366 Mr. 0. Heaviside on Electromagnetic Waves, and the 



38. The electric force at the origin due tofv at r = a. — Return- 

 ing to the case of impressed electric force, the differential 

 equation of F, the radial component of electric force inside 

 the sphere on whose surface r=a the vorticity of e is situated, 

 is, by § 21, 



-p. 2a cos 6 /. 1\/ , sinhorX,, , nn ^ 

 F = ^- 6 " ( 1+ Ya)( C0Shqr p V> < 232 > 



At the centre, therefore, the intensity of the full force, which 

 call F , whose direction is parallel to the axis, is 



F =i(l + qa)6-«"f=%(l-a-Qe-i°f. . . (233) 



Unless otherwise specified, I may repeat that the forces re- 

 ferred to are always those of the fluxes, thus doing away 

 with any consideration of the distribution of the impressed 

 force, and of scalar potential, of varying form, which it 

 involves. (233) is equivalent to 



F =i€-P t {l+av- 1 (p 2 -a 2 )i\€- aV ~ 1 (P 2 -^ i (/6P t ) . (234) 



Let /be constant, and p=<r, or g=0. Then (234) becomes 



F„=t/6-'g(^ + <r)(^)*+l] e -»- 1 (,W)V), . (235) 



of which the complete solution is, by (221), 



F = (f/)[e-^- 1 (p + <r)Jo{^- I (« 2 -^ 2 )n +XJ, (236) 

 where, subject to #=0, 



€-^(l)=X a ; (236a) 



or, solved, 



^='--["(^4')i(")'l i (M) 



in which i=(— l)t, and all the J's operate upon ati. This 

 solution (236) begins when t=a/v. The value of a is 

 4tt&/2c. 



In a good conductor a is immense. Then assume c=0, or 

 do away with the elastic displacement, and reduce (236) to 

 the pure diffusion formula, which is 



where y=(Airfika 2 /2t)K The relation of X a in (236) to the 

 preceding terms is explained by equations (233) or (235). 



