\&\;\;+i(k_{1}-k_{2})\{k_{2}i+k_{3}(i+1)\}a_{1}^{2i+1}]A_{1}.\end{aligned}}\right\}$

(6)

The value of the potential in the outer medium depends partly on the external sources of electricity, which produce currents independently of the existence of the sphere of heterogeneous matter within, and partly on the disturbance caused by the introduction of the heterogeneous sphere.

The first part must depend on solid harmonics of positive degrees only, because it cannot have infinite values within the sphere. The second part must depend on harmonics of negative degrees, because it must vanish at an infinite distance from the centre of the sphere.

Hence the potential due to the external electromotive forces must be expanded in a series of solid harmonics of positive degree. Let $A_{3}$ be the coefficient of one these, of the form

$A_{3}S_{i}r^{i}$

Then we can find $A_{1},$ the corresponding coefficient for the inner sphere by equation (6), and from this deduce $A_{2},B_{2}$ and $B_{3}.$ Of these $B_{3}$ represents the effect on the potential in the outer medium due to the introduction of the heterogeneous spheres.

Let us now suppose $k_{3}=k_{1},$ so that the case is that of a hollow shell for which $k=k_{2},$ separating an inner from an outer portion of the same medium for which $k=k_{1}.$

If we put

$C={\frac {1}{(2i+1)^{2}k_{1}k_{2}+i(i+1)(k_{2}-k_{1})^{2}\left(1-({\frac {a_{1}}{a_{2}}})^{2i+1}\right)}},$

then

$\left.{\begin{array}{l}A_{1}=k_{1}k_{2}(2i+1)^{2}CA_{3},\\A_{2}=k_{2}(2i+1)(k_{1}(i+1)+k_{2}i)CA_{3},\\B_{2}=k_{2}i(2i+1)(k_{1}-k_{2})a_{1}^{2i+1}CA_{3},\\B_{3}=i(k_{2}-k_{1})(k_{1}(i+1)+k_{2}i)(a_{2}^{2i+1}-a_{1}^{2i+1})CA_{3}.\end{array}}\right\}$

(7)

The difference between $A_{3}$ the undisturbed coefficient, and $A_{1}$ its value in the hollow within the spherical shell, is

$A_{3}-A_{1}=(k_{2}-k_{1})^{2}i(i+1)\left(1-({\frac {a_{1}}{a_{2}}})^{2i+1}\right)CA_{3}.$

(8)

Since this quantity is always positive whatever be the values of $k_{1}$ and $k_{2},$ it follows that, whether the spherical shell conducts better or worse than the rest of the medium, the electrical action within the shell is less than it would otherwise be. If the shell is a better conductor than the rest of the medium it tends to equalize the potential all round the inner sphere. If it is a worse conductor, it tends to prevent the electrical currents from reaching the inner sphere at all.

The case of a solid sphere may be deduced from this by making $a_{1}=0,$ or it may be worked out independently.

313.] The most important term in the harmonic expansion is that in which $i=1,$ for which

$\left.{\begin{array}{c}C={\frac {1}{9k_{1}k_{2}+2(k_{1}-k_{2})^{2}\left(1-({\frac {a_{1}}{a_{2}}})^{3}\right)}},\\A_{1}=9k_{1}k_{2}CA_{3},\quad \quad A_{2}=3k_{2}(2k_{1}+k_{2})CA_{3},\\B_{2}=3k_{2}(k_{1}-k_{2})a_{1}^{3}CA_{3},\quad B_{3}=(k_{2}-k_{1})(2k_{1}+k_{2})(a_{2}^{3}-a_{1}^{3})CA_{3}.\end{array}}\right\}$

(9)

The case of a solid sphere of resistance $k_{2}$ may be deduced from this by making $a_{1}=0.$ We then have

$\left.{\begin{array}{l}A_{2}={\frac {3k_{2}}{k_{1}+2k_{2}}}A_{3},\quad \quad B_{2}=0,\\B_{3}={\frac {k_{2}-k_{1}}{k_{1}+2k_{2}}}a_{2}^{3}A_{3}.\end{array}}\right\}$

(10)

It is easy to shew from the general expressions that the value of $B_{3}$ in the case of a hollow sphere having a nucleus of resistance $k_{1}$ surrounded by a shell of resistance $k_{2},$ is the same as that of a uniform solid sphere of the radius of the outer surface, and of resistance $K$, where

$K={\frac {(2k_{1}+k_{2})a_{2}^{3}+(k_{1}-k_{2})a_{1}^{3}}{(2k_{1}+k_{2})a_{2}^{3}-2(k_{1}-k_{2})a_{1}^{3}}}k_{2}.$

(11)

314.] If there are $n$ spheres of radius $a_{1}$ and resistance $k_{1},$ placed in a medium whose resistance is $k_{2},$ at such distances from each other that their effects in disturbing the course of the current may be taken as independent of each other, then if these spheres are all contained within a sphere of radius $a_{2},$ the potential at a great distance from the centre of this sphere will be of the form

$V=\left(A+nB{\frac {1}{r^{2}}}\right)\cos \theta ,$

(12)

where the value of $B$ is

$B={\frac {k_{1}-k_{2}}{2k_{1}+k_{2}}}a_{1}^{3}A.$

(13)

The ratio of the volume of the $n$ small spheres to that of the sphere which contains them is

$p={\frac {na_{1}^{3}}{a_{2}^{3}}}.$

(14)

The value of the potential at a great distance from the sphere may therefore be written

$V=\left(A+pa_{2}^{3}{\frac {k_{1}-k_{2}}{2k_{1}+k_{2}}}{\frac {1}{r^{2}}}\right)\cos \theta .$

(15)

Now if the whole sphere of radius $a_{2}$ had been made of a material of specific resistance $K,$ we should have had

$V=\left\{A+a_{2}^{3}{\frac {K-k_{2}}{2K+k_{2}}}{\frac {1}{r^{2}}}\right\}\cos \theta .$

(16)

That the one expression should be equivalent to the other,

$K={\frac {2k_{1}+k_{2}+p(k_{1}-k_{2})}{2k_{1}+k_{2}-2p(k_{1}-k_{2})}}k_{2}.$

(17)

This, therefore, is the specific resistance of a compound medium consisting of a substance of specific resistance $k_{2},$ in which are disseminated small spheres of specific resistance $k_{1},$ the ratio of the volume of all the small spheres to that of the whole being $p.$ In order that the action of these spheres may not produce effects depending on their interference, their radii must be small compared with their distances, and therefore $p$ must be a small fraction.

This result may be obtained in other ways, but that here given involves only the repetition of the result already obtained for a single sphere.

When the distance between the spheres is not great compared with their radii, and when ${\frac {k_{1}-k_{2}}{2k_{1}+k_{2}}}$ is considerable, then other terms enter into the result, which we shall not now consider. In consequence of these terms certain systems of arrangement of the spheres cause the resistance of the compound medium to be different in different directions.

Application of the Principle of Images.

315.] Let us take as an example the case of two media separated by a plane surface, and let us suppose that there is a source $S$ of electricity at a distance a from the plane surface in the first medium, the quantity of electricity flowing from the source in unit of time being $S.$

If the first medium had been infinitely extended the current at any point $P$ would have been in the direction $SP,$ and the potential at $P$ would have been ${\frac {E}{r_{1}}}$ where $E={\frac {Sk_{1}}{4\pi }}$ and $r_{1}=SP.$

In the actual case the conditions may be satisfied by taking a point $I,$ the image of $S$ in the second medium, such that $IS$ is normal to the plane of separation and is bisected by it. Let $r_{2}$ be the distance of any point from $I,$ then at the surface of separation

$r_{1}=r_{2},$

(1)

${\frac {dr_{1}}{d\nu }}=-{\frac {dr_{2}}{d\nu }}.$

(2)

Let the potential $V_{1}$ at any point in the first medium be that due to a quantity of electricity $E$ placed at $S,$ together with an imaginary quantity $E_{2}$ at $I,$ and let the potential $V_{2}$ at any point of the second medium be that due to an imaginary quantity $E_{1}$ at $S,$ then if

$V_{1}={\frac {E}{r_{1}}}+{\frac {E_{2}}{r_{1}}}\quad \quad \mathrm {and} \quad \quad V_{2}={\frac {E_{1}}{r_{1}}},$

(3)

the superficial condition $V_{1}=V_{2}$ gives

$E+E_{2}=E_{1},$

(4)

and the condition

${\frac {1}{k_{1}}}{\frac {dV_{1}}{d\nu }}={\frac {1}{k_{2}}}{\frac {dV_{2}}{d\nu }}$

(5)

gives

${\frac {1}{k_{1}}}(E-E_{2})={\frac {1}{k_{2}}}E_{1},$

(6)

whence

$E_{1}={\frac {2k_{2}}{k_{1}+k_{2}}}E,\quad \quad E_{2}={\frac {k_{2}-k_{1}}{k_{1}+k_{2}}}E.$

(7)

The potential in the first medium is therefore the same as would be produced in air by a charge $E$ placed at $S,$ and a charge $E_{1}$ at $I$ on the electrostatic theory, and the potential in the second medium is the same as that which would be produced in air by a charge $E_{1}$ at $S.$

The current at any point of the first medium is the same as would have been produced by the source $S$ together with a source ${\frac {k_{2}-k_{1}}{k_{1}+k_{2}}}S$ placed at $I$ if the first medium had been infinite, and the current at any point of the second medium is the same as would have been produced by a source ${\frac {2k_{1}S}{(k_{1}+k_{2})}}$ placed at $S$ if the second medium had been infinite.

We have thus a complete theory of electrical images in the case of two media separated by a plane boundary. Whatever be the nature of the electromotive forces in the first medium, the potential they produce in the first medium may be found by combining their direct effect with the effect of their image.

If we suppose the second medium a perfect conductor, then $k_{2}=0,$ and the image at $I$ is equal and opposite to the course at $S.$ This is the case of electric images, as in Thomson's theory in electrostatics.

If we suppose the second medium a perfect insulator, then $k_{2}=\infty ,$ and the image at $I$ is equal to the source at $S$ and of the same sign. This is the case of images in hydrokinetics when the fluid is bounded by a rigid plane surface.

316.] The method of inversion, which is of so much use in electrostatics when the bounding surface is supposed to be that of a perfect conductor, is not applicable to the more general case of the surface separating two conductors of unequal electric resistance. The method of inversion in two dimensions is, however, applicable, as well as the more general method of transformation in two dimensions given in Art. 190^[1].

Conduction through a Plate separating Two Media.

Fig. 23.

317.] Let us next consider the effect of a plate of thickness $AB$ of a medium whose resistance is $k_{2},$ and separating two media whose resistances are $k_{1}$ and $k_{2},$ in altering the potential due to a source $S$ in the first medium.

The potential will be equal to that due to a system of charges placed in air at certain points along the normal to the plate through $S.$

Make

$AI=SA,\quad BI_{1}=SB,\quad AJ_{1}=I_{1}A\quad BI_{2}=J_{1}B,\quad AJ_{2}=I_{2}A,\;\mathrm {\&c.;}$

then we have two series of points at distances from each other equal to twice the thickness of the plate.

318.] The potential in the first medium at any point $P$ is equal to

${\frac {E}{PS}}+{\frac {I}{PI}}+{\frac {I_{1}}{PI_{1}}}+{\frac {I_{2}}{PI_{2}}}+\;\mathrm {\&c.}$

(8)

that at a point $P'$ in the second

${\begin{array}{l}{\frac {E'}{P'S}}+{\frac {I'}{P'I}}+{\frac {I_{1}'}{P'I_{1}}}+{\frac {I_{2}'}{P'I_{2}}}+\;\mathrm {\&c.} \\\;\;\;+{\frac {J_{1}'}{P'J_{1}}}+{\frac {J_{2}'}{P'J_{2}}}+\;\;\mathrm {\&c.,} \end{array}}$

(9)

and that at a point $P''$ in the third

${\frac {E''}{P''S}}+{\frac {J_{1}}{P''J_{1}}}+{\frac {J_{2}}{P''J_{2}}}+\;\mathrm {\&c.,}$

(10)

where $I,\,I',$ &c. represent the imaginary charges placed at the points $I,$ &c., and the accents denote that the potential is to be taken within the plate.

Then, by the last Article, for the surface through $A$ we have,

$I={\frac {k_{2}-k_{1}}{k_{2}+k_{1}}}E,\quad \quad E'={\frac {2k_{2}}{k_{2}+k_{1}}}E.$

(11)

For the surface through $B$ we find

$I_{1}'={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}}E',\quad \quad E'={\frac {2k_{3}}{k_{2}+k_{1}}}E'.$

(12)

Similarly for the surface through $A$ again,

$J_{1}'={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}I_{1}',\quad \quad I_{1}={\frac {2k_{1}}{k_{1}+k_{2}}}I_{1}',$

(13)

and for the surface through $B,$

$J_{2}'={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}}J_{1}',\quad \quad J_{1}={\frac {2k_{3}}{k_{3}+k_{2}}}J_{1}'.$

(14)

If we make

$\rho ={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}\quad \mathrm {and} \quad \rho '={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}},$

we find for the potential in the first medium,

${\begin{array}{rl}V={\frac {E}{PS}}-\rho {\frac {E}{PI}}+(1-\rho ^{2})\rho '{\frac {E}{PI_{1}}}&+\rho '(1-\rho ^{2})\rho \rho '{\frac {E}{PI_{2}}}+\;\mathrm {\&c.} \\&\quad +\rho '(1-\rho ^{2})(\rho \rho ')^{n-1}{\frac {E}{PI_{n}}}.\end{array}}$

(15)

For the potential in the third medium we find

$V=(1+\rho ')(1-\rho )E\left\{{\frac {1}{PS}}+{\frac {\rho \rho '}{PJ_{1}}}+\,\mathrm {\&c.} +{\frac {(\rho \rho ')^{n}}{PJ_{n}}}\right\}.$

(16)

If the first medium is the same as the third, then $k_{1}=k_{3}$ and $\rho =\rho ',$ and the potential on the other side of the plate will be

$V=(1-\rho ^{2})E\left\{{\frac {1}{PS}}+{\frac {\rho ^{2}}{PJ_{1}}}+\,\mathrm {\&c.} +{\frac {\rho ^{2n}}{PJ_{n}}}\right\}.$

(17)

If the plate is a very much better conductor than the rest of the medium, $\rho$ is very nearly equal to 1 . If the plate is a nearly perfect insulator, $\rho$ is nearly equal to -1, and if the plate differs little in conducting power from the rest of the medium, $\rho$ is a small quantity positive or negative.

The theory of this case was first stated by Green in his 'Theory of Magnetic Induction (Essay, p. 65). His result, however, is correct only when $\rho$ is nearly equal to 1^[2]. The quantity $g$ which he uses is connected with $\rho$ by the equations

$g={\frac {2\rho }{3-\rho }}={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}},\quad \quad \rho ={\frac {3g}{2+g}}={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}.$

If we put $\rho ={\frac {2\pi \kappa }{1+2\pi \kappa }},$ we shall have a solution of the problem of the magnetic induction excited by a magnetic pole in an infinite plate whose coefficient of magnetization is $\kappa$.

On Stratified Conductors.

319.] Let a conductor be composed of alternate strata of thickness $c$ and $c'$ of two substances whose coefficients of conductivity are different. Required the coefficients of resistance and conductivity of the compound conductor.

Let the plane of the strata be normal to $Z.$ Let every symbol relating to the strata of the second kind be accented, and let every symbol relating to the compound conductor be marked with a bar thus, ${\overline {X}}$. Then

${\begin{array}{rl}{\overline {X}}=X=X',&(c+c'){\overline {u}}=cu+c'u',\\{\overline {Y}}=Y=Y',&(c+c'){\overline {v}}=cv+c'v';\\(c+c'){\overline {Z}}=cZ+c'Z',&\quad {\overline {w}}=w=w'.\end{array}}$

We must first determine $u,\,u',\,v,\,v',\,Z$ and $Z'$ in terms of ${\overline {X}},\,{\overline {Y}}$ and ${\overline {w}}$ from the equations of resistance, Art. 297, or those of conductivity, Art. 298. If we put $D$ for the determinant of the coefficients of resistance, we find

${\begin{array}{ll}ur_{3}D&=R_{2}{\overline {X}}-Q_{3}{\overline {Y}}+{\overline {w}}q_{2}D,\\vr_{3}D&=R_{1}{\overline {Y}}-P_{3}{\overline {X}}+{\overline {w}}p_{1}D,\\Zr_{3}&=-p_{2}{\overline {X}}-q_{1}{\overline {Y}}+{\overline {w}}.\end{array}}$

Similar equations with the symbols accented give the values of $u',\,v'$ and $z'.$ Having found ${\overline {u}},\,{\overline {v}}$ and ${\overline {w}}$ in terms of ${\overline {X}},\,{\overline {Y}}$ and ${\overline {Z}},$ we may write down the equations of conductivity of the stratified conductor. If we make $h={\frac {c}{r_{3}}}$ and $h'={\frac {c'}{r_{3}'}},$ we find

${\begin{array}{l}{\overline {p}}_{1}={\frac {hp_{1}+h'p_{1}'}{h+h'}},\quad \quad {\overline {q}}_{1}={\frac {hq_{1}+h'q_{1}'}{h+h'}},\\{\overline {p}}_{2}={\frac {hp_{2}+h'p_{2}'}{h+h'}},\quad \quad {\overline {q}}_{2}={\frac {hq_{2}+h'q_{2}'}{h+h'}},\\{\overline {p}}_{3}={\frac {cp_{3}+c'p_{3}'}{c+c'}}-{\frac {hh'(q_{1}-q_{1}')(q_{2}-q_{2}')}{(h+h')(c+c')}},\\{\overline {q}}_{3}={\frac {cq_{3}+c'q_{3}'}{c+c'}}-{\frac {hh'(p_{1}-p_{1}')(p_{2}-p_{2}')}{(h+h')(c+c')}},\\{\overline {r}}_{1}={\frac {cr_{1}+c'r_{1}'}{c+c'}}-{\frac {hh'(p_{2}-p_{2}')(q_{2}-q_{2}')}{(h+h')(c+c')}},\\{\overline {r}}_{2}={\frac {cr_{2}+c'r_{2}'}{c+c'}}-{\frac {hh'(p_{1}-p_{1}')(q_{1}-q_{1}')}{(h+h')(c+c')}},\end{array}}$

${\begin{array}{c}{\overline {r}}_{3}={\frac {c+c'}{h+h'}}.\end{array}}$

320.] If neither of the two substances of which the strata are formed has the rotatory property of Art. 303, the value of any $P$ or $p$ will be equal to that of its corresponding $Q$ or $q.$ From this it follows that in the stratified conductor also

${\overline {p}}_{1}={\overline {q}}_{1},\quad \quad {\overline {p}}_{2}={\overline {q}}_{2},\quad \quad {\overline {p}}_{3}={\overline {q}}_{3},$

or there is no rotatory property developed by stratification, unless it exists in the materials.

321.] If we now suppose that there is no rotatory property, and also that the axes of $x,\,y$ and $z$ are the principal axes, then the $p$ and $q$ coefficients vanish, and

${\overline {r}}_{1}={\frac {cr_{1}+c'r_{1}'}{c+c'}},\quad \quad {\overline {r}}_{2}={\frac {cr_{2}+c'r_{2}'}{c+c'}},\quad \quad {\overline {r}}_{3}={\frac {c+c'}{{\frac {c}{r_{3}}}+{\frac {c'}{r_{3}'}}}}$

If we begin with both substances isotropic, but of different conductivities, then the result of stratification will be to make the resistance greatest in the direction of a normal to the strata, and the resistance in all directions in the plane of the strata will be equal.

322.] Take an isotropic substance of conductivity $r,$ cut it into exceedingly thin slices of thickness $a,$ and place them alternately with slices of a substance whose conductivity is $s,$ and thickness $k_{1}a.$

Let these slices be normal to $x.$ Then cut this compound conductor into thicker slices, of thickness $b,$ normal to $y,$ and alternate these with slices whose conductivity is $s$ and thickness $k_{2}b.$

Lastly, cut the new conductor into still thicker slices, of thickness $c,$ normal to $z.$ and alternate them with slices whose conductivity is $s$ and thickness $k_{3}c.$

The result of the three operations will be to cut the substance whose conductivity is $r$ into rectangular parallelepipeds whose dimensions are $a,\,b$ and $c,$ where $b$ is exceedingly small compared with $c,$ and $a$ is exceedingly small compared with $b,$ and to embed these parallelepipeds in the substance whose conductivity is $s,$ so that they are separated from each other $k_{1}a$ in the direction of $x,\,k_{2}b$ in that of $y,$ and $k_{3}c$ in that of $z.$ The conductivities of the conductor so formed in the directions of $x,\,y$ and $z$ are

${\begin{array}{l}r_{1}={\frac {\{1+k_{1}(1+k_{2})(1+k_{3})\}r+(k_{2}+k_{3}+k_{2}k_{3})s}{(1+k_{2})(1+k_{3})(k_{1}r+s)}}s,\\r_{2}={\frac {(1+k_{2}+k_{2}k_{3})r+(k_{1}+k_{3}+k_{1}k_{2}+k_{1}k_{3}+k_{1}k_{2}k_{3})s}{(1+k_{3})\{k_{2}r+(1+k_{1}+k_{1}k_{2})s\}}}s,\\r_{3}={\frac {(1+k_{3})(r+(k_{1}+k_{2}+k_{1}k_{2})s)}{k_{3}r+(1+k_{1}+k_{2}+k_{2}k_{3}+k_{3}k_{1}+k_{1}k_{2}+k_{1}k_{2}k_{3})s}}s.\end{array}}$

The accuracy of this investigation depends upon the three dimensions of the parallelepipeds being of different orders of magnitude, so that we may neglect the conditions to be fulfilled at their edges and angles. If we make $k_{l},\,k_{2}$ and $k_{3}$ each unity, then

$r_{1}={\frac {5r+3s}{4r+4s}}s,\quad \quad r_{2}={\frac {3r+5s}{2r+6s}}s,\quad \quad r_{3}={\frac {2r+6s}{r+7s}}s.$

If $r=0,$ that is, if the medium of which the parallelepipeds are made is a perfect insulator, then

$r_{1}={\frac {3}{4}}s,\quad \quad r_{2}={\frac {5}{6}}s,\quad \quad r_{3}={\frac {6}{7}}s.$

If $r=\infty$, that is, if the parallelepipeds are perfect conductors,

$r_{1}={\frac {5}{4}}s,\quad \quad R_{2}={\frac {3}{2}}s,\quad \quad r_{3}=2s.$

In every case, provided $k_{1}=k_{2}=k_{3},$ it may be shewn that $r_{l},\,r_{2}$ and $r_{3}$ are in ascending order of magnitude, so that the greatest conductivity is in the direction of the longest dimensions of the parallelepipeds, and the greatest resistance in the direction of their shortest dimensions.

323.] In a rectangular parallelepiped of a conducting solid, let there be a conducting channel made from one angle to the opposite, the channel being a wire covered with insulating material, and let the lateral dimensions of the channel be so small that the conductivity of the solid is not affected except on account of the current conveyed along the wire.

Let the dimensions of the parallelepiped in the directions of the coordinate axes be $a,\,b,\,c,$ and let the conductivity of the channel, extending from the origin to the point $(abc),$ be $abcK.$

The electromotive force acting between the extremities of the channel is

$aX+bY+cZ,$

and if $C'$ be the current along the channel

$C'=Kabc(aX+bY+cZ).$

The current across the face be of the parallelepiped is $bcu,$ and this is made up of that due to the conductivity of the solid and of that due to the conductivity of the channel, or

or

${\begin{array}{rl}bcu=&bc(r_{1}X+p_{3}Y+q_{2}Z)+Kabc(aX+bY+cZ),\\u=&(r_{1}+Ka^{2})X+(p_{3}+Kab)Y+(q_{2}+Kca)Z.\end{array}}$

In the same way we may find the values of $v$ and $w.$ The coefficients of conductivity as altered by the effect of the channel will be

${\begin{array}{lll}r_{1}+Ka^{2},&r_{2}+Kb^{2},&r_{3}+Kc^{2},\\p_{1}+Kbc,&p_{2}+Kca,&p_{3}+Kab,\\q_{1}+Kbc,&q_{2}+Kca,&q_{3}+Kab.\end{array}}$

In these expressions, the additions to the values of $p_{1},$ &c., due to the effect of the channel, are equal to the additions to the values of $q_{1},$ &c. Hence the values of $p_{1}$ and $q_{1}$ cannot be rendered unequal by the introduction of linear channels into every element of volume of the solid, and therefore the rotatory property of Art. 303, if it does not exist previously in a solid, cannot be introduced by such means.

324.] To construct a framework of linear conductors which shall have any given coefficients of conductivity forming a symmetrical system.

Fig. 24.

Let the space be divided into equal small cubes, of which let the figure represent one. Let the coordinates of the points $0,\,L,\,M,\,N,$ and their potentials be as follows:

${\begin{array}{lllll}\quad \quad &x&y&z\quad &Potential.\\O&0&0&0&0\\L&0&1&1&0+Y+Z,\\M&1&0&1&0+Z+X,\\N&1&1&0&0+X+Y.\end{array}}$

Let these four points be connected by six conductors,

$OL,\quad \quad OM,\quad \quad ON,\quad \quad MN,\quad \quad NL,\quad \quad LM,$

of which the conductivities are respectively

$A,\quad \quad B,\quad \quad C,\quad \quad P,\quad \quad Q,\quad \quad R.$

The electromotive forces along these conductors will be

$Y+Z,\quad \quad Z+X,\quad \quad X+Y,\quad \quad Y-Z,\quad \quad Z-X,\quad \quad X-Y,$

and the currents

$A(Y+Z),\quad B(Z+X),\quad C(X+Y),\quad P(Y-Z),\quad Q(Z-X),\quad R(X-Y).$

Of these currents, those which convey electricity in the positive direction of $x$ are those along $LM,\,LN,\,OM$ and $ON,$ and the quantity conveyed is

$u=(B+C+Q+R)X+(C-R)Y\quad \quad +(B-Q)Z.$

Similarly

${\begin{array}{lll}v=(C-R)X&+(C+A+R+P)Y&+(A-P)Y,\\w=(B-Q)X&+(A-P)Y&+(A+B+P+Q)Z;\end{array}}$

whence we find by comparison with the equations of conduction, Art. 298,

${\begin{array}{ll}4A=r_{2}+r_{3}-r_{1}+2p_{1},&4P=r_{2}+r_{3}-r_{1}-2p_{1}\\4B=r_{3}+r_{1}-r_{2}+2p_{2},&4Q=r_{3}+r_{1}-r_{2}-2p_{2}\\4C=r_{1}+r_{2}-r_{3}+2p_{3},&4R=r_{1}+r_{2}-r_{3}-2p_{3}\end{array}}$

1. See Kirchhoff, Pogg. Ann. lxiv. 497, and lxvii. 344; Quincke, Pogg. xcvii. 382; and Smith, Proc. R. S. Edin., 1869-70, p. 79.
2. See Sir W. Thomson's 'Note on Induced Magnetism in a Plate,' Camb. and Dub. Math. Journ., Nov. 1845, or Reprint, art. ix. § 156.