Consider a system of two equal points charges, each `Q = 8 muC`, which are fixed at points (2m, 0 ) and (-2m, 0). Another charge `mu` is held at a point (0,0.1m) on the y-axis. Mass of the charge `mu` is 91 mg . At t= 0 , `mu` is released from rest and it is observed to oscillate along y-axis in a simple harmonic manner. It is also observed that at t = 0 , the force experienced by it is ` 9 xx 10^(-3)N`.
Amplitude of motion is

Options

A. `10 cm`
B. `20 cm`
C. `30 cm`
D. `40 cm`

Explanation

To solve the problem, we need to analyze the forces acting on the charge \( \mu \) when it is released from rest at the point (0, 0.1 m) on the y-axis. The two fixed charges \( Q \) are located at (2 m, 0) and (-2 m, 0).

### Step-by-Step Solution:

1. **Identify the forces acting on charge \( \mu \)**:
- The charge \( \mu \) experiences electrostatic forces due to both point charges \( Q \). Since both charges are equal and positioned symmetrically along the x-axis, the forces exerted by these charges will have components in both the x and y directions.

2. **Calculate the force experienced by \( \mu \)**:
- The force experienced by \( \mu \) at the initial position (0, 0.1 m) is given as \( F = 9 \times 10^{-3} \, \text{N} \).

3. **Determine the nature of the charge \( \mu \)**:
- Since the force is attractive (as indicated by the direction of the force), we can conclude that charge \( \mu \) must be negative. If it were positive, the forces from both charges would repel it, leading to a net force away from the origin.

4. **Analyze the motion**:
- When released, the charge \( \mu \) will move towards the origin due to the attractive forces from the two charges. As it passes through the origin, it will continue to oscillate back and forth along the y-axis.

5. **Identify the amplitude of motion**:
- The problem states that the charge oscillates in a simple harmonic manner. The maximum displacement from the equilibrium position (origin) is the amplitude of the oscillation.
- Given that the charge starts at \( (0, 0.1 \, \text{m}) \) and moves towards the origin, it will reach the maximum negative displacement at \( (0, -0.1 \, \text{m}) \) before reversing direction.

6. **Calculate the amplitude**:
- The amplitude of motion \( A \) is the distance from the equilibrium position to the maximum displacement. Therefore, the amplitude is:
\[ A = 0.1 \, \text{m} = 10 \, \text{cm} \]

### Final Answer:
The amplitude of motion is \( 10 \, \text{cm} \).