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 determine the charge \( q \) that is held at the point (0, 0.1 m) on the y-axis, given the forces acting on it due to two fixed point charges \( Q \).

### Step-by-Step Solution:

1. **Identify the Charges and Their Positions**:
- We have two equal point charges, each \( Q = 8 \, \mu C = 8 \times 10^{-6} \, C \), located at points \( (2 \, m, 0) \) and \( (-2 \, m, 0) \).
- The charge \( q \) is held at the point \( (0, 0.1 \, m) \).

2. **Calculate the Distance from Charge \( Q \) to Charge \( q \)**:
- The distance from either charge \( Q \) to charge \( q \) can be calculated using the distance formula:
\[ r = \sqrt{(2 - 0)^2 + (0 - 0.1)^2} = \sqrt{4 + 0.01} = \sqrt{4.01} \, m \]

3. **Calculate the Force on Charge \( q \)**:
- The force \( F \) experienced by charge \( q \) due to one charge \( Q \) is given by Coulomb's Law:
\[ F = k \frac{|Q \cdot q|}{r^2} \]
- Here, \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) and \( r^2 = 4.01 \).

4. **Calculate the Components of the Force**:
- Since there are two charges \( Q \), the total force \( F_{\text{total}} \) acting on charge \( q \) will have components from both charges.
- The horizontal components will cancel out, and only the vertical components will add up:
\[ F_{\text{total}} = 2 F \cos(\theta) \]
- Where \( \cos(\theta) = \frac{0.1}{\sqrt{4.01}} \).

5. **Substituting the Values**:
- We know that at \( t = 0 \), the force experienced by \( q \) is \( 9 \times 10^{-3} \, N \):
\[ 9 \times 10^{-3} = 2 \cdot k \frac{|Q \cdot q|}{(4.01)} \cdot \frac{0.1}{\sqrt{4.01}} \]

6. **Rearranging to Solve for \( q \)**:
- Rearranging the equation gives:
\[ q = \frac{9 \times 10^{-3} \cdot 4.01 \cdot \sqrt{4.01}}{2 \cdot k \cdot Q \cdot 0.1} \]

7. **Substituting Values**:
- Substitute \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) and \( Q = 8 \times 10^{-6} \, C \):
\[ q = \frac{9 \times 10^{-3} \cdot 4.01 \cdot \sqrt{4.01}}{2 \cdot (9 \times 10^9) \cdot (8 \times 10^{-6}) \cdot 0.1} \]

8. **Calculating \( q \)**:
- Calculate the values step by step to find \( q \):
\[ q \approx -5 \, \mu C = -5 \times 10^{-6} \, C \]

### Final Answer:
The charge \( q \) is approximately \( -5 \, \mu C \).