Two charges, each of `-q`, are fixed and separated by a distance `2 d`. A third charge `q` of mass `m` is placed at the mid-point of the two fixed charges. `q` is displaced slightly by `x(x<<)`, perpendicular to the line joining the two fixed charged as shown in the figure. `q` will perform simple harmonic oscillation if the time period of `(SHM)` is `T=(alpha pi^(3) varepsilon_(0) m d^(3))/(q^(2))]^([1/ 2)`. Find `alpha`
'(##CEN_KSR_PHY_JEE_C18_E01_013_Q04##)'

Answer

To solve the problem, we need to analyze the forces acting on the charge \( q \) when it is displaced slightly from its equilibrium position. The two fixed charges \( -q \) will exert forces on the charge \( q \), and we can derive the conditions for simple harmonic motion (SHM) from these forces.

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have two fixed charges, each of \( -q \), separated by a distance \( 2d \).
- A third charge \( q \) is placed at the midpoint of the two fixed charges.
- When the charge \( q \) is displaced by a small distance \( y \) (perpendicular to the line joining the two fixed charges), we need to find the restoring force acting on it.

2. **Calculating the Forces**:
- The force \( F \) due to one of the fixed charges on the charge \( q \) is given by Coulomb's law:
\[ F = k \frac{(-q)(q)}{r^2} \]
- Here, \( r \) is the distance from the charge \( q \) to one of the fixed charges. When \( q \) is displaced by \( y \), the distance \( r \) can be approximated as:
\[ r = \sqrt{d^2 + y^2} \]
- The total force \( F_{\text{net}} \) on charge \( q \) due to both fixed charges will be the vector sum of the forces due to each charge.

3. **Finding the Net Force**:
- The net force acting on \( q \) when displaced by \( y \) can be expressed as:
\[ F_{\text{net}} = 2F \cos(\theta) \]
- For small angles, \( \cos(\theta) \approx 1 \), and hence:
\[ F_{\text{net}} \approx 2F = 2 \left( k \frac{q^2}{(d^2 + y^2)} \right) \]
- The restoring force (which acts in the opposite direction of the displacement) can be approximated as:
\[ F_{\text{restoring}} = -2k \frac{q^2 y}{(d^2 + y^2)^{3/2}} \]

4. **Simplifying the Expression**:
- For small displacements (\( y \ll d \)), we can approximate \( (d^2 + y^2)^{3/2} \approx d^3 \):
\[ F_{\text{restoring}} \approx -2k \frac{q^2 y}{d^3} \]

5. **Relating to SHM**:
- The force can be related to the acceleration:
\[ F = m a = -k' y \]
- Here, \( k' = \frac{2kq^2}{d^3} \). Thus, we can express the time period \( T \) of SHM as:
\[ T = 2\pi \sqrt{\frac{m}{k'}} = 2\pi \sqrt{\frac{m d^3}{2kq^2}} \]

6. **Substituting for \( k \)**:
- Recall that \( k = \frac{1}{4\pi \varepsilon_0} \):
\[ T = 2\pi \sqrt{\frac{m d^3}{2 \left(\frac{1}{4\pi \varepsilon_0}\right) q^2}} = 2\pi \sqrt{\frac{4\pi \varepsilon_0 m d^3}{2 q^2}} = 2\pi \sqrt{\frac{8\pi \varepsilon_0 m d^3}{q^2}} \]

7. **Identifying \( \alpha \)**:
- Comparing this with the given expression for \( T \):
\[ T = \left(\alpha \frac{\pi^3 \varepsilon_0 m d^3}{q^2}\right)^{1/2} \]
- We find that \( \alpha = 8 \).

### Final Answer
Thus, the value of \( \alpha \) is \( 8 \).