Method of treating Electrostatic Theorems. 21 



or 



When r=co , p is evidently zero ; 



E.Q 



3. If we consider the case of a " charge " imparted to a 

 conductor of any shape whatever, it will not be ordinarily 

 possible to determine the pressure at any point in the sur- 

 rounding medium. The same difficulty arises of course in 

 the ordinary method. But it is possible, as in the ordinary 

 method, to map out the nature of the strain in the medium by 

 drawing lines of displacement and surfaces of equal pressure. 



A surface of equal pressure corresponds to a surface of 

 equal potential. 



Since there can be no displacement along a surface of equal 

 pressure (otherwise the pressure must vary along the surface) , 

 displacements always take place at right angles to such surfaces. 



Hence a line drawn so that at every point on it the dis- 

 placement is along the line is always perpendicular to the 

 surfaces of equal pressure which it meets. 



Also if a number of such lines be taken forming a tube, 

 there will be no displacement across the walls of the tube ; 

 and hence if S be the area of any section, and x the displace- 

 ment at right angles to it, S . on = constant along the tube. 



Such a tube can only, of course, proceed from a place where 

 the dielectric is pushed back from the conductor to a place 

 where it is drawn in ; and if Q = total amount of charge at 

 one end of the tube, — Q = charge at the other end. 



Suppose that the surfaces of equal pressure are so drawn 

 that the pressure on any one differs by unity from the pres- 

 sure on either of the adjacent surfaces ; and tubes of displace- 

 ment are drawn to fill all space, and each of such size that 

 the flux along it, S . x, is unity. Then it may be shown that 

 each of the cells into which space is divided contains half a 

 unit of energy. 



For consider the cell in the figure (fig. 2). Let its bound- 

 aries be a tube of flux Sf, and a pair of surfaces of equal 

 pressure, the difference of the two pressures being Sp. Both 

 8/ and hp are to be so small that the cell Fig. 2. 



may be considered cylindrical. 



Let d be the perpendicular distance rrrrj ) 



tween the two surfaces, s the area of y ~>" — 



3tion of the tube, x the displacement. ' ■ 



If a unit volume be displaced a certain distance x, the force 



