Question

Two large insulating plates having surface charge densities $+\sigma $ and $-\sigma $ 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

• A. $v=\frac{q \sigma }{k \epsilon _{0}}\sqrt{\frac{k}{m}}$
• B. $v=\frac{q \sigma }{k \epsilon _{0}}\sqrt{\frac{k}{2 m}}$
• C. $v=\frac{q \sigma }{k \epsilon _{0}}\sqrt{\frac{2 k}{m}}$
• D. $v=\frac{q \sigma }{2 k \epsilon _{0}}\sqrt{\frac{k}{m}}$

Answer 1

Answer: (B)

The electric field between plates is
$E=\frac{\sigma }{\epsilon _{0}}$ (Constant)
The angular frequency of SHM doesn't change by a constant force, only the equilibrium position will be affected.
$\omega =\sqrt{\frac{2 k}{m}}$
The amplitude of SHM can be calculated by finding the distance of the equilibrium position from the rest position (starting position) of the particle.
$2kA=\frac{q \sigma }{\epsilon _{0}}$
$\Rightarrow A=\frac{q \sigma }{2 k \epsilon _{0}}$
So the particle will cross its equilibrium position with a speed
$v=\frac{q \sigma }{k \epsilon _{0}}\sqrt{\frac{k}{2 m}}$