Page:On Faraday's Lines of Force.pdf/22

176 be proportional to the resultant attraction of all the particles. Now we may ﬁnd the resultant pressure at any point by adding the pressures due to the given sources, and therefore we may ﬁnd the resultant velocity in a given direction from the rate of decrease of pressure in that direction, and this will be proportional to the resultant attraction of the particles resolved in that direction.

Since the resultant attraction in the electrical problem is proportional to the decrease of pressure in the imaginary problem, and since we may select any values for the constants in the imaginary problem, we may assume that the resultant attraction in any direction is numerically equal to the decrease of pressure in that direction, or

$X=-\frac{dp}{dx}$.

By this assumption we ﬁnd that if $$V$$ be the potential, $ dV=Xdx+Ydy+Zdz=-dp$, or since at an inﬁnite distance $$V=0$$ and $$p$$=0, $$V=-p$$.

In the electrical problem we have

$V= -\Sigma\left(\frac{dm}{r}\right)$, In the ﬂuid $$p= \Sigma\left(\frac{k}{4\pi}\frac{S}{r}\right)$$;

$\therefore S=\frac{4\pi}{k}dm$.

If k be supposed very great, the amount of ﬂuid produced by each source in order to keep up the pressures will be very small.

The potential of any system of electricity on itself will be

$\Sigma(pdm)=\frac{k}{4\pi}, \Sigma(pS)=\frac{k}{4\pi}W$.

If $$\Sigma(dm), \Sigma(dm')$$ be two systems of electrical particles and p, p'  the potentials due to them respectively, then by (32) $\Sigma(pdm')=\frac{k}{4\pi}\Sigma(pS')=\frac{k}{4\pi}\Sigma(p'S)=\Sigma(p'dm)$.

or the potential of the ﬁrst system on the second is equal to that of the second system on the ﬁrst.