Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/254

Two Spheres not Intersecting.

171.] When a space is bounded by two spherical surfaces which do not intersect, the successive images of an influencing point within this space form two infinite series, all of which lie beyond the spherical surfaces, and therefore fulfil the condition of the applicability of the method of electrical images.

Any two non-intersecting spheres may be inverted into two concentric spheres by assuming as the point of inversion either of the two common inverse points of the pair of spheres.

We shall begin, therefore, with the case of two uninsulated concentric spherical surfaces, subject to the induction of an electrified point placed between them.

Let the radius of the first be $b$, and that of the second be $be^{\varpi }$, and let the distance of the influencing point from the centre be $r=be^{u}$.

Then all the successive images will be on the same radius as the influencing point.

Let $Q_{0}$, Fig. 14, be the image of $P$ in the first sphere, $P_{1}$ that of $Q_{0}$ in the second sphere, $Q_{1}$ that of $P_{1}$ in the first sphere, and so on; then

Fig. 14

${\begin{array}{rll}&&OP_{s}\cdot OQ_{s}=b^{2},\\\mathrm {and} &&OP_{s}\cdot OQ_{s-1}=b^{2}e^{2\varpi },\\\mathrm {also} &&OQ_{0}=be^{-u}\\&&OP_{1}=be^{u+2\varpi },\\&&OQ_{1}=be^{-(u+2\varpi )},\ etc.\\\mathrm {Hence} &&OP_{s}=be^{(u+2s\varpi )},\\&&OQ_{s}=be^{-(u+2s\varpi )}.\end{array}}$

If the charge of $P$ is denoted by $P$, then

$P_{s}=Pe^{s\varpi },\ Q_{s}=-Pe^{-(u+s\varpi )}.$

Next, let $Q_{1}'$ be the image of $P$ in the second sphere, $P_{1}'$ that of $Q_{1}'$ in the first, &c.,

${\begin{array}{rlcrl}OQ_{1}'=&be^{2\varpi -u},&&OP_{1}'=&be^{u-2\varpi },\\OQ_{2}'=&be^{4\varpi -u},&&OP_{2}'=&be^{u-4\varpi },\\OP_{s}'=&be^{u-2s\varpi },&&OQ_{s}'=&be^{2s\varpi -u},\\P_{s}'=&Pe^{s\varpi },&&Q_{s}'=&Pe^{s\varpi -u}.\end{array}}$

Of these images all the $P's$ are positive, and all the $Q's$ negative, all the $P'$'s and $Q'$'s belong to the first sphere, and all the $P$'s and $Q$'s to the second.