Two equal negative charges `-q` are fixed at points `(0, -a)` and `(0,a)` on y-axis. A poistive charge Q is released from rest at point `(2a, 0)` on the x-axis. The charge Q will

Options

A. (a)execute simple harmonic motion about the origin move to the origin and remain at rest
B. (b)move to infinity
C. (c)execute oscillatory but not simple of the harmonic motion
D. (d) None of these

Answer

To solve the problem, we will analyze the forces acting on the positive charge \( Q \) when it is released from the point \( (2a, 0) \) and determine its motion.

### Step-by-Step Solution

1. **Understanding the Setup**:
- We have two negative charges \( -q \) located at \( (0, -a) \) and \( (0, a) \) on the y-axis.
- A positive charge \( Q \) is released from rest at the point \( (2a, 0) \) on the x-axis.

2. **Calculating the Forces**:
- The force acting on charge \( Q \) due to each negative charge can be calculated using Coulomb's law:
\[ F = k \frac{|q_1 q_2|}{r^2} \]
- Here, \( q_1 = Q \) and \( q_2 = -q \), and \( r \) is the distance between the charges.

3. **Distance Calculation**:
- The distance from charge \( Q \) at \( (2a, 0) \) to each negative charge is:
\[ r = \sqrt{(2a - 0)^2 + (0 - y)^2} = \sqrt{(2a)^2 + a^2} = \sqrt{4a^2 + a^2} = \sqrt{5a^2} = a\sqrt{5} \]

4. **Force Components**:
- The force \( F \) due to one negative charge is:
\[ F = k \frac{Qq}{(a\sqrt{5})^2} = k \frac{Qq}{5a^2} \]
- Since both negative charges exert forces in the same direction (towards the y-axis), we need to consider the x-component of the forces.

5. **Net Force Calculation**:
- The net force \( F_{\text{net}} \) acting on charge \( Q \) will be the sum of the x-components of the forces due to both negative charges.
- The x-component of the force from one charge is:
\[ F_x = F \cdot \frac{2a}{a\sqrt{5}} = k \frac{Qq}{5a^2} \cdot \frac{2a}{a\sqrt{5}} = k \frac{2Qq}{5a\sqrt{5}} \]
- Since there are two charges, the total x-component is:
\[ F_{\text{net}} = 2 \cdot F_x = 2 \cdot k \frac{2Qq}{5a\sqrt{5}} = k \frac{4Qq}{5a\sqrt{5}} \]

6. **Direction of the Force**:
- The force acts towards the origin (negative x-direction) since both charges are negative and attract the positive charge \( Q \).

7. **Motion Analysis**:
- As \( Q \) moves towards the origin, the force will continue to act towards the origin until \( Q \) reaches the origin where the net force becomes zero.
- After reaching the origin, due to inertia, \( Q \) will continue moving in the negative x-direction until the forces from the negative charges reverse its motion.

8. **Conclusion**:
- The charge \( Q \) will not execute simple harmonic motion (SHM) because the net force is not directly proportional to its displacement from the origin. However, it will perform oscillatory motion about the origin.

### Final Answer:
The positive charge \( Q \) will perform oscillatory motion about the origin but not simple harmonic motion.