Page:On Faraday's Lines of Force.pdf/13

Rh it are the same in both distributions, the pressure at the surface in the third distribution would be zero, and all the sources within the surface would vanish, by (15).

Then by (16) the pressure at every point in the third distribution must be zero; but this is the difference of the pressures in the two former cases, and therefore these cases are the same, and there is only one distribution of pressure possible.

(18) Let us next determine the pressure at any point of an inﬁnite body of ﬂuid in the centre of which a unit source is placed, the pressure at an inﬁnite distance from the source being supposed to be zero.

The ﬂuid will ﬂow out from the centre symmetrically, and since unity of volume ﬂows out of every spherical surface surrounding the point in unit of time, the velocity at a distance r from the source will be

The rate of decrease of pressure is therefore kv or $$ \frac{k}{4\pi r^2} $$, and since the pressure=0 when r is inﬁnite, the actual pressure at any point will be

The pressure is therefore inversely proportional to the distance from the source. It is evident that the pressure due to a unit sink will be negative and

equal to $$ - \frac{k}{4\pi r} $$

If we have a source formed by the coalition of $$ S $$ unit sources, then the resulting pressure will be $$ p = \frac{kS}{4\pi r} $$, so that the pressure at a given distance varies as the resistance and number of sources conjointly.

(19) If a number of sources and sinks coexist in the ﬂuid, then in order to determine the resultant pressure we have only to add the pressures which each source or sink produces. For by (15) this will be a solution of the problem, and by (17) it will be the only one. By this method we can determine the pressures due to any distribution of sources, as by the method