Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/417

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)