336 Mr. 0. Heaviside on the Electromagnetic Effects due 



The whole displacement, of amount q, should therefore lie 

 within this cone. And since the moving charge is a convec- 

 tion-current qu, the displacement-current should he towards 

 the apex in the axial portion of the cone, and change sign at 

 some unknown distance, so as to be away from the apex either 

 in the outer part of the cone or else upon its boundary. The 

 pulling back of the charge by the electric stress would require 

 the continued application of impressed force to keep up the 

 motion, and its activity would be accounted for by the con- 

 tinuous addition made to the energy in the cone ; for the 

 transfer of energy on its boundary is perpendicularly outward, 

 and the field at the apex is being continuously renewed. 



The above general reasoning seems plausible enough, but I 

 cannot find any solution to correspond that will satisfy all the 

 necessary conditions. It is clear that (29) will not do when 

 u > V. Nor is it of any use to change the sign of the quantity 

 under the radical, when needed, to make real. It is suggested 

 that whilst there should be a definite solution, there cannot 

 be one representing a stead?/ condition of E and H with respect 

 to the moving charge. As regards physical possibility, in 

 connexion with the structure of the sether, that is not in 

 question. 



16. Let us now derive from (29), or from (27), the results 

 in some cases of distributed electrification^ in steady rectilinear 

 motion. The integrations to be effected being all of an ele- 

 mentary character, it is not necessary to give the working. 



First, let a straight line 

 A B be charged to linear ^^ -P 



density q, and be in motion 

 at speed u in its own line 



from left to right. Then at ^^-'^s , , , 



P we shall have ^ b 



A = g^.logfi-^^-Kl-'->W:), . . (39) 

 from which H= —dKjdh gives 



^=Hi-S[iar; 



Vl 



-ri ( 1 — vi i^lv^) + 7-1/^1 ( 1 — Vi^ w7«^) ^ 



— same function of r^^ fx^, v^ . (40) 



where fjb= cos 9, v= sin 0. 



When P is vertically over B, and A is at an infinite 

 distance, we shall find 



B. = qu/h, (41) 



