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 systematically, we can break it down into several steps:

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

### Step 2: Analyze the Forces Acting on Charge -q
When the charge \( -q \) is released, it will experience forces due to both positive charges \( Q \). The forces can be denoted as \( F_1 \) (from the charge at \( (a, 0) \)) and \( F_2 \) (from the charge at \( (-a, 0) \)).

### Step 3: Calculate the Distance from Charge -q to Charges Q
The distance \( r \) from the charge \( -q \) to either charge \( Q \) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{a^2 + a^2} = \sqrt{2a^2} = a\sqrt{2} \]

### Step 4: Calculate the Magnitude of Forces
The magnitude of the electrostatic force \( F \) between two charges is given by Coulomb's law:
\[ F = k \frac{|Q \cdot (-q)|}{r^2} \]
Since both forces \( F_1 \) and \( F_2 \) are equal, we can write:
\[ F_1 = F_2 = k \frac{Qq}{(a\sqrt{2})^2} = k \frac{Qq}{2a^2} \]

### Step 5: Resolve Forces into Components
The forces \( F_1 \) and \( F_2 \) can be resolved into components. The horizontal components will cancel each other out due to symmetry, while the vertical components will add up:
\[ F_{1y} = F_1 \sin \theta \quad \text{and} \quad F_{2y} = F_2 \sin \theta \]
Where \( \theta \) is the angle between the line connecting the charges and the x-axis. Since \( \sin \theta = \frac{a}{r} = \frac{a}{a\sqrt{2}} = \frac{1}{\sqrt{2}} \):
\[ F_{1y} = F_{2y} = F_1 \sin \theta = k \frac{Qq}{2a^2} \cdot \frac{1}{\sqrt{2}} = \frac{kQq}{2a^2\sqrt{2}} \]

### Step 6: Calculate the Net Force
The net vertical force \( F_{net} \) acting on charge \( -q \) is:
\[ F_{net} = F_{1y} + F_{2y} = 2 F_{1y} = 2 \cdot \frac{kQq}{2a^2\sqrt{2}} = \frac{kQq}{a^2\sqrt{2}} \]

### Step 7: Relate Force to Acceleration
Using Newton's second law, \( F = ma \):
\[ ma = \frac{kQq}{a^2\sqrt{2}} \]
Thus, the acceleration \( a \) of charge \( -q \) is:
\[ a = \frac{kQq}{ma^2\sqrt{2}} \]

### Step 8: Identify the Nature of Motion
The acceleration \( a \) is directly proportional to the displacement from the equilibrium position (which is the origin). This indicates that the motion of charge \( -q \) is simple harmonic motion (SHM) about the origin.

### Conclusion
The charge \( -q \) will undergo simple harmonic motion (SHM) about the origin.

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