Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/300

 But by putting the inductor $$A$$ in communication with the receiver $$D$$, and the inductor $$C$$ with the receiver $$B$$, the potentials of the inductors will be continually increased, and the quantity of electricity communicated to the receivers in each revolution will continually increase.

For instance, let the potential of $$A$$ and $$O$$ be $$U$$, and that of and $$C$$, $$V$$, and when the carrier is within $$A$$ let the charge on $$A$$ and $$C$$ be $$x$$, and that on the carrier $$z$$, then, since the potential of the carrier is zero, being in contact with earth, its charge is $$z=-QU $$. The carrier enters $$B$$ with this charge and communicates it to $$B$$. If the capacity of $$B$$ and $$C$$ is $$B$$, their potential will be changed from $$V$$ to $$V-\tfrac{Q}{B}U $$.

If the other carrier has at the same time carried a charge $$-QV$$ from $$C$$ to $$D$$, it will change the potential of $$A$$ and $$O$$ from $$U$$ to $$U-\tfrac{Q'}{A}V $$, if $$Q'$$ is the coefficient of induction between the carrier and $$C$$, and $$A$$ the capacity of $$A$$ and $$D$$. If, therefore, $$U_n$$ and $$V_n$$ be the potentials of the two inductors after $$n$$ half revolutions, and $$U_{n+1} $$ and $$V_{n+1} $$ after $$n+1$$ half revolutions,

$\begin{array}{c} U_{n+1}=U_{n}-\frac{Q'}{A}V_{n},\\ \\ V_{n+1}=V_{n}-\frac{Q}{B}U_{n}. \end{array} $

If we write $$p^{2}=\tfrac{Q}{B} $$ and $$q^{2}=\tfrac{Q'}{A} $$, we find

$\begin{array}{l} pU_{n+1}+qV_{n+1}=\left(pU_{n}+qV_{n}\right)(1-pq)=\left(pU_{0}+qV_{0}\right)(1-pq)^{n+1},\\ pU_{n+1}-qV_{n+1}=\left(pU_{n}-qV_{n}\right)(1+pq)=\left(pU_{0}-qV_{0}\right)(1+pq)^{n+1}. \end{array} $

Hence

$\begin{array}{l} U_{n}=U_{0}\left((1-pq){}^{n}+(1+pq){}^{n}\right)+\frac{q}{p}V_{0}\left((1-pq){}^{n}-(1+pq){}^{n}\right),\\ \\ V_{n}=\frac{p}{q}U_{0}\left((1-pq){}^{n}-(1+pq){}^{n}\right)+V_{0}\left((1-pq){}^{n}+(1+pq){}^{n}\right). \end{array} $

It appears from these equations that the quantity $$pU+qV $$ continually diminishes, so that whatever be the initial state of electrification the receivers are ultimately oppositely electrified, so that the potentials of $$A$$ and $$B$$ are in the ratio of $$p$$ to $$-q$$.

On the other hand, the quantity $$pU-qV $$ continually increases, so that, however little $$pU$$ may exceed or fall short of $$qV$$ at first, the difference will be increased in a geometrical ratio in each