Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/250

 At the points of intersection, $$D, D'$$, the density is zero.

If one of the spheres is very much larger than the other, the density at the vertex of the smaller sphere is ultimately three times that at the vertex of the larger sphere.

(2) The lens $$P'DQ'D'$$ formed by the two smaller segments of the spheres, charged with a quantity of electricity $$=-\frac{\alpha\beta}{\sqrt{\alpha^{2}+\beta^{2}}}$$, and acted on by points $$A$$ and $$B$$, charged with quantities $$\alpha$$ and $$\beta$$, is also at potential unity, and the density at any point is expressed by the same formulae.

(3) The meniscus $$DPD'Q'$$ formed by the difference of the segments charged with a quantity $$\alpha$$, and acted on by points $$B$$ and $$C$$, charged respectively with quantities $$\beta$$ and $$\frac{-\alpha\beta}{\sqrt{\alpha^{2}+\beta^{2}}}$$, is also in equilibrium at potential unity.

(4) The other meniscus $$QDP'D'$$ under the action of $$A$$ and $$C$$.

We may also deduce the distribution of electricity on the following internal surfaces.

The hollow lens $$P'DQ'D$$ under the influence of the internal electrified point $$C$$ at the centre of the circle $$DD'$$.

The hollow meniscus under the influence of a point at the centre of the concave surface.

The hollow formed of the two larger segments of both spheres under the influence of the three points $$A, B, C$$.

But, instead of working out the solutions of these cases, we shall apply the principle of electrical images to determine the density of the electricity induced at the point $$P$$ of the external surface of the conductor $$PDQD'$$ by the action of a point at $$O$$ charged with unit of electricity.

Let

$\begin{array}{lllllll} OA=a, & & OB=b, & & OP=r, &  & BP=p,\\ \\AD=\alpha, & & BD=\beta, &  & AB=\sqrt{\alpha^{2}+\beta^{2}}.\end{array}$|undefined

Invert the system with respect to a sphere of radius unity and centre $$O$$.

The two spheres will remain spheres, cutting each other orthogonally, and having their centres in the same radii with $$A$$ and $$B$$. If we indicate by accented letters the quantities corresponding to the inverted system,

$\begin{array}{l} a'=\frac{a}{a^{2}-\alpha^{2}},\ b'=\frac{b}{b^{2}-\beta^{2}},\ \alpha'=\frac{\alpha}{a^{2}-\alpha^{2}},\ \beta'=\frac{\beta}{b^{2}-\beta^{2}},\\ \\r'=\frac{1}{r},\ p'^{2}=\frac{\beta^{2}r^{2}+\left(b^{2}-\beta^{2}\right)\left(p^{2}-\beta^{2}\right)}{r^{2}\left(b^{2}-\beta^{2}\right)^{2}}.\end{array}$|undefined