Two charges, each equal to q, aer 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. −y
B. `(1)/(y)`
C. `-(1)/(y)`
D. y

Answer

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: Understand the Configuration
We have two charges \( q \) located at \( x = -a \) and \( x = a \) on the x-axis. A charge \( q_0 = \frac{q}{2} \) is placed at the origin (0,0).

### Step 2: Determine the Forces Acting on \( q_0 \)
When the charge \( q_0 \) is displaced a small distance \( y \) along the y-axis, it experiences forces due to both charges \( q \). The forces from each charge will have both x and y components.

### Step 3: Calculate the Distance to Each Charge
The distance \( r \) from the charge \( q_0 \) to either charge \( q \) can be calculated using the Pythagorean theorem:
\[ r = \sqrt{a^2 + y^2} \]

### Step 4: Calculate the Magnitude of the Force
The force \( F \) exerted on \( q_0 \) by one of the charges \( q \) is given by Coulomb's law:
\[ F = k \frac{q \cdot q_0}{r^2} = k \frac{q \cdot \frac{q}{2}}{(a^2 + y^2)} = \frac{k q^2}{2(a^2 + y^2)} \]

### Step 5: Resolve the Forces into Components
The force \( F \) can be resolved into x and y components. The angle \( \theta \) can be found using:
\[ \cos \theta = \frac{a}{\sqrt{a^2 + y^2}} \quad \text{and} \quad \sin \theta = \frac{y}{\sqrt{a^2 + y^2}} \]
Thus, the y-component of the force from one charge is:
\[ F_y = F \sin \theta = \frac{k q^2}{2(a^2 + y^2)} \cdot \frac{y}{\sqrt{a^2 + y^2}} \]

### Step 6: Calculate the Total Force
Since there are two charges, the total y-component of the force acting on \( q_0 \) is:
\[ F_{net} = 2 F_y = 2 \cdot \frac{k q^2}{2(a^2 + y^2)} \cdot \frac{y}{\sqrt{a^2 + y^2}} = \frac{k q^2 y}{(a^2 + y^2)^{3/2}} \]

### Step 7: Analyze the Limit for Small Displacement
For small displacements where \( y \) is much less than \( a \) (i.e., \( y \ll a \)), we can approximate:
\[ F_{net} \approx \frac{k q^2 y}{a^3} \]
This shows that the net force acting on the particle is proportional to \( y \).

### Conclusion
Thus, the net force acting on the particle \( q_0 \) when displaced along the y-axis is proportional to \( y \).