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`.
Charge q is

Options

A. `-8muC`
B. `-6.5muC`
C. `-5muC`
D. `+6.5muC`

Explanation

To solve the problem, we need to analyze the forces acting on the charge \( \mu \) due to the two fixed charges \( Q \) located at points (2m, 0) and (-2m, 0).

### Step-by-Step Solution:

1. **Identify the System**:
We have two equal point charges \( Q = 8 \, \mu C = 8 \times 10^{-6} \, C \) located at (2m, 0) and (-2m, 0). The charge \( \mu \) is at (0, 0.1m) on the y-axis.

2. **Calculate the Distance**:
The distance \( r \) from each charge \( Q \) to the charge \( \mu \) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{(2^2 + 0.1^2)} = \sqrt{4 + 0.01} = \sqrt{4.01} \approx 2.0025 \, m \]

3. **Calculate the Force on Charge \( \mu \)**:
The force \( F \) on charge \( \mu \) due to one charge \( Q \) is given by Coulomb's law:
\[ F = \frac{1}{4 \pi \epsilon_0} \frac{|Q \mu|}{r^2} \]
where \( \epsilon_0 \) is the permittivity of free space, approximately \( 8.85 \times 10^{-12} \, C^2/(N \cdot m^2) \).

4. **Total Force Calculation**:
Since there are two charges \( Q \), the total force \( F_{total} \) in the y-direction (considering symmetry) is:
\[ F_{total} = 2 F \cos(\theta) \]
where \( \theta \) is the angle between the line connecting the charges and the y-axis. The cosine of the angle can be calculated as:
\[ \cos(\theta) = \frac{0.1}{r} = \frac{0.1}{\sqrt{4.01}} \approx \frac{0.1}{2.0025} \approx 0.04998 \]

5. **Force Expression**:
The force experienced by charge \( \mu \) is given as \( 9 \times 10^{-3} \, N \). Thus:
\[ 9 \times 10^{-3} = 2 \cdot \frac{1}{4 \pi \epsilon_0} \frac{|Q \mu|}{r^2} \cdot \cos(\theta) \]

6. **Substituting Values**:
Substitute \( r = 2.0025 \, m \) and \( \cos(\theta) \):
\[ 9 \times 10^{-3} = 2 \cdot \frac{9 \times 10^9}{4 \pi} \cdot \frac{|Q \mu|}{(2.0025)^2} \cdot 0.04998 \]

7. **Solve for Charge \( \mu \)**:
Rearranging gives:
\[ |Q \mu| = \frac{9 \times 10^{-3} \cdot (2.0025)^2}{2 \cdot \frac{9 \times 10^9}{4 \pi} \cdot 0.04998} \]

8. **Finding \( \mu \)**:
Given that the mass of \( \mu \) is \( 91 \, mg = 91 \times 10^{-3} \, kg \), we can find \( \mu \) by substituting the known values and solving for \( Q \).

9. **Final Calculation**:
After substituting and simplifying, we find:
\[ Q = -5 \, \mu C \]
(The negative sign indicates that the charge \( Q \) is opposite to the charge \( \mu \) to allow for oscillation.)