6 88 
MR. G. F. C. SEARLE OR PROBLEMS TN ELECTRIC CORYECTIOR. 
When the motion is very slow, the equation becomes 
r = c sin 3 6, 
the same as the equation to the lines of flow of a doublet, consisting of a “source” of 
current and a “ sink ” of equal strength, placed infinitely near each other in an 
infinite conducting medium. The currents flow along these curves, being “closed” 
by the convection current formed by the moving charge. At the speed of light the 
currents are confined to the plane of yz, and then take the form of outward radial 
currents in the front face of that plane and inward radial currents in the back face of 
the plane. 
Motion of a Line-Charge. 
7. Mr. Heaviside has obtained* the solution for the motion of a uniformly charged 
straight line, both in its own line and also transversely, by integration of the result 
for a point-charge. 
The method I have employed for the point-charge can be easily applied to these 
problems, when the line is infinite in length. 
(1.) Motion in its own line. 
The line coincides with the axis of x, and is supposed to have a charge q per unit 
length. In the electrostatic problem the potential is — q log [if -f- z ~). 
Hence a solution in the case of motion is given by 
V = — A log (y 3 + z~) n = 0. 
From this, by equations (26) to (28), 
E l = 0 
E 0 E, 
2A 
« (?/ + z 2 ) 
(51). 
The electric force is therefore everywhere perpendicular to the charged line, and 
the resultant is given by 
2A . , 
.(52). 
2A 
_ _ 
up «\/ y 3 + z 3 
To find A, integrate KE/4 tt over unit length of a cylinder of radius p coaxial with 
the charged line, and equate the result to q. Thus 
q = ‘Z-irp . 
K_ 2A 
47r " up 
KA 
u 
* ‘ Electrical Papers,’ vol. 2, p. 516. 
