24 ON POTENTIAL. 



or, from Poisson's theorem, 

 (15) 



f f 5V f 



- Vp<fo= - V a?S+ F 2 <fo. 



J J 3 J 



35. TUBES OF FORCE. We have seen that a tube of force is an 

 orthogonal canal with equipotential surfaces, bounded laterally by 

 lines of force. 



Let us consider a tube of force terminated by any two surfaces 

 S and S' (Fig. 7), and let us apply equation (7). to the volume thus 

 denned. The lateral surface does not enter into the integral, for 

 the perpendicular component of the force at each point is zero ; 

 the integral is reduced therefore to the two terms furnished by the 

 terminal surfaces. 



Let us first suppose that the tube contains no electrical mass. 



Fig. 7- 



The integral reduced to the two terms of the bases must be zero, 

 which gives 



F^S + 

 J 



or, in absolute values, 



J 



n </S= f 



that is to say that the flow of force is then the same at both ends of 

 the tube. 



It thus appears that the flow of force is the same across any 

 section of the tube. This flow is kept up like the supply of a 

 moving liquid, whose velocity at each point is equal and parallel 

 to the direction of the force. 



If the section of this tube is infinitely narrow, and if the 

 terminal surfaces are perpendicular to the force, the equation 

 reduces to 



