Point charges `+4q, -q` and `+4q` are kept on the `x`-axis at points `x=0,x=a` and `X=2a` respectively, then

Options

A. Only `q` is in stable equilibrium
B. None of the charges are in equilibrium
C. All the charges are in unstable equilibrium
D. All the charges are in stable equlibrium

Explanation

To solve the problem, we need to analyze the forces acting on each of the point charges placed on the x-axis at specified positions. The charges are as follows:

- Charge 1: \( +4q \) at \( x = 0 \)
- Charge 2: \( -q \) at \( x = a \)
- Charge 3: \( +4q \) at \( x = 2a \)

### Step 1: Analyze the force on charge \( -q \) at \( x = a \)

The forces acting on charge \( -q \) are due to the two \( +4q \) charges:

1. **Force due to the charge at \( x = 0 \)**:
- The distance from \( +4q \) (at \( x = 0 \)) to \( -q \) (at \( x = a \)) is \( a \).
- The magnitude of the force \( F_1 \) is given by Coulomb's law:
\[ F_1 = \frac{k \cdot |4q \cdot (-q)|}{a^2} = \frac{4kq^2}{a^2} \]
- This force is attractive, directed towards the charge at \( x = 0 \).

2. **Force due to the charge at \( x = 2a \)**:
- The distance from \( -q \) (at \( x = a \)) to \( +4q \) (at \( x = 2a \)) is also \( a \).
- The magnitude of the force \( F_2 \) is:
\[ F_2 = \frac{k \cdot |4q \cdot (-q)|}{a^2} = \frac{4kq^2}{a^2} \]
- This force is repulsive, directed away from the charge at \( x = 2a \).

### Step 2: Determine the net force on charge \( -q \)

The net force \( F_{net} \) on charge \( -q \) is the sum of the forces acting on it:
\[ F_{net} = F_1 - F_2 = \frac{4kq^2}{a^2} - \frac{4kq^2}{a^2} = 0 \]
Since the net force on charge \( -q \) is zero, it is in equilibrium.

### Step 3: Analyze the forces on the charges \( +4q \) at \( x = 0 \) and \( x = 2a \)

1. **For the charge at \( x = 0 \)**:
- The force due to the charge at \( x = a \) (which is \( -q \)) is:
\[ F = \frac{4kq^2}{a^2} \quad \text{(attractive)} \]
- The force due to the charge at \( x = 2a \) is:
\[ F = \frac{16kq^2}{(2a)^2} = \frac{4kq^2}{a^2} \quad \text{(repulsive)} \]
- The net force on the charge at \( x = 0 \):
\[ F_{net} = -\frac{4kq^2}{a^2} + \frac{4kq^2}{a^2} = 0 \]
- Hence, the charge at \( x = 0 \) is also in equilibrium.

2. **For the charge at \( x = 2a \)**:
- The force due to the charge at \( x = a \) (which is \( -q \)) is:
\[ F = \frac{4kq^2}{a^2} \quad \text{(attractive)} \]
- The force due to the charge at \( x = 0 \) is:
\[ F = \frac{16kq^2}{(2a)^2} = \frac{4kq^2}{a^2} \quad \text{(repulsive)} \]
- The net force on the charge at \( x = 2a \):
\[ F_{net} = -\frac{4kq^2}{a^2} + \frac{4kq^2}{a^2} = 0 \]
- Hence, the charge at \( x = 2a \) is also in equilibrium.

### Step 4: Stability of the equilibrium

- **For charge \( -q \)**: If displaced slightly, the attractive force from \( +4q \) at \( x = 0 \) will increase while the repulsive force from \( +4q \) at \( x = 2a \) will decrease, leading to a net force in the direction of the displacement. Thus, it is in **unstable equilibrium**.

- **For charges \( +4q \)**: If either \( +4q \) charge is displaced slightly, the forces acting on them will also lead to a net force in the direction of the displacement, indicating that they are also in **unstable equilibrium**.

### Conclusion

All three charges are in **unstable equilibrium**.