Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/426

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,

1. Théorie de la Chaleur, art. 105.