Page:On Faraday's Lines of Force.pdf/59

Rh Let the value of the potential undisturbed by the presence of the sphere be

$p=Ix$.

Let the sphere produce an additional potential, which for external points is

$p'=A\frac{a^3}{r^3}x$,

and let the potential within the sphere be

$p_1=Bx$.

Let $$k'$$ he the coefﬁcient of resistance outside, and $$k$$ inside the sphere, then the conditions to be fulﬁlled are, that the interior and exterior potentials should coincide at the surface, and that the induction through the surface should be the same whether deduced from the external or the internal potential. Putting $$x=rcos\theta$$, we have for the external potential

$P= \left(Ir+A\frac{a^3}{r^2}\right) cos\theta $, and for the internal

$p_1 =Br cos \theta$,

and these must be identical when $$r=a$$, or

$I+A = B.$

The induction through the surface in the external medium is

$\frac{1}{k'}\frac{dp_1}{dr_{r=a}}=\frac{1}{k'}(I-2A)cos\theta$,|undefined

and that through the interior surface is

$\frac{1}{k}\frac{dp_1}{dr_{r=a}}=\frac{1}{k}B\ cos\theta$,|undefined

$and \therefore \ \frac{1}{k'}(I-2A)=\frac{1}{k}B.$

These equations give

$A=\frac{k-k'}{2k+k'}I, \ B=\frac{3k}{2k+k'} I .$

The effect outside the sphere is equal to that of a little magnet whose length is $$l$$ and moment $$ml,$$ provided

$ml=\frac{k-k'}{2k+k'}a^3 I.$