Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/253

and the charge at the intersection of the four perpendiculars is

$-{\frac {1}{\sqrt {{\frac {1}{\alpha ^{2}}}+{\frac {1}{\beta ^{2}}}+{\frac {1}{\gamma ^{2}}}+{\frac {1}{\delta ^{2}}}}}}.$

System of Four Spheres Intersecting at Right Angles under the Action of an Electrified Point.

170.] Let the four spheres be $A,B,C,D$, and let the electrified point be $O$. Draw four spheres $A_{1},B_{1},C_{1},D_{1}$, of which any one, $A_{1}$, passes through and cuts three of the spheres, in this case $B,C$, and $D$, at right angles. Draw six spheres $(ab),(ac),(ad),(be),(bd),(cd)$, of which each passes through and through the circle of intersection of two of the original spheres.

The three spheres $B_{1},C_{1},D_{1}$ will intersect in another point besides $O$. Let this point be called $A'$, and let $B',C'$, and $D'$ be the intersections of $C_{1},D_{1},A_{1}$ of $D_{1},A_{1},B_{1}$, and of $A_{1},B_{1},C_{1}$ respectively. Any two of these spheres, $A_{1},B_{1}$, will intersect one of the six ($cd$) in a point ($a'b'$). There will be six such points.

Any one of the spheres, $A_{1}$, will intersect three of the six $(ab),(ac),(ad)$ in a point $a'$. There will be four such points. Finally, the six spheres $(ab),(ac),(ad),(cd),(db),(bc)$, will intersect in one point $S$.

If we now invert the system with respect to a sphere of radius $R$ and centre $O$, the four spheres $A,B,C,D$ will be inverted into spheres, and the other ten spheres will become planes. Of the points of intersection the first four $A',B',C',D'$ will become the centres of the spheres, and the others will correspond to the other eleven points in the preceding article. These fifteen points form the image of $O$ in the system of four spheres.

At the point $A'$, which is the image of $O$ in the sphere $A$, we must place a charge equal to the image of $O$, that is, $-{\tfrac {\alpha }{a}}$, where $\alpha$ is the radius of the sphere $A$, and $a$ is the distance of its centre from $O$. In the same way we must place the proper charges at $B',C',D'$.

The charges for each of the other eleven points may be found from the expressions in the last article by substituting $\alpha ',\beta ',\gamma ',\delta '$ for $\alpha ,\beta ,\gamma ,\delta$, and multiplying the result for each point by the distance of the point from $O$, where

$\alpha '=-{\frac {\alpha }{a^{2}-\alpha ^{2}}},\ \beta '=-{\frac {\beta }{b^{2}-\beta ^{2}}},\ \gamma '=-{\frac {\gamma }{c^{2}-\gamma ^{2}}},\ \delta '=-{\frac {\delta }{d^{2}-\delta ^{2}}}.$