To analyze the situation described in the question, we will follow these steps:
### Step 1: Understand the Configuration of Charges
We have two positive point charges, each of magnitude \( q \), located at coordinates \( (0, a) \) and \( (0, -a) \) on the y-axis. A positively charged particle of charge \( q' \) and mass \( m \) is initially at the origin \( (0, 0) \).
### Step 2: Analyze the Forces at the Origin
At the origin, the net force acting on the charge \( q' \) is zero because the two charges \( q \) located at \( (0, a) \) and \( (0, -a) \) exert equal and opposite forces on it. Since both charges are positive, they repel the charge \( q' \) equally in the x-direction, resulting in no net force at the origin.
### Step 3: Displacement from the Origin
When the charge \( q' \) is slightly displaced in the positive x-direction, the distances to the two charges change. Let’s denote the new position of \( q' \) as \( (x, 0) \) where \( x \) is a small positive value.
### Step 4: Calculate the Forces After Displacement
After displacement, the forces acting on the charge \( q' \) will no longer be equal. The force exerted by the charge at \( (0, a) \) will have a component in the positive x-direction, while the force from the charge at \( (0, -a) \) will have a component in the negative x-direction.
1. The distance from \( q' \) to the charge at \( (0, a) \) is \( r_1 = \sqrt{x^2 + a^2} \).
2. The distance from \( q' \) to the charge at \( (0, -a) \) is \( r_2 = \sqrt{x^2 + a^2} \).
The forces due to each charge can be calculated using Coulomb's law:
- Force from the charge at \( (0, a) \):
\[
F_1 = k \frac{q q'}{r_1^2} = k \frac{q q'}{x^2 + a^2}
\]
- Force from the charge at \( (0, -a) \):
\[
F_2 = k \frac{q q'}{r_2^2} = k \frac{q q'}{x^2 + a^2}
\]
### Step 5: Determine the Direction of the Net Force
Since both charges are positive and repel \( q' \):
- The x-component of the force from the charge at \( (0, a) \) will push \( q' \) further in the positive x-direction.
- The x-component of the force from the charge at \( (0, -a) \) will also push \( q' \) in the positive x-direction.
Thus, the net force acting on \( q' \) after displacement is directed in the positive x-direction.
### Step 6: Conclusion on Equilibrium
Since the net force is directed away from the origin when \( q' \) is displaced, it will continue to move away from the origin and will not return to its original position. This indicates that the equilibrium at the origin is unstable.
### Final Answer
The charge \( q' \) will be in **unstable equilibrium** at the origin.
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