184 Prof. B. N. Mallik on Lines of 



This is, obviously, a curve, passing through 



x 2 = iqap, y = 0. 

 But since 2e 



X = qap, 



this is obviously the point of equilibrium. 



This curve also divides the lines of force into two sets, 

 those issuing from e, and those tbat do not issue from e. 



In the fig. IV., q = 14, p = 8a. 



The line of force through the point of equilibrium is 

 defined by r + s = 28. 



13. We may also obtain the equation (2) in a slightly 

 different manner. 



Let dooi be the solid angle of an elementary tube issuing 

 from <?!. 



Then F x <iS = e x dw\, 



where dS is the normal section of the tube at P. 



But dS = n dd x dz, 



if dz is an element of length, perpendicular to the plane APB, 

 and r u 0i are polar coordinates of P referred to e Y . 



Hence ^ m 



r i r 1 do 1 dz = e x dco^ 



.'. equation (1) of art. 2 reduces to 

 e 1 d(o l + e 2 d(o 2 = ; 

 i. e., e-i o>i 4 e 2 co 2 = constant. 



Let Nj = whole number of tubes issuing from e 1 ; then 

 N, = e 1 ; and if ?i 1 = number of tubes corresponding to g> 1} 



&) T n x 



i. e., ^w, = 4 . it ,n Y . 



.*. the equation can be written 



n i + n 2 = constant. 



14. Comparing this with the equation r + s = constant, 

 defining a line of force, we notice that the index numbers 

 enumerate the lines of force issuing from a charged point, on an 

 assumed scale, ranging from the initial line joining the charges, 

 in either direction, according to the signs of the charges. 



Maxwell's method of drawing the lines of force is based on 

 this principle. [See also Mascart and Joubert's treatise, 

 Yol. i. ch. vii.] 



