Page:A Treatise on Electricity and Magnetism - Volume 1.djvu/195

$V={\frac {Aa-Bb}{a-b}}+{\frac {A-B}{a^{-1}-b^{-1}}}r^{-1}.$ $R=-{\frac {dV}{dr}}={\frac {A-B}{a^{-1}-b^{-1}}}r^{-2}$

If $\sigma _{1},\ \sigma _{2}$ are the surface-densities on the opposed surfaces of a solid sphere of radius $a$, and a spherical hollow of radius $b$, then

$\sigma _{1}={\frac {1}{4\pi a^{2}}}{\frac {A-B}{a^{-1}-b^{-1}}},\ \sigma _{2}={\frac {1}{4\pi b^{2}}}{\frac {B-A}{a^{-1}-b^{-1}}}.$

If $E_{1}$ and $E_{2}$ be the whole charges of electricity on these surfaces,

$E_{1}=4\pi a^{2}\sigma _{1}={\frac {A-B}{a^{-1}-b^{-1}}}=-E_{2}.$

The capacity of the enclosed sphere is therefore ${\frac {ab}{b-a}}$.

If the outer surface of the shell be also spherical and of radius $c$, then, if there are no other conductors in the neighbourhood, the charge on the outer surface is

$E_{3}=Bc.\,$

Hence the whole charge on the inner sphere is

$E_{1}={\frac {ab}{b-a}}(A-b),$

and that of the outer

$E_{2}+E_{3}={\frac {ab}{b-a}}(B-A)+Bc$

If we put $b=\infty$, we have the case of a sphere in an infinite space. The electric capacity of such a sphere is $a$, or it is numerically equal to its radius.

The electric tension on the inner sphere per unit of area is

$p={\frac {1}{8\pi }}{\frac {b^{2}}{a^{2}}}{\frac {(A-B)^{2}}{(b-a)^{2}}}.$

The resultant of this tension over a hemisphere is $\pi a^{2}p=F$ normal to the base of the hemisphere, and if this is balanced by a surface tension exerted across the circular boundary of the hemisphere, the tension on unit of length being $T$, we have

$F=2\pi aT.$

Hence

${\begin{array}{l}F={\frac {b^{2}}{8}}{\frac {(A-B)^{2}}{(b-a)^{2}}}={\frac {E_{1}^{2}}{8a^{2}}},\\\\T={\frac {b^{2}}{16\pi a}}{\frac {(A-B)^{2}}{(b-a)^{2}}}.\end{array}}$