142 The State of the Medium in the Electrostatic Field [CH. vi 



158. These three equations ensure that the medium shall have no 

 motion of translation, but for equilibrium it is also necessary that there 

 should be no rotation. To a first approximation, the stress across any face 

 may be supposed to act at the centre of the face, and the force H, H, Z at 

 the centre of the parallelepiped. Taking moments about a line through the 

 centre parallel to the axis of Ox, we obtain as the equation of equilibrium 



P v ,-Zv = .............................. (80). 



This and the two similar equations obtained by taking moments about 

 lines parallel to Oy, Oz ensure that there shall be no rotation of the medium. 

 Thus the necessary and sufficient condition for the equilibrium of the medium 

 is expressed by three equations of the form of (79), and three equations of the 

 form of (80). 



159. Suppose next that we take a small area dS anywhere in the 

 medium. Let the direction cosines of the normal 



to dS be I, m, + n. Let the parts of the 

 medium close to dS and on the two sides of it be 

 spoken of as S + and $_, these being named so 

 that a line drawn from dS with direction cosines 

 + 1, +m, +n will be drawn into S + , and one 

 with direction cosines I, m, n will be drawn 

 into S-. Let the force exerted by 8 + on $_ 

 across the area dS have components 



FdS, GdS, HdS, Flo 



then the force exerted by SL on S+ will have 



components 



-FdS, -GdS, -HdS. 



The quantities F, G, H are spoken of as the components of stress across 

 a plane of direction cosines I, m, n. 



To find the values of F, G, H, let us draw a small tetrahedron having 

 three faces parallel to the coordinate planes and a fourth having direction 

 cosines I, m, n. If dS is the area of the last face, the areas of the other 

 faces are IdS, mdS, ndS and the volume of the parallelepiped is 

 %. Resolving parallel to Ox, we have, since the medium inside 



this tetrahedron is in equilibrium, 



H - ldSP xx - mdSP yx -ndSP zx 



giving, since dS is supposed vanishingly small, 



F=lP xx + mP yx + nP 2X ........................ (81) 



and there are two similar equations to determine G and H. 



