MR. G. E. C. SEARLE ON PROBLEMS IN ELECTRIC CONVECTION. 
689 
Hence 
A = qa/K 
(53). 
The state of the field is therefore given by the equations 
Ej = 0 
E 0 — 
1 
2qy 
Eo 
K (2/2 + y 
__ | 
K (y2 + ^ j 
(54). 
Hi 
Ho 
= 0 
'2quz 
1 
Ho = 
r 
Zquy 
T 
• (55). 
The resultant electric force is perpendicular to the charged line. Its value is 
E = 2q/Hp 
■ (56), 
the same as if the charge were at rest. 
The resultant magnetic force is in circles round the wire. 
Its value is 
H = 2 qujp .(57), 
the same as that due to a current qu. Thus the motion introduces a magnetic force 
without affecting the electric force at all. 
(2.) Motion perpendicular to its own line. 
Let the charged line coincide with the axis of 2 . The potential in the electro¬ 
static problem is — q log ( x 1 -f- ?/-). Hence a solution in the case of motion is 
given by 
T' =■ - A log (^— + ?/) 
From this, by equations (26) to (28), 
_ Ej _ _2A_ 
■ r ~ y ~+ 
The lines of electric force are thus everywhere straight and at right angles to the 
MDCCCXCVI.-A. 4 T 
n = o. 
Eg = 0.(58). 
