Question

A ring of radius 100 cm has a uniformly distributed charge Q and uniformly distributed mass m. On the axis a point charge -Q of mass 2m is placed at a distance 3 cm from it's centre. Both are released from rest. Assuming only electric interaction between the ring and particle. Find the amplitude (in cm) of SHM of particle.​

Answer 1

Answer:

component of the force cancels out in pairs due to opposite particles in the ring.

So, cosine component of force dominates and the net force is in −x direction when −q charged particle is +x direction and vice versa.

And there is symmetry in the electric force on the charge particle placed on the axis of ring due to charged ring. So charge particle will oscillate between +2R and −2R about the centre of the ring

Explanation:

i hope this helps..

Answer 2

Answer:

the amplitude of SHM of particle is given by 1 cm

Explanation:

• Component of the force cancels out in pairs due to opposite particles in the ring.
• So, cosine component of force dominates, and the net force is in −x direction when −q charged particle is +x direction and vice versa.
• And there is symmetry in the electric force on the charge particle placed on the axis of ring due to charged ring. So, charge particle will oscillate between +2R and −2R about the center of the ring
• Since there is no external force, CM will remain at rest. Displacement of the particle till it comes to the center of the ring is given by

$x=3 * \frac{m}{m+2} \\x=\frac{3*(m+2)*m}{m+2} \\x=\frac{(3m+6)*m}{m+2} \\x=\frac{3m^2+6m}{m+2}\\m+2=3m^2+6m\\3m^2+6m-m-2=0\\3m^2+5m-2=0\\m=-2 \ or \ m=\frac{1}{3}$

Applying the value of m in the equation

the amplitude of SHM of particle is given by 1 cm