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

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.