Problem: Two charges of magnitude Q are placed at 2a distance apart and a third charge q is placed at mid point. If q charge is slightly displaced from its position along the line then determine its time period (mass of q charge is m).

Solution:

To solve this problem, we can use the concept of simple harmonic motion. When the charge q is slightly displaced from its equilibrium position, it experiences a net force due to the electrostatic forces exerted by the two charges Q. This net force acts as a restoring force and causes the charge q to oscillate about its equilibrium position.

Let's break down the solution into the following steps:

1. Finding the force on the charge q:
- The two charges Q exert an electrostatic force on the charge q.
- By Coulomb's law, the magnitude of this force is given by F = kQq/(a/2)^2, where k is the Coulomb's constant.
- The direction of this force is towards the equilibrium position of the charge q.
- When the charge q is displaced from its equilibrium position by a small distance x, the force becomes F = kQq/(a/2 + x)^2 - kQq/(a/2 - x)^2.
- Expanding this expression using binomial theorem and neglecting higher order terms in x, we get F = -(2kQqx)/(a^3), which is the restoring force acting on the charge q.

2. Applying Newton's second law:
- According to Newton's second law, the acceleration of the charge q is given by a = F/m, where m is the mass of the charge q.
- Substituting the expression for F, we get a = -(2kQx)/(ma^3)q.

3. Finding the time period of oscillation:
- The time period of oscillation of the charge q is given by T = 2π√(m/k), where k is the effective spring constant of the restoring force.
- To find k, we can use the formula for the period of a simple pendulum, T = 2π√(l/g), where l is the length of the pendulum and g is the acceleration due to gravity.
- By analogy, we can write T = 2π√(m/k) as T = 2π√(a/g'), where g' = (2kQ)/(ma^2) is the effective acceleration due to the restoring force.
- Substituting the expression for k, we get T = 2π√((ma^2)/(2kQ)), which simplifies to T = 2π√((ma^2)/(4πεQ^2)), where ε is the permittivity of free space.

4. Final answer:
- Substituting the given values of Q, a, and m, we get T = 2π√((m/(8πε))(1/Q^2)).
- Thus, the time period of oscillation of the charge q is inversely proportional to the square of the charge Q and directly proportional to the square root of the mass m.
- Note that this formula assumes that the displacement of the charge q is small, so that we can neglect higher order terms in x. If the displacement