JEE Main2013PhysicsOscillationsActual
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 q 0 = - q 2 is placed at the origin. If charge q 0 is given a small displacement (y << a) along the y-axis, the net force acting on the particle is proportional to :
Options
- A1 y
- B- 1 y
- C-y
- Dy
Correct answer
C. -y
Step-by-step solution
⇒ F net = 2Fcos θ F net = 2 k q q 2 y 2 + a 2 2 y y 2 + a 2 F n e t = 2 k q q 2 y y 2 + a 2 3 / 2 ≈ k q 2 a 3 y ⇒ F ∝ - y Net force F is opposite the direction of displacement y . Hence, negative sign is used.