Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/256

 equal to $$Ce^u$$ where $$C$$ is a numerical quantity which for simplicity we may make equal to unity.

We therefore put

$u=\log\frac{O'P}{OP},\ \alpha=\log\frac{O'A}{OA},\ \beta=\log\frac{O'B}{OB}.$

Let

$\beta-\alpha=\varpi,\ u-a=\theta.$

Then all the successive images of $$P$$ will lie on the arc $$OAPBO'$$.

The position of the image of $$P$$ in $$A$$ is $$Q_0$$ where

$u\left(Q_{0}\right)=\log\frac{O'Q}{OQ}=2a-u.$

That of $$Q_0$$ in $$B$$ is $$P_1$$ where

$u\left(P_{1}\right)=\log\frac{O'P_{1}}{OP_{1}}=u+2\varpi.$|undefined

Similarly

$u\left(P_{s}\right)=u+2s\varpi,\ u\left(Q_{s}\right)=2a-u-2s\varpi.$

In the same way if the successive images of $$P$$ in $$B, A, B$$, &c. are $$Q_{0}',P_{1}',Q_{1}',$$, &c.,

$\begin{array}{lll} u\left(Q_{0}'\right)=2\beta-u, & & u\left(P_{0}'\right)=u-2\varpi;\\ u\left(P_{s}'\right)=u-2s\varpi, & & u\left(Q_{s}'\right)=2\beta-u+2s\varpi.\end{array}$

To find the charge of any image $$P_s$$ we observe that in the inverted figure its charge is

$P\sqrt{\frac{OP_{s}}{OP}}$|undefined

In the original figure we must multiply this by $$O'P_s$$. Hence the charge of $$P_s$$ in the dipolar figure is

$P\sqrt{\frac{OP_{s}\cdot O'P_{s}}{OP\cdot O'P}}.$|undefined

If we make $$\xi=\sqrt{OP\cdot O'P}$$, and call $$\xi$$ the parameter of the point $$P$$, then we may write

$P_{s}=\frac{\xi_{s}}{\xi}P,$|undefined

or the charge of any image is proportional to its parameter.

If we make use of the curvilinear coordinates $$u$$ and $$v$$, such that

$e^{u+\sqrt{-1}v}=\frac{x+\sqrt{-1}y-k}{x+\sqrt{-1}y+k},$

then

$x=-\frac{k\sin hu}{\cos hu-\cos v},\ y=\frac{k\sin v}{\cos hu-\cos v};$