and Magnetic Field Constants. 211 



Next we have 



-.*»( (1-A s sin*^)*<ty 





= Ftt-2PE (32) 



(Collecting these results we obtain 



I = |7r{& 3 (l + P)E-//(l-P)F} 



= | 7r {/>(/>2 + a 2 jE _ / ^2_ a 2) F | (33) 



For surface density cr we have to multiply this result 



by <*. 



It follows that, if to the disk just considered we add 

 a second disk of density cr' and radius a, potential energy 

 is exhausted of amount 



F = %7r*(r r b{(b 2 +a 2 )Fj-{b 2 -a 2 )F}. . . (34) 



This is apart from the exhaustion of potential energy involved 

 in the formation of the second disk, which is itself under the 

 potential of its own matter. As we have seen in (26) above, 

 this latter potential energy amounts to §7ra' 2 a 3 . 

 Now by (6) above the potential at the point P is 



V = 2™/J t J 1 (\?0J„(A/) < ^; .... (35) 



and the integral of this over a ring of radius /'and breadth df 

 is 



Hence the total surface integral of potential over a 

 concentric circular area of radius a is 



27T j Yfdf= 4ttV/> ^ J,(X&) I J J (VWJ Y' (36) 



But we have seen that 



J 



Jo(X/)/J/=^J 1 (Xa). 



