Method of treating Electrostatic Theorems, 33 



perpendiculars from the centre on the tangent planes at \ u \ 2 , 

 X 3 , then the sides of the section of the tube by the surface \ x are 



~ — and 77-^. The area of the section is therefore— r~ — ■• 

 2p 2 2p d 4^ 2 p 3 



The thickness, at the point X 1 \ 2 X 3 , of the shell bounded by 

 the ellipsoid X t and the similar ellipsoid into which \ x expands 

 under the given strain, is pp u the linear dimensions of \ being 

 supposed to expand in the ratio 1 to 1 + /0. Now the volume 

 of this shell is equal to the charge on the conductor, Q. 



Hence we obtain easily 



Q 



Now 



47r*/(a 2 + \ 1 )(6 ,2 + \ 1 )(c 2 + \; 



V X, 2 X 3 ■ 



and p 2 p s have similar values. 



Hence the flux along the tube, which is equal to 



oX 2 • 0X3 



is independent of \ 1? and therefore constant along the tube. 

 Our suppositions are then possible. 



Perhaps it may make things clearer to state them in another 

 way. If we suppose that tubes, like the one mentioned and 

 occupying all space, act as guides to the particles as they are 

 displaced outwards, and if the conductor itself expand 

 " similarly," then by what we have just proved the particles 

 on any confocal ellipsoid are displaced in such a way that the 

 ellipsoid expands "similarly." 



We have now to show that the supposed strain is one which 

 ensures mechanical equilibrium. 



Let P be the point of intersection of X. l5 \ 2 , an d ^3? P' of 

 Xj + SXi, X 2 , an d X 3 . Let o\ ± be small. 



Then the difference of pressures at P and P' 



E . x PP' x displacement of medium at P. 

 = E p -^ 



= Ex 2 -x m 



2 ' 



which is the same for all points on \ v 



Phil. Mag. S. 5. Vol. 34. No. 206. July 1892. D 



