346 Dr. E. B. Rosa on the Specific Inductive 



we have the case of a single charged wire, where a = constant 

 is the equation of the equipotential surfaces and /3 = constant 

 is the equation of the lines of force. Hence 



gives the case of two oppositely charged wires. Transferring 

 the origin to a point equidistant from the wires and on the 

 line joining them, we find 



B«= Vab, R/j= \/A 2 + B 2 . 



a represents the equipotential lines, and /3 the lines of force ; 

 a and b are the distances from the points M and N respectively 

 to the points c l9 c 2 , c 3 . . . along the axis of X. For given 

 values of a and 6, a. is constant and so also is R a , the radius of 

 the equipotential lines, which are therefore circles with 

 c l9 c 3 , c 3 . . . as centres. A is the constant distance ON. B is 

 the distance from to the points S 1? S 2 , S 3 . . . For a given 

 value of B, j3 is constant and also R^, the radius of the corre- 

 sponding line of force, which is therefore a circle. These 

 two systems of circles divide up the plane into areas which 

 are in the limit squares, and the figure shows how the force 

 varies from point to point. To get the force at any point 



along the axis of X, find the corresponding value of — > 



ax 



y being zero ; and for the force at points along the axis of Y 

 (the force of course being perpendicular to the axis of Y), 



since t- = -r-i find the corresponding values of ~- , x being 

 dx dy r dy ° 



zero. Thus, 



d*_d*db _Ap 4A 2 1 



dx ~ db'dx ~ 2 Lr 2 (A 2 -^ 2 ) 2 J' 



dff 2A 



dy ~ */ 2 + A 2 * 



At the origin x=.y = 0, and R a = ^ ; hence 



