NTA Abhyas JEE Main2020PhysicsOscillationsPractice
Two large insulating plates having surface charge densities + σ and - σ are fixed some distance apart in a gravity-free region and two ideal insulating springs of force constant k are connected to the plates as shown in the figure. A particle of charge q and mass m which is attached to the junction of the springs is released from rest, then the particle will cross its equilibrium position with a speed
Options
- Av = q σ ε o k m
- Bv = q σ ε 0 k 2 m
- Cv = q σ ε 0 2 k m
- Dv = q σ 2 ε 0 k m
Correct answer
B. v = q σ ε 0 k 2 m
Step-by-step solution
The electric field between plates is E = σ ε 0 (Constant) The angular frequency of SHM doesn't change by a constant force, only the equilibrium position will be affected. ω = 2 k m The amplitude of SHM can be calculated by finding the distance of the equilibrium position from the rest position (starting position) of the particle. 2 k A = q σ ε 0 ⇒ A = q σ 2 k ε 0 So the particle will cross its equilibrium position with a speed v = q σ ε 0 k 2 m