Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/245

 If we now superpose on this system a charge $$e$$ at $$A$$, where

the potential on the spherical surface, and at all points on the same side as $$B$$, will be reduced to zero. At all points on the same side as $$A$$ the potential will be that due to a charge $$e$$ at $$A$$, and a charge $$e'\frac{R}{f'}$$ at $$B$$.

But

as we found before for the charge of the image at $$B$$.

To find the density at any point of the first sphere we have

Substituting for the value of $$\sigma'$$ in terms of the quantities be longing to the first sphere, we find the same value as in Art. 158,

On Finite Systems of Successive Images.

165.] If two conducting planes intersect at an angle which is a submultiple of two right angles, there will be a finite system of images which will completely determine the electrification.

For let $$AOB$$ be a section of the two conducting planes per pendicular to their line of inter section, and let the angle of intersection $$AOB=\frac{\pi}{n},$$, let $$P$$ be an electrified point, and let $$PO=r$$, and $$POB=\theta$$. Then, if we draw a circle with centre and radius $$OP$$, and find points which are the successive images of $$P$$ in the two planes beginning with $$OB$$, we shall find $$Q_1$$ for the image of $$P$$ in $$OB$$, $$P_2$$ for the image of $$Q_1$$ in $$OA$$, $$Q_3$$ for that of $$P_2$$ in $$OB$$, $$P_3$$ for that of $$Q_3$$ in $$OA$$, and $$Q_2$$ for that of $$P_3$$ in $$OB$$.

If we had begun with the image of $$P$$ in $$AO$$ we should have found the same points in the reverse order $$Q_{2},P_{3},Q_{3},P_{2},Q_{1}$$, provided $$AOB$$ is a submultiple of two right angles.