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