Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/303

 carrier when it is at the middle point of its passage, but $$C, C'$$ do not cover it so much.

We shall suppose $$A, B, C$$ to be connected with a Leyden jar of great capacity at potential $$V$$, and $$A', B', C'$$ to be connected with another jar at potential $$-V'$$.

$$P$$ is one of the carriers moving in a circle from $$A$$ to $$C'$$, &c., and touching in its course certain springs, of which a and a are connected with $$A$$ and $$A'$$ respectively, and $$e, e'$$ are connected with the earth.

Let us suppose that when the carrier $$P$$ is in the middle of $$A$$ the coefficient of induction between $$P$$ and $$A$$ is $$-A$$. The capacity of $$P$$ in this position is greater than $$A$$, since it is not completely surrounded by the receiver $$A$$. Let it be $$A + a$$.

Then if the potential of $$P$$ is $$U$$, and that of $$A, V$$, the charge on $$P$$ will be $$(A+a)U-AV $$.

Now let $$P$$ be in contact with the spring $$a$$ when in the middle of the receiver $$A$$, then the potential of $$P$$ is $$V$$, the same as that of $$A$$, and its charge is therefore $$aV$$.

If $$P$$ now leaves the spring $$a$$ it carries with it the charge $$aV$$. As $$P$$ leaves $$A$$ its potential diminishes, and it diminishes still more when it comes within the influence of $$C'$$, which is negatively electrified.

If when $$P$$ comes within $$C$$ its coefficient of induction on $$C$$ is $$-C'$$, and its capacity is $$C'+c' $$, then, if $$U$$ is the potential of $$P$$ the charge on $$P$$ is

$(C'+c')U+C'V'=aV. $

If

$C'V'=aV $

then at this point $$U$$ the potential of $$P$$ will be reduced to zero.

Let $$P$$ at this point come in contact with the spring $$e'$$ which is connected with the earth. Since the potential of $$P$$ is equal to that of the spring there will be no spark at contact.

This conductor $$C'$$, by which the carrier is enabled to be connected to earth without a spark, answers to the contrivance called a regenerator in heat-engines. We shall therefore call it a Regenerator.

Now let $$P$$ move on, still in contact with the earth-spring $$e'$$, till it comes into the middle of the inductor $$B$$, the potential of which is $$V$$. If $$-B$$ is the coefficient of induction between $$P$$ and $$B$$ at this point, then, since $$U=0$$ the charge on $$P$$ will be $$-BV$$.

When $$P$$ moves away from the earth-spring it carries this charge with it. As it moves out of the positive inductor $$B$$ towards the