To determine the stability of the equilibrium of the charge `-q` at the origin when displaced slightly along the x-axis and y-axis, we can analyze the forces acting on it due to the other two charges `+q` and `-q` fixed at `(a, 0, 0)` and `(-a, 0, 0)` respectively.
### Step-by-Step Solution:
1. **Understanding the System**:
- We have two fixed charges: `+q` at `(a, 0, 0)` and `+q` at `(-a, 0, 0)`.
- A third charge `-q` is placed at the origin `(0, 0, 0)`.
- We need to analyze the stability of the equilibrium of the charge `-q` when it is slightly displaced.
2. **Displacement along the x-axis**:
- Let’s consider a small displacement `s` along the x-axis, making the position of `-q` at `(s, 0, 0)`.
- The distance from `-q` to the charge at `(a, 0, 0)` becomes `|s - a|` and to the charge at `(-a, 0, 0)` becomes `|s + a|`.
3. **Calculating Forces**:
- The force due to the charge at `(a, 0, 0)` (let’s call it `F1`) is attractive and given by:
\[
F_1 = \frac{k \cdot |q \cdot (-q)|}{(s - a)^2}
\]
- The force due to the charge at `(-a, 0, 0)` (let’s call it `F2`) is also attractive and given by:
\[
F_2 = \frac{k \cdot |q \cdot (-q)|}{(s + a)^2}
\]
4. **Direction of Forces**:
- When `-q` is displaced to the right (positive x direction), `F1` (towards `+q` at `(a, 0, 0)`) becomes greater than `F2` (towards `+q` at `(-a, 0, 0)`), pulling `-q` further away from the equilibrium position.
- Conversely, if `-q` is displaced to the left (negative x direction), `F2` becomes greater than `F1`, again pulling `-q` back towards the direction of the charge at `(a, 0, 0)`.
5. **Conclusion for x-axis Displacement**:
- Since any small displacement leads to a net force that pushes the charge further away from the equilibrium position, we conclude that the equilibrium is **unstable** along the x-axis.
6. **Displacement along the y-axis**:
- Now consider a small displacement `s'` along the y-axis, making the position of `-q` at `(0, s', 0)`.
- The distances to both charges remain the same, as they are still at `(a, 0, 0)` and `(-a, 0, 0)` respectively, which means the distances are:
\[
r_1 = \sqrt{(a^2 + (s')^2)} \quad \text{and} \quad r_2 = \sqrt{((-a)^2 + (s')^2)}
\]
7. **Calculating Forces**:
- The forces `F1` and `F2` due to both charges are equal in magnitude since the distances are equal:
\[
F_1 = F_2 = \frac{k \cdot |q \cdot (-q)|}{(a^2 + (s')^2)}
\]
8. **Direction of Forces**:
- Since both forces are equal and opposite, the net force acting on the charge `-q` when displaced along the y-axis is zero.
9. **Conclusion for y-axis Displacement**:
- Since the net force is zero for any small displacement along the y-axis, the equilibrium is **neutral** along the y-axis.
### Final Answer:
- (a) Along the x-axis: Unstable equilibrium.
- (b) Along the y-axis: Neutral equilibrium.
