2 Lord Rayleigh on the Correction to the 



simply the double of that due to the end of a rod infinitely 

 long. 



At an interior node of an infinitely long rod the electric 

 forces, giving rise (we may suppose) to potential energy, 

 are a maximum, while the magnetic forces representing 

 kinetic energy are evanescent. The end of a terminated rod 

 corresponds, approximately at any rate, to a node. The 

 complications due to the end thus tell mainly upon the 

 electric forces *, and the problem is reduced to the electro- 

 statical one of finding the capacity of the terminated rod as 

 enclosed in the infinite cylindrical case at potential zero. 

 But this simplified form of the problem still presents 

 difficulties. 



Taking cylindrical coordinates z, r, we identify the axis of 

 symmetry with that of z, supposing also that the origin of z 

 coincides with the flat end of the interior conducting rod 

 which extends from — qo to 0. The enclosing case on the 

 other hand extends from — go to -foe . At a distance from 

 the end on the negative side the potential Y, which is sup- 

 posed to be unity on the rod and zero on the case, has the 

 form 



and the capacity per unit length is l/(21og5/a). 



On the plane z = the value of V from r=0 to r = a is 

 unity. If we knew T also the value of Y from r=ator=6,we 

 could treat separately the problems arising on the positive 

 and negative sides. On the positive side we could express 

 the solution by means of the functions appropriate to the 

 complete cylinder r<b, and on the negative side by those 

 appropriate to the annular cylindrical space b>r>a. Jf 

 we assume an arbitrary value for Y over the part in question 

 of the plane 2 = 0, the criterion of its suitability may be 

 taken to be the equality of the resulting values of dY/dz 

 on the two sides. 



We may begin by supposing that (1) holds good on the 

 negative side throughout ; and we have then to form for 

 the positive side a function which shall agree with this at 

 2 = 0. The general expression for a function which shall 

 vanish when r — b and when sr=+«o, and also satisfy 

 Laplace's equation, is 



Ai J (hr) e~ k S + A 2 J ( k 2 r) e -**+..., . (2) 

 where k h k 2i &c. are the roots of J (kb)=0 ; and this is to 



* Compare the analogous acoustical questions in ' Theory of Sound,' 

 §§265,317. 



