Question

to point chaarges +q & +4q are placed at a distance of 6a ona horizontal plane. find the locus of the point on the line joining the two charges where electric field is zero.

Answer 1

_____________________________________________________________
Suppose the charge "q " is placed at origin and the charge "4q"is placed at 
(6a,0).

i.e. the "4q" is on the x - axis at a distance of  "6a" from  the origin  



Let P be the point on the x - axis at the distance of "x" from the origin where  electric field is zero  .

Electric field  at P  due to  the charge  "q" is E(q)  = Kq/x²

Electric field at P due to  the charge "4Q"is E(4q) = K(4q)/(6a - x ) ²

Let , E be the  resultant of these two Electric fields and the angle between  them is  180°

E² = [E(q)]²+ [E(4q)]²+ 2[E(q)] [E(4q)]cos 180 

⇒ 0 =  [E(q)]² + [E(4q)]² - 2[E(q)] [E(4q)]

⇒ [ E(q) -  E(4q) ]² = 0  

⇒ E(q) - E(4q)  = 0 

⇒ E(q) = E(4q) 

⇒ Kq/x² = K(4q)/(6a-x)²

⇒ 1/x² =4/(6a-x)²

⇒ 1/x = 2(6a-x)

⇒ 6a = 2 x +x 

⇒ 6a = 3x 

⇒ x = 2a 

So the coordinates of point P 
= (2a,0)