To solve the problem, we need to analyze the forces acting on a point charge \( q \) that can move along the x-axis due to the presence of two other point charges \( Q \) placed at coordinates (0, y) and (0, -y). Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the Setup
We have two point charges \( Q \) located at (0, y) and (0, -y). A third charge \( q \) of the same polarity can move along the x-axis.
### Step 2: Analyze the Forces on Charge \( q \)
The force on charge \( q \) due to each charge \( Q \) can be calculated using Coulomb's law. The distance from charge \( q \) at position \( (x, 0) \) to either charge \( Q \) is given by:
\[
r = \sqrt{x^2 + y^2}
\]
### Step 3: Calculate the Electric Field
The electric field \( E \) at position \( (x, 0) \) due to each charge \( Q \) is given by:
\[
E = \frac{kQ}{r^2} = \frac{kQ}{x^2 + y^2}
\]
where \( k \) is Coulomb's constant.
### Step 4: Determine the Direction of the Forces
Both charges \( Q \) exert forces on \( q \) that are directed away from themselves since all charges are of the same polarity. The net electric field \( E \) at point \( (x, 0) \) will be the vector sum of the fields due to both charges.
### Step 5: Analyze Equilibrium at the Origin
At the origin (0, 0), the distances to both charges are equal, and the magnitudes of the electric fields due to both charges are equal but in opposite directions. Therefore, the net electric field at the origin is zero:
\[
E_{net} = E_{Q1} + E_{Q2} = 0
\]
Thus, the charge \( q \) is in equilibrium at the origin.
### Step 6: Determine the Nature of Motion
For any position of charge \( q \) along the x-axis (other than the origin), the net force will always point away from the origin. This means that if \( q \) is displaced from the origin, it will experience a force that pushes it further away from the origin, indicating that oscillatory motion is impossible.
### Step 7: Find the Position of Maximum Force
To find where the force on charge \( q \) is maximized, we need to differentiate the expression for the net force \( F \) acting on \( q \) and set the derivative to zero. The force \( F \) can be expressed as:
\[
F = 2 \cdot \frac{kQq}{(x^2 + y^2)^{3/2}} \cdot x
\]
Differentiating this with respect to \( x \) and setting it to zero will yield the condition for maximum force.
### Step 8: Solve for Maximum Force Position
After differentiating and simplifying, we find that the maximum force occurs at:
\[
y^2 = 2x^2 \quad \Rightarrow \quad x = \pm \frac{y}{\sqrt{2}}
\]
### Conclusion
Based on the analysis:
- The charge \( q \) is in equilibrium at the origin.
- The force on \( q \) is directed away from the origin for any position other than the origin.
- The force is maximized at \( x = \pm \frac{y}{\sqrt{2}} \).
### Final Answer
The correct options are:
- Option A: True (Force is maximum at \( x = \pm \frac{y}{\sqrt{2}} \))
- Option B: True (Charge \( q \) is in equilibrium at the origin)
- Option D: True (For any position of \( q \) other than the origin, the force is directed away from the origin)
