To solve the problem, we need to analyze the motion of a positive charge \( Q \) released from rest on the x-axis, in the presence of two equal negative charges fixed on the y-axis. Here’s a step-by-step breakdown of the solution:
### Step 1: Understand the Setup
We have two equal negative charges, let's denote them as \( -q \), fixed at points \( (0, a) \) and \( (0, -a) \) on the y-axis. A positive charge \( Q \) is released from rest at a point on the x-axis, say \( (x, 0) \).
**Hint**: Visualize the scenario by sketching the positions of the charges on a coordinate system.
### Step 2: Determine the Forces Acting on Charge \( Q \)
The positive charge \( Q \) will experience electrostatic forces due to both negative charges. The forces will be directed towards each negative charge.
- The distance from charge \( Q \) to each negative charge is given by:
\[
r = \sqrt{x^2 + a^2}
\]
- The magnitude of the electrostatic force \( F \) exerted by one negative charge on \( Q \) is given by Coulomb's law:
\[
F = k \frac{|qQ|}{r^2} = k \frac{qQ}{x^2 + a^2}
\]
**Hint**: Remember that both negative charges exert forces in the same direction towards themselves.
### Step 3: Calculate the Net Force on Charge \( Q \)
Since both forces have the same magnitude and are directed towards the negative charges, we can find the net force \( F_{net} \) on charge \( Q \) by considering the components of the forces.
- The net force will have both x and y components, but due to symmetry, we can focus on the x-component:
\[
F_{net,x} = 2F \cdot \cos(\theta)
\]
where \( \cos(\theta) = \frac{x}{\sqrt{x^2 + a^2}} \).
Thus,
\[
F_{net,x} = 2 \cdot k \frac{qQ}{x^2 + a^2} \cdot \frac{x}{\sqrt{x^2 + a^2}} = \frac{2kqQx}{(x^2 + a^2)^{3/2}}
\]
**Hint**: The net force depends on the position \( x \) and the distances involved.
### Step 4: Analyze the Motion of Charge \( Q \)
The force \( F_{net,x} \) is proportional to \( x \), but it is not a linear relationship because of the \( (x^2 + a^2)^{3/2} \) term in the denominator. This means the force does not satisfy the criteria for simple harmonic motion (SHM), which requires a linear restoring force proportional to displacement.
### Step 5: Determine the Type of Motion
As charge \( Q \) moves towards the origin, it will accelerate due to the attractive forces from the negative charges. However, when it reaches the origin, the forces will change direction, causing it to oscillate back and forth around the origin.
- The charge will not settle at the origin because it will have velocity upon reaching it, leading to oscillatory motion.
**Hint**: Recognize that while the motion is periodic, it does not conform to the strict definition of SHM.
### Conclusion
Based on the analysis, the correct answer is:
**D. Execute oscillatory but not simple harmonic motion.**
