Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/267

 image of $$EF$$. Bisect the arc $$F'CE'$$ in $$D'$$, so that $$F'D'=D'E'$$, and draw $$D'QD$$ to meet the sphere in $$D$$. $$D$$ is the point required. Also through $$O$$, the centre of the sphere, and $$Q$$ draw $$HOQH'$$ meeting the sphere in $$H$$ and $$H'$$. Then if $$P$$ be any point in the bowl, the surface-density at $$P$$ on the side which is separated from $$Q$$ by the completed spherical surface, induced by a quantity $$q$$ of electricity at $$Q$$, will be

$\sigma=\frac{q}{2\pi^{2}}\frac{QH\cdot QH'}{HH'\cdot PQ^{3}}\left\{ \frac{PQ}{DQ}\left(\frac{CD^{2}-a^{2}}{a^{2}-CP^{2}}\right)^{\frac{1}{2}}-\tan^{-1}\left[\frac{PQ}{DQ}\left(\frac{CD^{2}-a^{2}}{a^{2}-CP^{2}}\right)^{\frac{1}{2}}\right]\right\} ,$|undefined

where $$a$$ denotes the chord drawn from $$C$$, the pole of the bowl, to the rim of the bowl.

On the side next to $$Q$$ the surface-density is

$\sigma+\frac{q}{2\pi^{2}}\frac{QH\cdot QH'}{HH'\cdot PQ^{3}}.$|undefined