Page:On Faraday's Lines of Force.pdf/56

210 If there be two shells, one giving out ﬂuid at a rate $$4\pi Pa^2$$, and the other absorbing at the rate of $$4\pi P'a'^2$$, then the expression for the pressure will be, outside the shells,

$p=4\pi P\frac{a^2}{r}-4\pi P'\frac{a'^2}{r'}$

where r and r' are the distances from the centres of the two shells. Equating this expression to zero we have, as the surface of no pressure, that for which

$\frac{r'}{r} = \frac{P'a'^2}{Pa^2}$.

Now the surface, for which the distances to two ﬁxed points have a given ratio, is a sphere of which the centre $$O$$ is in the line joining the centres of the shells $$CC'$$ produced, so that

$C'O=CC'\frac{\overline{P'a'^2}\vert^2}{\overline{Pa^2}\vert^2-\overline{P'a'^2}\vert^2}$,

and its radius

$=CC'\frac{Pa^2.P'a'^2}{\overline{Pa^2}\vert^2-\overline{P'a'^2}\vert^2}$,

If at the centre of this sphere we place another source of the ﬂuid, then the pressure due to this source must be added to that due to the other two; and since this additional pressure depends only on the distance from the centre, it will be constant at the surface of the sphere, where the pressure due to the two other sources is zero.

We have now the means of arranging a system of sources within a given sphere, so that when combined with a given system of sources outside the sphere, they shall produce a given constant pressure at the surface of the sphere.

Let $$a$$ be the radius of the sphere, and $$p$$ the given pressure, and let the given sources be at distances $$b_1, b_2 ,$$ &c. from the centre, and let their rates of production be $$4\pi P_1, 4\pi P_2$$,&c.

Then if at distances $$\frac{a^2}{b_1}, \frac{a^2}{b_2}$$, &c. (measured in the same direction as $$b_1, b_2 ,$$ &c. from the centre) we place negative sources whose rates are

$-4\pi P_1\frac{a}{b_1}, -4\pi_2\frac{a}{b_2}$, &c.