Two equal positive charges Q are fixed at points (a, 0) and `(-a,0)` on the x-axis. An opposite charge -q at rest is relased from point (0, a) on the y-axis. The charge -q will

Options

A. Move to infinity
B. Move to origin and rest there
C. Undergo SHM about the origin
D. Execute oscillatory motion but not SHM

Answer

To solve the problem, we need to analyze the motion of the charge -q that is released from the point (0, a) on the y-axis, in the presence of two equal positive charges Q fixed at points (a, 0) and (-a, 0) on the x-axis.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two positive charges \( Q \) located at \( (a, 0) \) and \( (-a, 0) \).
- A negative charge \( -q \) is released from the point \( (0, a) \) on the y-axis.

2. **Calculating Forces**:
- The distance from the charge \( -q \) to each positive charge \( Q \) can be calculated using the distance formula. The distance \( r \) from \( -q \) to either \( Q \) is:
\[ r = \sqrt{(0 - a)^2 + (a - 0)^2} = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

3. **Force on Charge -q**:
- The force \( F \) exerted on charge \( -q \) by one of the positive charges \( Q \) is given by Coulomb's law:
\[ F = k \frac{|Q \cdot (-q)|}{r^2} = k \frac{Qq}{(a\sqrt{2})^2} = k \frac{Qq}{2a^2} \]
- Since there are two positive charges, the total force \( F_{\text{net}} \) on charge \( -q \) is:
\[ F_{\text{net}} = 2F = 2 \left(k \frac{Qq}{2a^2}\right) = k \frac{Qq}{a^2} \]

4. **Direction of the Force**:
- The force on charge \( -q \) is directed towards the origin (0,0) because both positive charges repel the negative charge.

5. **Analyzing Motion**:
- As the charge \( -q \) moves towards the origin, it will accelerate due to the force acting on it. However, as it approaches the origin, the forces will change because the distance to each charge will decrease.

6. **Oscillatory Motion**:
- The charge \( -q \) will oscillate about the origin. It will gain speed as it approaches the origin, reach maximum speed at the origin, and then slow down as it moves away from the origin due to the repulsive forces from the positive charges.

7. **Conclusion**:
- The charge \( -q \) does not settle at the origin but oscillates between the two positive charges. The motion is periodic but not simple harmonic motion (SHM) because the restoring force is not directly proportional to the displacement from the equilibrium position.

### Final Answer:
The charge \( -q \) will oscillate between the two positive charges \( Q \) and will not settle at the origin.

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