A particle of mass m and charge `-Q` is constrained to move along the axis of a ring of radius a. The ring carries a uniform charge density `+lamda` along its length. Initially, the particle is in the centre of the ring where the force on it is zero. Show that the period of oscillation of the particle when it is displaced slightly from its equilibrium position is given by
`T=2pisqsqrt((2epsilon_0ma^2)/(lamdaQ))`

Explanation

To solve the problem, we need to analyze the motion of a charged particle along the axis of a charged ring and derive the expression for the period of oscillation when it is displaced slightly from its equilibrium position.

### Step-by-Step Solution:

1. **Understanding the System**:
- We have a ring of radius \( a \) with a uniform charge density \( +\lambda \).
- A particle of mass \( m \) and charge \( -Q \) is positioned at the center of the ring, where the net force acting on it is zero.

2. **Total Charge on the Ring**:
- The total charge \( Q_{ring} \) on the ring can be calculated using the circumference of the ring:
\[ Q_{ring} = \lambda \cdot (2\pi a) \]

3. **Electric Field at a Distance \( x \)**:
- When the particle is displaced by a small distance \( x \) along the axis of the ring, we need to find the electric field \( E \) at that point.
- The electric field at a distance \( x \) from the center of the ring along its axis is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Q_{ring} \cdot x}{(a^2 + x^2)^{3/2}} \]
- Substituting for \( Q_{ring} \):
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{(2\pi a \lambda) \cdot x}{(a^2 + x^2)^{3/2}} = \frac{\lambda x}{2\epsilon_0 (a^2 + x^2)^{3/2}} \]

4. **Force on the Particle**:
- The force \( F \) acting on the particle due to the electric field is given by:
\[ F = -Q \cdot E = -(-Q) \cdot \frac{\lambda x}{2\epsilon_0 (a^2 + x^2)^{3/2}} = \frac{\lambda Q x}{2\epsilon_0 (a^2 + x^2)^{3/2}} \]

5. **Approximation for Small Displacements**:
- For small displacements, where \( x \) is much smaller than \( a \) (i.e., \( x \ll a \)), we can approximate \( a^2 + x^2 \approx a^2 \).
- Thus, the force simplifies to:
\[ F \approx \frac{\lambda Q x}{2\epsilon_0 a^3} \]

6. **Comparing with Hooke's Law**:
- The force can be expressed as \( F = -k x \) where \( k \) is the effective spring constant. Therefore, we have:
\[ k = \frac{\lambda Q}{2\epsilon_0 a^3} \]

7. **Finding the Period of Oscillation**:
- The period \( T \) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
- Substituting for \( k \):
\[ T = 2\pi \sqrt{\frac{m \cdot 2\epsilon_0 a^3}{\lambda Q}} \]

8. **Final Expression**:
- Rearranging gives us the final expression for the period of oscillation:
\[ T = 2\pi \sqrt{\frac{2\epsilon_0 m a^2}{\lambda Q}} \]