Page:A Dynamical Theory of the Electromagnetic Field.pdf/37

Rh {{MathForm2|(51)|$$\left.\begin{array}{lll} e_{1}=-f_{1}, & & \frac{de_{1}}{dt}=-p_{1},\\ \\e_{2}=f_{1}-f_{2}, & & \frac{de_{2}}{dt}=p_{1}-p_{2},\\ etc. & & etc.\end{array}\right\} $$}}

But by equations (E) and (F),

{{MathForm2|(52)|$$\left.\begin{array}{lll} \Psi_{1}-\Psi_{2} & =a_{1}k_{1}f_{1} & =-r_{1}p_{1},\\ \Psi_{2}-\Psi_{3} & =a_{2}k_{2}f_{2} & =-r_{2}p_{2},\\ etc. & etc. & etc.\end{array}\right\} $$}}

After the electromotive force has been kept up for a sufficient time the current becomes the same in each layer, and

$p_{1}=p_{2}=etc.=p=\frac{\Psi}{r}$

where $$\Psi$$ is the total difference of potentials between the extreme layers. We have then

{{MathForm2|(53)|$$\left.\begin{array}{lllcl} & f_{1}=- & \frac{\Psi}{r}\frac{r_{1}}{a_{1}k_{1}}, & & f_{2}=-\frac{\Psi}{r}\frac{r_{2}}{a_{2}k_{2}},\ etc.\\ \mathrm{and}\\ & e_{1}= & \frac{\Psi}{r}\frac{r_{1}}{a_{1}k_{1}}, & & e_{2}=\frac{\Psi}{r}\left(\frac{r_{2}}{a_{2}k_{2}}-\frac{r_{1}}{ak_{1}}\right),\ etc.\end{array}\right\} $$}}

These are the quantities of electricity on the different surfaces.

(87) Now let the condenser be discharged by connecting the extreme surfaces through a perfect conductor so that their potentials are instantly rendered equal, then the electricity on the extreme surfaces will be altered, but that on the internal surfaces will not have time to escape. The total difference of potentials is now

whence if $$e'_{1}$$ is what $$e_{1}$$ becomes at the instant of discharge,

The instantaneous discharge is therefore $$\frac{\Psi}{ak}$$, or the quantity which would be discharged by a condenser of air of the equivalent thickness $$a$$, and it is unaffected by the want of perfect insulation.

(88) Now let us suppose the connexion between the extreme surfaces broken, and the condenser left to itself, and let us consider the gradual dissipation of the internal charges. Let $$\Psi'$$ be the difference of potential of the extreme surfaces at any time $$t$$; then

but