THE FORCE ON A SMALL CHARGED BODY 71 



(1) V 2 V = ; (2) V = a constant for each conductor; 



(3) / / ^- - dS = 4-7T X given charge on each conductor. This 



function is the only one satisfying the conditions, and tells us all 

 about the field. 



Green's theorem states that if U and V are two functions, 



r AT ^ dS - 

 = // 



where the surface integrals are taken over the conductors and the 

 volume integrals are taken through the intervening air dielectric. 

 Make V the function considered above, and make U the potential 

 according to the inverse square law of a positive unit of charge 

 ibuted uniformly with density p through a very small sphere 

 of radius a and centre at any point P in the field. Then at a 

 distance r from P, greater than a, U = 1 /r, and at a distance r 



Q 



less than a, U = guyjr 1 . Then within the small sphere V 2 U = 4-rrp 

 and without it V*U = 0. 



On the conducting surfaces + 47nr = 0, and in the dielectric 



V*V = 0. Since V is constant over each conductor J J 



dS = ^ Gauss ' 8 theorem. Also ///VV 2 Udo%<fcr 



only differs from zero in the small sphere round P and there it may 

 be put \pf J f tirpdxdydz = 4-7rV P , since the total charge round 

 P = 1. Then 



Vp=yy^<js 



or the potential is that which would be found by calculation on 

 the inverse square law with direct action. In other words, the 

 inverse square law is the only one which would give the strain 

 along a tube constant, and a constant potential in the conductors. 



