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 of finding the period of oscillation of a charged particle constrained to move along the axis of a charged ring, we will follow these steps:

### Step 1: Determine the total charge on the ring
The ring has a uniform charge density `λ` and a radius `a`. The total charge `Q` on the ring can be calculated using the formula for the circumference of the ring:
\[ Q = \lambda \cdot (2\pi a) \]

### Step 2: Analyze the equilibrium position
Initially, the particle of mass `m` and charge `-Q` is placed at the center of the ring. At this position, the net force acting on the particle is zero due to the symmetry of the electric field produced by the ring.

### Step 3: Calculate the electric field at a distance `x` from the center
When the particle is displaced a small distance `x` along the axis of the ring, we need to calculate the electric field `E` at that point. The electric field due to a ring of charge at a distance `x` along its axis is given by:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{Qx}{(a^2 + x^2)^{3/2}} \]
Substituting `Q = \lambda \cdot (2\pi a)`, we get:
\[ E = \frac{1}{4\pi \epsilon_0} \cdot \frac{(2\pi a \lambda)x}{(a^2 + x^2)^{3/2}} \]

### Step 4: Simplify the electric field for small displacements
For small displacements where \( x \ll a \), we can approximate \( (a^2 + x^2)^{3/2} \approx a^3 \). Thus, the electric field simplifies to:
\[ E \approx \frac{1}{4\pi \epsilon_0} \cdot \frac{(2\pi a \lambda)x}{a^3} = \frac{\lambda}{2\epsilon_0 a^2} x \]

### Step 5: Calculate the force acting on the particle
The force `F` acting on the particle due to the electric field is given by:
\[ F = -Q \cdot E = -(-Q) \cdot \left(\frac{\lambda}{2\epsilon_0 a^2} x\right) = \frac{Q \lambda}{2\epsilon_0 a^2} x \]

### Step 6: Relate the force to the restoring force
The force can be expressed in the form of Hooke's law:
\[ F = -kx \]
where \( k = \frac{Q \lambda}{2\epsilon_0 a^2} \).

### Step 7: Find the period of oscillation
The time period \( T \) of oscillation for a mass-spring system is given by:
\[ T = 2\pi \sqrt{\frac{m}{k}} \]
Substituting the value of \( k \):
\[ T = 2\pi \sqrt{\frac{m \cdot 2\epsilon_0 a^2}{Q \lambda}} \]

### Step 8: Final expression for the period
Thus, the period of oscillation is:
\[ T = 2\pi \sqrt{\frac{2\epsilon_0 m a^2}{\lambda Q}} \]