178 Prof. D. N. Mallik on Lines of 



We have, accordingly, 



ei sin X d0 x + e 2 sin 2 d0 2 — 

 or e\ cos X + <? 2 cos 2 = constant (2) 



If 0, 0\ are the interior angles of the triangle APB, we get 



e 1 cos 0—e 2 cos 0' = constant. 



This form of the equation enables us to give a simple 

 geometrical construction for these lines. 



3. Case I. — Two equal and opposite charges. 

 Here, e 2 = — e x . 



The equation is cos + cos 6' = const. 



Divide AB into any number of equal parts //,, say, and let 

 AB = fia. 



Produce AB to C, and BA to D, so that AC = BD = qa, 

 where q is as much larger than //,, as possible. 



With A, B as centres and radii equal to AC, BD describe 

 two circles. 



Divide BC, AD into parts, each equal to a, and draw 

 perpendiculars so as to meet these circles. 



Number the points of intersection of these lines with either 

 circle, viz. 1, 2, 3 &c, beginning with the lowest, on the side 

 remote from the centre of the corresponding circle, and join 

 these points to the same centre. 



Calling now these numbers, the index numbers of these 

 lines, we have the following rule : — 



The points of intersection of the two sets of lines, such 

 that the sum of their index numbers is constant, lie on a line 

 of force. 



For let r and s be the index numbers of any two lines from 

 A and B respectively, intersecting at P. 



And let 0, 0' define the point P (measured from the initial 

 line in opposite directions — one counter-clockwise, (say) the 

 other clockwise). 



Then, confining ourselves to the points on one side of the 

 axis, 



ga cos = (q — r)a, 

 qa cos & — (q — s)a. 



cos + cos 0' = 2 = const., if r + s is constant. 



