291021

Question : Two charges each equal to q, are kept at x=-a and x=+a on the X-axis. A particle of mass m and charge q₀=q/2 is placed at the origin. If charge q₀ is given a small displacement y (y<<a) along the Y-axis, the net force acting on the particle is proportional to 

a) y
b) -y
c) 1/y
d) (-1/y)

Doubt by Sonali

[JEE Mains 2013]

Solution : 

In figure, 

AO = OB = a
OP = y
AP = r = (a²+y²)^1/2

Force of repulsion experienced by charger q₀ due to charge q at A = Force of repulsion experienced by charge q₀ due to charge q at B = F

Resolving F into rectangular components Fcosθ and Fsinθ

The component Fcosθ will cancel out each other. So the total Force of the charge q₀ will be 2Fsinθ

= 2Fsinθ

Clearly F_Net∝y

Hence, a) would be the correct option.