Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/127

Hence

${\frac {dV_{2}}{dv}}={\frac {K_{1}}{K_{2}}}{\frac {dV_{1}}{dv}}$, and ${\frac {K_{1}-K_{2}}{K_{2}}}{\frac {dV_{1}}{dv}}+4\pi \rho '=0$,

${\frac {dV_{1}}{dv}}={\frac {4\pi \rho 'K_{2}}{K_{1}-K_{2}}}$, ${\color {White}xxxxx}{\frac {dV_{2}}{dv}}={\frac {4\pi \rho 'K_{1}}{K_{1}-K_{2}}}$.

The surface-density $\sigma '$ is that of the apparent electrification produced at the surface of the solid dielectric by induction. It disappears entirely when the inducing force is removed, but if during the action of the inducing force the apparent electrification of the surface is discharged by passing a flame over the surface, then, when the inducing force is taken away, there will appear an electrification opposite to $\sigma '$ ^[1].

In a heterogeneous dielectric in which $K$ varies continuously, if $\rho '$ be the apparent volume-density,

${\frac {d^{2}V}{dx^{2}}}+{\frac {d^{2}V}{dy^{2}}}+{\frac {d^{2}V}{dz^{2}}}+4\pi \rho '=0$

Comparing this with the equation above, we find

$4\pi (\rho -K\rho ')+{\frac {dK}{dx}}{\frac {dV}{dx}}+{\frac {dK}{dy}}{\frac {dV}{dy}}+{\frac {dK}{dz}}{\frac {dV}{dz}}=0$

The true electrification, indicated by $\rho$, in the dielectric whose variable inductive capacity is denoted by $K$, will produce the same potential at every point as the apparent electrification, indicated by $\rho '$, would produce in a dielectric whose inductive capacity is every where equal to unity.

1. See Faraday's 'Remarks on Static Induction', Proceedings of the Royal Institution, Feb. 12, 1858.