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

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

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

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