Question

Two charges $ -q $ each are fixed, separated by distance $ 2d $. A third charge $ q $ of mass $ m $ placed at the mid-point is displaced slightly by $ x' (x \ll d) $ perpendicular to the line joining the two fixed charges as shown in the figure. The time period of oscillation of $ q $ will be:

• A. \( T = \sqrt{\frac{8 \epsilon_0 m}{q^2}} x^2 \)
• B. \( T = \sqrt{\frac{8 \epsilon_0 m}{q^2}} x^3 \)
• C. \( T = \sqrt{\frac{4 \epsilon_0 m}{q^2}} x^3 \)
• D. \( T = \sqrt{\frac{8 \epsilon_0 m}{q^2}} x^2 \)

Hint:

When solving for the time period of oscillation of a charge in an electric field, the restoring force is analogous to the force in a spring system, and the time period is determined by the effective spring constant.

Answer 1

The Correct Option is D

The problem involves three charges: two charges \( -q \) and one charge \( +q \), with the charge \( +q \) displaced slightly from the equilibrium position. 
The forces acting on the charge \( q \) due to the two fixed charges \( -q \) will create an electric field, and if we displace the charge \( q \) by a small distance \( x' \), it will experience a restoring force. 
We calculate the force acting on the charge \( q \) due to the electric field of the fixed charges. Using Coulomb’s law and the approximation for small displacements, the force \( F \) on \( q \) is: \[ F = -kx' \] Where \( k \) is the effective spring constant related to the electric force. 
For the oscillations to occur, the time period \( T \) of the charge is given by: \[ T = 2\pi \sqrt{\frac{m}{k}} \] Since \( k \) is related to the charges and the separation distance, the expression for \( T \) becomes: \[ T = \sqrt{\frac{8 \epsilon_0 m}{q^2}} x^2 \] Thus, the time period of oscillation for the charge \( q \) is \( \sqrt{\frac{8 \epsilon_0 m}{q^2}} x^2 \).