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

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.
$V_{1}$
$V_{2}$
| $V_{1}=V_{2},$ | (1) |
$u_{1},v_{1},w_{1}$
$u_{2},v_{2},w_{2}$
$l,m,n$
| $u_{1}l+v_{1}m+w_{1}n=u_{2}l+v_{2}m+w_{2}n.$ | (2) |
$u,v,w$
$V,$
| $\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 $r.$
$X',Y',Z'$
$X,Y,Z$
| $V=V',$ | (4) |
| or | $X\,dx+Y\,dy+Z\,dz=X'\,dx+Y'\,dy+Z'\,dz,$ | (5) |
| when | $l\,dx+m\,dy+n\,dz=0.$ | (6) |
| $X'=X+4\pi \sigma l,\quad \quad Y'=Y+4\pi \sigma m,\quad \quad Z'=Z+4\pi \sigma n,$ | (7) |
where $\sigma$ is the surface-density.
| $u'=rX',\quad \quad v'=rY',\quad \quad w'=rZ',$ | (8) |
| $u'l+v'm+w'n=ul+vm+wn,$ | (9) |
| ${\begin{aligned}&\qquad r(lX+mY+nZ+4\pi \sigma )\\&l(r_{1}X+p_{3}Y+q_{2}Z)+m(q_{3}X+r_{2}Y+p_{1}Z)+n(p_{2}X+q_{1}Y+r_{3}Z),\end{aligned}}$ | (10) |
| ${\begin{aligned}4\pi \sigma r=(l(r_{1}-r)+mq_{3}+np_{2})X+(lp_{3}+m(r_{2}-r)+nq_{1})Y\qquad \\+(lq_{2}+mp_{1}+n(r_{3}-r))Z.\end{aligned}}$ | (11) |
$\sigma$
$p$
$q$
$r$
| $4\pi \sigma =({\frac {r_{1}}{r}}-1)(lX+mY+nZ),$ | (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 $r$.
$k$
| $u=-{\frac {1}{k}}{\frac {dV}{dx}},\quad \quad v=-{\frac {1}{k}},\quad \quad w=-{\frac {1}{k}}{\frac {dV}{dz}},$ | (13) |
$\nu$
| ${\frac {1}{k_{1}}}{\frac {dV_{1}}{d\nu }}={\frac {1}{k_{2}}}{\frac {dV_{2}}{d\nu }}.$ | (14) |
$\theta _{1}$
$\theta _{2}$
| $k_{1}\tan \theta _{1}=k_{2}\tan \theta _{2}.$ | (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.
$S_{i}$
| $V_{1}=\left(A_{1}r^{i}+B_{1}r^{-(i+1)}\right),$ | (1) |
| $V_{2}=\left(A_{2}r^{i}+B_{2}r^{-(i+1)}\right)S_{i}$ | (2) |
within and without the sphere respectively.
$r=a$
| $V_{1}=V_{2}\quad \quad {\mbox{and}}\quad \quad {\frac {1}{k_{1}}}{\frac {dV_{1}}{dr}}={\frac {1}{k_{2}}}{\frac {dV_{2}}{dr}}.$ | (3) |
| $\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.
$A_{1}$
$B_{1}$
$A_{2}$
$B_{2}$
| $\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 $B_{1}=0.$
$A_{3}$
$B_{3},$
| $\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.
$A_{3}$
* $A_{3}S_{i}r^{i}$
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 $k=k_{1}.$
* $C={\frac {1}{(2i+1)^{2}k_{1}k_{2}+i(i+1)(k_{2}-k_{1})^{2}\left(1-({\frac {a_{1}}{a_{2}}})^{2i+1}\right)}},$
| $\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) |
$A_{3}$
$A_{1}$
| $A_{3}-A_{1}=(k_{2}-k_{1})^{2}i(i+1)\left(1-({\frac {a_{1}}{a_{2}}})^{2i+1}\right)CA_{3}.$ | (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.
$i=1,$
| $\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) |
$k_{2}$
$a_{1}=0.$
| $\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) |
$B_{3}$
$k_{1}$
$k_{2},$
$K$
| $K={\frac {(2k_{1}+k_{2})a_{2}^{3}+(k_{1}-k_{2})a_{1}^{3}}{(2k_{1}+k_{2})a_{2}^{3}-2(k_{1}-k_{2})a_{1}^{3}}}k_{2}.$ | (11) |
$n$
$a_{1}$
$k_{1},$
$k_{2},$
$a_{2},$
| $V=\left(A+nB{\frac {1}{r^{2}}}\right)\cos \theta ,$ | (12) |
$B$
| $B={\frac {k_{1}-k_{2}}{2k_{1}+k_{2}}}a_{1}^{3}A.$ | (13) |
$n$
| $p={\frac {na_{1}^{3}}{a_{2}^{3}}}.$ | (14) |
| $V=\left(A+pa_{2}^{3}{\frac {k_{1}-k_{2}}{2k_{1}+k_{2}}}{\frac {1}{r^{2}}}\right)\cos \theta .$ | (15) |
$a_{2}$
$K,$
| $V=\left\{A+a_{2}^{3}{\frac {K-k_{2}}{2K+k_{2}}}{\frac {1}{r^{2}}}\right\}\cos \theta .$ | (16) |
| $K={\frac {2k_{1}+k_{2}+p(k_{1}-k_{2})}{2k_{1}+k_{2}-2p(k_{1}-k_{2})}}k_{2}.$ | (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 $r_{1}=SP.$
$I,$
$S$
$IS$
$r_{2}$
$I,$
| $r_{1}=r_{2},$ | (1) |
| ${\frac {dr_{1}}{d\nu }}=-{\frac {dr_{2}}{d\nu }}.$ | (2) |
$V_{1}$
$E$
$S,$
$E_{2}$
$I,$
$V_{2}$
$E_{1}$
$S,$
| $V_{1}={\frac {E}{r_{1}}}+{\frac {E_{2}}{r_{1}}}\quad \quad \mathrm {and} \quad \quad V_{2}={\frac {E_{1}}{r_{1}}},$ | (3) |
$V_{1}=V_{2}$
| $E+E_{2}=E_{1},$ | (4) |
| ${\frac {1}{k_{1}}}{\frac {dV_{1}}{d\nu }}={\frac {1}{k_{2}}}{\frac {dV_{2}}{d\nu }}$ | (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) |
$E$
$S,$
$E_{1}$
$I$
$E_{1}$
$S.$
​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.

### Conduction through a Plate separating Two Media.

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 $S.$
* $AI=SA,\quad BI_{1}=SB,\quad AJ_{1}=I_{1}A\quad BI_{2}=J_{1}B,\quad AJ_{2}=I_{2}A,\;\mathrm {\&c.;}$
then we have two series of points at distances from each other equal to twice the thickness of the plate.
$P$
| ${\frac {E}{PS}}+{\frac {I}{PI}}+{\frac {I_{1}}{PI_{1}}}+{\frac {I_{2}}{PI_{2}}}+\;\mathrm {\&c.}$ | (8) |
$P'$
| ${\begin{array}{l}{\frac {E'}{P'S}}+{\frac {I'}{P'I}}+{\frac {I_{1}'}{P'I_{1}}}+{\frac {I_{2}'}{P'I_{2}}}+\;\mathrm {\&c.} \\\;\;\;+{\frac {J_{1}'}{P'J_{1}}}+{\frac {J_{2}'}{P'J_{2}}}+\;\;\mathrm {\&c.,} \end{array}}$ | (9) |
$P''$
| ${\frac {E''}{P''S}}+{\frac {J_{1}}{P''J_{1}}}+{\frac {J_{2}}{P''J_{2}}}+\;\mathrm {\&c.,}$ | (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.
$A$
| $I={\frac {k_{2}-k_{1}}{k_{2}+k_{1}}}E,\quad \quad E'={\frac {2k_{2}}{k_{2}+k_{1}}}E.$ | (11) |
$B$
| $I_{1}'={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}}E',\quad \quad E'={\frac {2k_{3}}{k_{2}+k_{1}}}E'.$ | (12) |
$A$
| $J_{1}'={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}I_{1}',\quad \quad I_{1}={\frac {2k_{1}}{k_{1}+k_{2}}}I_{1}',$ | (13) |
$B,$
| $J_{2}'={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}}J_{1}',\quad \quad J_{1}={\frac {2k_{3}}{k_{3}+k_{2}}}J_{1}'.$ | (14) |
* $\rho ={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}\quad \mathrm {and} \quad \rho '={\frac {k_{3}-k_{2}}{k_{3}+k_{2}}},$
| ${\begin{array}{rl}V={\frac {E}{PS}}-\rho {\frac {E}{PI}}+(1-\rho ^{2})\rho '{\frac {E}{PI_{1}}}&+\rho '(1-\rho ^{2})\rho \rho '{\frac {E}{PI_{2}}}+\;\mathrm {\&c.} \\&\quad +\rho '(1-\rho ^{2})(\rho \rho ')^{n-1}{\frac {E}{PI_{n}}}.\end{array}}$ | (15) |
| $V=(1+\rho ')(1-\rho )E\left\{{\frac {1}{PS}}+{\frac {\rho \rho '}{PJ_{1}}}+\,\mathrm {\&c.} +{\frac {(\rho \rho ')^{n}}{PJ_{n}}}\right\}.$ | (16) |
$k_{1}=k_{3}$
$\rho =\rho ',$
| $V=(1-\rho ^{2})E\left\{{\frac {1}{PS}}+{\frac {\rho ^{2}}{PJ_{1}}}+\,\mathrm {\&c.} +{\frac {\rho ^{2n}}{PJ_{n}}}\right\}.$ | (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.
$\rho$
$g$
$\rho$
* $g={\frac {2\rho }{3-\rho }}={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}},\quad \quad \rho ={\frac {3g}{2+g}}={\frac {k_{1}-k_{2}}{k_{1}+k_{2}}}.$
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.
$Z.$
${\overline {X}}$
* ${\begin{array}{rl}{\overline {X}}=X=X',&(c+c'){\overline {u}}=cu+c'u',\\{\overline {Y}}=Y=Y',&(c+c'){\overline {v}}=cv+c'v';\\(c+c'){\overline {Z}}=cZ+c'Z',&\quad {\overline {w}}=w=w'.\end{array}}$
$u,\,u',\,v,\,v',\,Z$
$Z'$
${\overline {X}},\,{\overline {Y}}$
${\overline {w}}$
$D$
* ${\begin{array}{ll}ur_{3}D&=R_{2}{\overline {X}}-Q_{3}{\overline {Y}}+{\overline {w}}q_{2}D,\\vr_{3}D&=R_{1}{\overline {Y}}-P_{3}{\overline {X}}+{\overline {w}}p_{1}D,\\Zr_{3}&=-p_{2}{\overline {X}}-q_{1}{\overline {Y}}+{\overline {w}}.\end{array}}$
$u',\,v'$
$z'.$
${\overline {u}},\,{\overline {v}}$
${\overline {w}}$
${\overline {X}},\,{\overline {Y}}$
${\overline {Z}},$
$h={\frac {c}{r_{3}}}$
$h'={\frac {c'}{r_{3}'}},$
${\begin{array}{l}{\overline {p}}_{1}={\frac {hp_{1}+h'p_{1}'}{h+h'}},\quad \quad {\overline {q}}_{1}={\frac {hq_{1}+h'q_{1}'}{h+h'}},\\{\overline {p}}_{2}={\frac {hp_{2}+h'p_{2}'}{h+h'}},\quad \quad {\overline {q}}_{2}={\frac {hq_{2}+h'q_{2}'}{h+h'}},\\{\overline {p}}_{3}={\frac {cp_{3}+c'p_{3}'}{c+c'}}-{\frac {hh'(q_{1}-q_{1}')(q_{2}-q_{2}')}{(h+h')(c+c')}},\\{\overline {q}}_{3}={\frac {cq_{3}+c'q_{3}'}{c+c'}}-{\frac {hh'(p_{1}-p_{1}')(p_{2}-p_{2}')}{(h+h')(c+c')}},\\{\overline {r}}_{1}={\frac {cr_{1}+c'r_{1}'}{c+c'}}-{\frac {hh'(p_{2}-p_{2}')(q_{2}-q_{2}')}{(h+h')(c+c')}},\\{\overline {r}}_{2}={\frac {cr_{2}+c'r_{2}'}{c+c'}}-{\frac {hh'(p_{1}-p_{1}')(q_{1}-q_{1}')}{(h+h')(c+c')}},\end{array}}$
${\begin{array}{c}{\overline {r}}_{3}={\frac {c+c'}{h+h'}}.\end{array}}$
$P$
$p$
$Q$
$q.$
* ${\overline {p}}_{1}={\overline {q}}_{1},\quad \quad {\overline {p}}_{2}={\overline {q}}_{2},\quad \quad {\overline {p}}_{3}={\overline {q}}_{3},$
or there is no rotatory property developed by stratification, unless it exists in the materials.
$x,\,y$
$z$
$p$
$q$
* ${\overline {r}}_{1}={\frac {cr_{1}+c'r_{1}'}{c+c'}},\quad \quad {\overline {r}}_{2}={\frac {cr_{2}+c'r_{2}'}{c+c'}},\quad \quad {\overline {r}}_{3}={\frac {c+c'}{{\frac {c}{r_{3}}}+{\frac {c'}{r_{3}'}}}}$
​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 $k_{3}c.$
$r$
$a,\,b$
$c,$
$b$
$c,$
$a$
$b,$
$s,$
$k_{1}a$
$x,\,k_{2}b$
$y,$
$k_{3}c$
$z.$
$x,\,y$
$z$
* ${\begin{array}{l}r_{1}={\frac {\{1+k_{1}(1+k_{2})(1+k_{3})\}r+(k_{2}+k_{3}+k_{2}k_{3})s}{(1+k_{2})(1+k_{3})(k_{1}r+s)}}s,\\r_{2}={\frac {(1+k_{2}+k_{2}k_{3})r+(k_{1}+k_{3}+k_{1}k_{2}+k_{1}k_{3}+k_{1}k_{2}k_{3})s}{(1+k_{3})\{k_{2}r+(1+k_{1}+k_{1}k_{2})s\}}}s,\\r_{3}={\frac {(1+k_{3})(r+(k_{1}+k_{2}+k_{1}k_{2})s)}{k_{3}r+(1+k_{1}+k_{2}+k_{2}k_{3}+k_{3}k_{1}+k_{1}k_{2}+k_{1}k_{2}k_{3})s}}s.\end{array}}$
$k_{l},\,k_{2}$
$k_{3}$
* $r_{1}={\frac {5r+3s}{4r+4s}}s,\quad \quad r_{2}={\frac {3r+5s}{2r+6s}}s,\quad \quad r_{3}={\frac {2r+6s}{r+7s}}s.$
$r=0,$
* $r_{1}={\frac {3}{4}}s,\quad \quad r_{2}={\frac {5}{6}}s,\quad \quad r_{3}={\frac {6}{7}}s.$
$r=\infty$
* $r_{1}={\frac {5}{4}}s,\quad \quad R_{2}={\frac {3}{2}}s,\quad \quad r_{3}=2s.$
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 $abcK.$
* $aX+bY+cZ,$
$C'$
* $C'=Kabc(aX+bY+cZ).$
$bcu,$
| 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
${\begin{array}{lll}r_{1}+Ka^{2},&r_{2}+Kb^{2},&r_{3}+Kc^{2},\\p_{1}+Kbc,&p_{2}+Kca,&p_{3}+Kab,\\q_{1}+Kbc,&q_{2}+Kca,&q_{3}+Kab.\end{array}}$
$p_{1},$
$q_{1},$
$p_{1}$
$q_{1}$
​324.] To construct a framework of linear conductors which shall have any given coefficients of conductivity forming a symmetrical system.
$0,\,L,\,M,\,N,$
${\begin{array}{lllll}\quad \quad &x&y&z\quad &Potential.\\O&0&0&0&0\\L&0&1&1&0+Y+Z,\\M&1&0&1&0+Z+X,\\N&1&1&0&0+X+Y.\end{array}}$
* $OL,\quad \quad OM,\quad \quad ON,\quad \quad MN,\quad \quad NL,\quad \quad LM,$
* $A,\quad \quad B,\quad \quad C,\quad \quad P,\quad \quad Q,\quad \quad R.$
* $Y+Z,\quad \quad Z+X,\quad \quad X+Y,\quad \quad Y-Z,\quad \quad Z-X,\quad \quad X-Y,$
* $A(Y+Z),\quad B(Z+X),\quad C(X+Y),\quad P(Y-Z),\quad Q(Z-X),\quad R(X-Y).$
$x$
$LM,\,LN,\,OM$
$ON,$
* $u=(B+C+Q+R)X+(C-R)Y\quad \quad +(B-Q)Z.$
* ${\begin{array}{lll}v=(C-R)X&+(C+A+R+P)Y&+(A-P)Y,\\w=(B-Q)X&+(A-P)Y&+(A+B+P+Q)Z;\end{array}}$
whence we find by comparison with the equations of conduction, Art. 298,
${\begin{array}{ll}4A=r_{2}+r_{3}-r_{1}+2p_{1},&4P=r_{2}+r_{3}-r_{1}-2p_{1}\\4B=r_{3}+r_{1}-r_{2}+2p_{2},&4Q=r_{3}+r_{1}-r_{2}-2p_{2}\\4C=r_{1}+r_{2}-r_{3}+2p_{3},&4R=r_{1}+r_{2}-r_{3}-2p_{3}\end{array}}$
* ↑ See Kirchhoff, Pogg. Ann. lxiv. 497, and lxvii. 344; Quincke, Pogg. xcvii. 382; and Smith, Proc. R. S. Edin., 1869-70, p. 79.
* ↑ See Sir W. Thomson's 'Note on Induced Magnetism in a Plate,' Camb. and Dub. Math. Journ., Nov. 1845, or Reprint, art. ix. § 156.