Page:On Faraday's Lines of Force.pdf/57

Rh the pressure at the surface $$r=a$$ will be reduced to zero. Now placing a source $$4\pi\frac{pa}{k}$$ at the centre, the pressure at the surface will be uniform and equal to $$p$$.

The whole amount of ﬂuid emitted by the surface $$r=a$$ may be found byadding the rates of production of the sources within it. The result is

$4\pi a \left\{\frac{p}{k}-\frac{P_1}{b_1}-\frac{P_2}{b_2}-\&c.\right\}$.

To apply this result to the case of a conducting sphere, let us suppose the external sources $$4\pi P_1 ,4\pi P_2$$, to be small electriﬁed bodies, containing $$e_1, e_2 ,$$ of positive electricity. Let us also suppose that the whole charge of the conducting sphere is $$=E$$ previous to the action of the external points. Then all that is required for the complete solution of the problem is, that the surface of the sphere shall be a surface of equal potential, and that the total charge of the surface shall be $$E$$.

If by any distribution of imaginary sources within the spherical surface we can effect this, the value of the corresponding potential outside the sphere is the true and only one. The potential inside the sphere must really be constant and equal to that at the surface.

We must therefore ﬁnd the images of the external electriﬁed points, that is, for every point at distance $$b$$ from the centre we must find a point on the same radius at a distance $$\frac{a^2}{b_1}$$, and at that point we must place a quantity $$= -e\frac{a}{b_1}$$ of imaginary electricity.

At the centre we must put a quantity $$E'$$ such that

$E' = E+e_1\frac{a}{b_1}+e_2\frac{a}{b_2} + \&c.$;

then if $$R$$ be the distance from the centre, $$r_1, r_2, \&c.$$ the distances from the electriﬁed points, and $$r'_1, r'_2 , \&c.$$ the distances from their images at any point outside the sphere, the potential at that point will be

$p=\frac{E'}{R}+e_1\left(\frac{1}{r_1}-\frac{a}{b_1}\frac{1}{r'_1}\right)+e_2\left(\frac{1}{r_2}-\frac{a}{b_2}\frac{1}{r'_2}\right)+\&c.$

$=\frac{E}{R}+\frac{e_1}{b_1}\left(\frac{a}{R}+\frac{b_1}{r_1}-\frac{a}{r'_1}\right)+\frac{e_2}{b_2}\left(\frac{a}{R}+\frac{b_2}{r_2}-\frac{a}{r'_2}\right)+\&c.$