Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/249

images, $n'$ in each series. We shall obtain in this way, besides the influencing point, $2nn'-1$ positive and $2nn'$ negative images. These $4nn'$ points are the intersections of $n$ circles with $n'$ other circles, and these circles belong to the two systems of lines of curvature of a cyclide.

If each of these points is charged with the proper quantity of electricity, the surface whose potential is zero will consist of $n+n'$ spheres, forming two series of which the successive spheres of the first set intersect at angles ${\tfrac {\pi }{n}}$, and those of the second set at angles ${\tfrac {\pi }{n'}}$, while every sphere of the first set is orthogonal to every sphere of the second set.

Case of Two Spheres cutting Orthogonally. See Fig. IV at the end of this volume.

Fig. IV

Fig. 12

168.] Let $A$ and $B$, Fig. 12, be the centres of two spheres cutting each other orthogonally in $D$ and $D'$, and let the straight line $DD'$ cut the line of centres in $C$. Then $C$ is the image of $A$ with respect to the sphere $B$, and also the image of $B$ with respect to the sphere whose centre is $A$. If $AD=\alpha$, $BD=\beta$, then $AB={\sqrt {\alpha ^{2}+\beta ^{2}}}$, and if we place at $A,B,C$ quantities of electricity equal to $\alpha ,\beta$, and $-{\tfrac {\alpha \beta }{\sqrt {\alpha ^{2}+\beta ^{2}}}}$ respectively, then both spheres will be equipotential surfaces whose potential is unity.

We may therefore determine from this system the distribution of electricity in the following cases:

(1) On the conductor $PDQD'$ formed of the larger segments of both spheres. Its potential is 1, and its charge is

$\alpha +\beta -{\frac {\alpha \beta }{\sqrt {\alpha ^{2}+\beta ^{2}}}}=AD+BD-CD$

This quantity therefore measures the capacity of such a figure when free from the inductive action of other bodies.

The density at any point $P$ of the sphere whose centre is $A$, and the density at any point $Q$ of the sphere whose centre is $B$, are respectively

${\frac {1}{4\pi \alpha }}\left(1-\left({\frac {\beta }{BP}}\right)^{3}\right)$ and ${\frac {1}{4\pi \beta }}\left(1-\left({\frac {\alpha }{AQ}}\right)^{3}\right)$