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

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

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

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