Force due to given Static Charges, 1 83 



10. Put 2e 



^r = qap, 



the equation then becomes 



V 2 

 qa cos 6 — '— = constant. 



P 



This leads to the following construction : — 



Draw the lines from A, having the charge e, with index 

 numbers 1, 2, 3, &c. as before. 



Along the axis of y, lay off lengths si pa, \l2pa, \/3pa, 



and through their extremities draw parallels to the axis of x y 

 and let these have index numbers 1 , 2, &c. Then the points 

 of intersections of these two sets of lines, such that the sum 

 of their index numbers is constant, lie along a line of force. 



For, if P is a point defined by index numbers r and s, 

 we have 



qa cos 6 = (q — r)a, 



y = v' pas. 



qa cos 6— — = {q — r + s \ a = constant, 

 since r±s is constant. 



11. A practical method of laying off lengths sjpa, ^2 pa] 

 <fec, along the axis of y, will obviously be to describe the 

 parabola y 2 = px (say on the negative side of the axis 

 of x) and measure off lengths — a, -2a, along this axis. 

 Then, the corresponding ordinates will be of lengths sj'pa, 

 \l2pa, &c. 



12. Putting y = 0, we have 



cos0 = l-- + -, 

 9 



where 6, as before, is the inclination of the tangent to a line 

 of force at A, to the axis of x ; and 6 is imaginary, if 

 r+s>2q: J 



i. e., lines of force corresponding to r + s>2q do not issue 

 from e. 



If r + s = 2q, the equation of the corresponding line of 

 force becomes 



V 2 

 qa cos — •— = — qa ; 



P 



and changing into Cartesians 



