"INVERSE SQUARE" SYSTEMS 39 



>mall, </ also is small, and the volume between the surfaces is 



-d. Hence 



v = 





That is, the displacement of any particle varies inversely as the 

 square of its distance from the source. 



Now imagine an electrical system in which a charge q is put 

 at O. The intensity will be numerically equal to the fluid dis- 



placement if we make q = . If the point O is a sink instead 



4?r 



of a source, and if volume v of fluid is removed, the displacement 

 is reversed and is equal to the intensity due to a quantity of nega- 



tive electricity q r placed at the sink. 



TJ7T 



Evidently if the quantities of fluid introduced or removed at 

 the various points of a system are exceedingly small, we may com- 

 pound the emplacement! due to each separately, according to the 

 \ector law, in order to obtain the resultant displacement. We 

 in iv, therefore, imagine a fluid system corresponding to any 



uded gravitathe, magnetic, or electric system, the matter in a 

 gravilat: in being replaced by a tenet of sinks, while North- 



magncti>m and positive electric-it v are replaced by sources, 

 and South -seeking magnetism and negative electricity are replaced 

 1>\ sinks. The quantities of fluid introduced or withdrawn are 

 proportional to the quantities of matter, magnetism, or electricity 

 which they represent. Lines and tubes of flow in the fluid system 

 will then follow the same course as lines and tubes of force in the 

 correspond}!. ijTSteOM. Drawing a displacement tube in the 



fluid system, since no fluid passes out through its walls, equal 

 quantities must pass across every section in any part of the tube not 

 containing sources or sinks. If, then, d l d 2 be the displacements 

 at two sections S t S r r/ x Sj = r/ 2 S 2 , or displacement X cross-section 

 i^ constant. Replacing displacements by the corresponding and 

 proportional intensities, we obtain for a tube of force the 

 corresponding property that intensity X cross-section is constant. 



We may also prove easily the other properties of tubes of force, 

 but these ire le.-ive to the reader. 



\\ shall now find the value of the intensity in some simple 



Spherical shell uniformly charged. Intensity out- 

 side the shell. Ix.-t S be a sphere, l'ig. 33, over which a 

 <li;i uniformly spread. Let S t be a concentric sphere 



outride S and of radius / ,. The intensity over Sj is, from the 

 sMimietry of the system, everywhere perpendicular to Sj and 

 everywhere of the same magnitude. Let it be equal to I. Then 



