1913] TERMINATED RODS IN ELECTRICAL PROBLEMS 149 (15) represents the capacity. A calculation of capacity founded upon an approximate value of V in (14) is thus always an overestimate. In the present case we may substitute (15) for (14), if we consider the positive and negative sides separately, since it is only at 2 = 0 that Laplace's equation fails to receive satisfaction. The complete expression for V on the right is given by combination of (2) and (11), and the surface of integration is composed of the cylindrical wall r = b from z = 0 to z — GO , and of the plane z = 0 from r = 0 to r = b *. The cylindrical wall contributes nothing, since F vanishes along it. At z = 0 F= % (A + B) J0 (kr}, - d Vfdz = ^k (A + B) JQ (Jcr) ; and (15) = %b-*Łk(A + B)'* J^ (kb)...................(16) On the left the complete value of Fincludes (1) and (8). There are here two cylindrical surfaces, but r=b contributes nothing for the same reason as before. On r = a we have F = 1 and _, mi clV 1 ^7 Ł7* .J' / 7 \ Ji ? dr a log bja "* ^ ' so that this part of the surface, extending to a great distance z= — l, contributes to (15) 21og&/a 2^ '........................ There remains to be considered the annular area at z — 0. Over this .........................(19) The integrals required are f </>(/ir)r^r=-/i-1{i<j!)/(/^)-a0'(M}) ..................(20) J a, rb logr6(hr)rdr=~h~l{blogb<l>'(hb) — alQga<f)'(ha)}, ...(21) Jo ' -fra»{0'(M}a; ...............(22) and we get for this part of the surface Thus for the whole surface on the left i •as(j>'a(ha)], .........(24) * The surface at z = + <x> may evidently be disregarded.nts, even though only a limited number included. Every fresh coefficient that is included renders the approximate closer, and as near an approach as we please to the truth may be arrived by continuing the process. The true value of (14) is equal by Gree theorem to