Consider two positive point charges, each of magnitude q, placed on the y-axis at the points (0, a) and (0, -a). A positively charged particle of charge q' and mass m is displaced slightly from the origin in the direction of the positive x-axis.
At the origin, charge q' will be in

Options

A. stable equilibrium
B. unstable equilibrium
C. neutral equilibrium
D. cannot determine

Answer

To solve the problem, we need to analyze the forces acting on the positively charged particle \( q' \) when it is displaced slightly from the origin in the direction of the positive x-axis. Here's the step-by-step solution:

### Step 1: Understand the Configuration
We have two positive point charges, each of magnitude \( q \), located at the points \( (0, a) \) and \( (0, -a) \) on the y-axis. The charge \( q' \) is initially at the origin \( (0, 0) \).

**Hint:** Visualize the setup with a coordinate system where the y-axis has the two charges and the x-axis is where the charge \( q' \) is displaced.

### Step 2: Analyze the Forces at the Origin
At the origin, the charge \( q' \) experiences forces due to both charges \( q \). Since both charges are positive, they will repel the charge \( q' \).

**Hint:** Remember that like charges repel each other. Calculate the force exerted by each charge on \( q' \).

### Step 3: Calculate the Force Components
When \( q' \) is displaced a small distance \( x \) along the x-axis, we can denote its new position as \( (x, 0) \). The distance from \( q' \) to each charge \( q \) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{x^2 + a^2} \]
The force \( F \) exerted by each charge \( q \) on \( q' \) is given by Coulomb's law:
\[ F = k \frac{q \cdot q'}{r^2} \]
where \( k \) is Coulomb's constant.

**Hint:** Break down the forces into their x and y components. The y-components will cancel out due to symmetry.

### Step 4: Determine the Resultant Force
The x-component of the force from each charge \( q \) on \( q' \) can be found using:
\[ F_x = F \cdot \frac{x}{\sqrt{x^2 + a^2}} \]
Since there are two charges, the total x-component of the force acting on \( q' \) is:
\[ F_{net} = 2F \cdot \frac{x}{\sqrt{x^2 + a^2}} \]

**Hint:** Consider how the net force behaves as \( x \) increases.

### Step 5: Analyze Stability
As \( q' \) is displaced slightly along the x-axis, the net force \( F_{net} \) will act in the same direction as the displacement. This indicates that if \( q' \) is displaced, it will continue to move away from the origin.

**Hint:** Recall the conditions for stability: if the net force acts in the direction of displacement, the equilibrium is unstable.

### Conclusion
Since the net force acting on the charge \( q' \) is directed away from the origin when displaced, it indicates that the equilibrium at the origin is unstable.

**Final Answer:** The charge \( q' \) will be in unstable equilibrium at the origin.