Two charges, each equal to q, are kept at `x=-a` and `x=a` on the x-axis. A particle of mass m and charge `q_0=q/2` is placed at the origin. If charge `q_0` is given a small displacement `(ylt lt a)` along the y-axis, the net force acting on the particle is proportional to

Options

A. (a) y
B. (b) `-y`
C. (c) `1/y`
D. (d) `-1/y`

Explanation

To solve the problem step by step, we will analyze the forces acting on the charge \( q_0 \) when it is displaced along the y-axis.

### Step 1: Understanding the Setup
We have two charges, each equal to \( q \), located at \( x = -a \) and \( x = a \) on the x-axis. A particle with charge \( q_0 = \frac{q}{2} \) is placed at the origin \( (0, 0) \).

### Step 2: Displacement of the Charge
The charge \( q_0 \) is displaced a small distance \( y \) along the y-axis. The new position of \( q_0 \) is \( (0, y) \).

### Step 3: Calculating Forces on \( q_0 \)
The forces acting on \( q_0 \) due to the two charges can be calculated using Coulomb's law. The forces will be directed away from the charges because they are like charges.

1. **Force due to the charge at \( x = -a \)**:
\[ F_1 = k \frac{q \cdot q_0}{r_1^2} \]
where \( r_1 = \sqrt{a^2 + y^2} \).

2. **Force due to the charge at \( x = a \)**:
\[ F_2 = k \frac{q \cdot q_0}{r_2^2} \]
where \( r_2 = \sqrt{a^2 + y^2} \).

Since \( q_0 = \frac{q}{2} \), we can substitute this into the equations for \( F_1 \) and \( F_2 \):
\[ F_1 = k \frac{q \cdot \frac{q}{2}}{(a^2 + y^2)} = \frac{k q^2}{2(a^2 + y^2)} \]
\[ F_2 = k \frac{q \cdot \frac{q}{2}}{(a^2 + y^2)} = \frac{k q^2}{2(a^2 + y^2)} \]

### Step 4: Resolving Forces into Components
Next, we need to resolve these forces into their x and y components.

- The x-components of \( F_1 \) and \( F_2 \) will cancel each other out because they are equal in magnitude and opposite in direction.
- The y-components will add up.

### Step 5: Finding the Y-components
The angle \( \theta \) can be found using the triangle formed by the charges and the displaced charge:
\[ \sin \theta = \frac{y}{\sqrt{a^2 + y^2}} \]

Thus, the y-component of the forces is:
\[ F_{1y} = F_1 \sin \theta = \frac{k q^2}{2(a^2 + y^2)} \cdot \frac{y}{\sqrt{a^2 + y^2}} = \frac{k q^2 y}{2(a^2 + y^2)^{3/2}} \]
\[ F_{2y} = F_2 \sin \theta = \frac{k q^2}{2(a^2 + y^2)} \cdot \frac{y}{\sqrt{a^2 + y^2}} = \frac{k q^2 y}{2(a^2 + y^2)^{3/2}} \]

### Step 6: Total Y-component of the Force
The total y-component of the force \( F_y \) is:
\[ F_y = F_{1y} + F_{2y} = 2 \cdot \frac{k q^2 y}{2(a^2 + y^2)^{3/2}} = \frac{k q^2 y}{(a^2 + y^2)^{3/2}} \]

### Step 7: Considering Small Displacement
Since \( y \) is small compared to \( a \) (i.e., \( y \ll a \)), we can approximate \( a^2 + y^2 \approx a^2 \).

Thus, the force simplifies to:
\[ F_y \approx \frac{k q^2 y}{a^3} \]

### Conclusion
The net force acting on the particle \( q_0 \) is proportional to \( y \).

### Final Answer
The net force acting on the particle is proportional to \( y \).

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