Consider two positive point charges, each of magnitude q, placed on the y-axis at the points (0, a) and (0, -a). A positively charged particle of charge q' and mass m is displaced slightly from the origin in the direction of the positive x-axis.
Speed of charge q' at infinity is

Options

A. `sqrt((qq')/(piepsilon_(0)m))`
B. `sqrt((q')/(piepsilon_(0)am))`
C. `sqrt((qq')/(epsilon_(0)am))`
D. `sqrt((qq')/(piepsilon_(0)am))`

Answer

To solve the problem, we need to analyze the forces acting on the charged particle \( q' \) and apply the principle of conservation of energy.

### Step-by-step Solution:

1. **Understanding the Setup**:
- We have two positive point charges \( q \) located at coordinates \( (0, a) \) and \( (0, -a) \) on the y-axis.
- A positively charged particle \( q' \) with mass \( m \) is initially at the origin \( (0, 0) \) and is displaced slightly in the positive x-direction.

2. **Initial Conditions**:
- The initial kinetic energy \( KE_i \) of the charge \( q' \) is zero since it starts from rest.
- The initial potential energy \( PE_i \) can be calculated using the formula for the electric potential energy between point charges.

3. **Calculating Initial Potential Energy**:
- The distance from the charge \( q' \) at the origin to each charge \( q \) is \( a \).
- The potential energy \( PE \) due to one charge \( q \) is given by:
\[ PE = \frac{1}{4 \pi \epsilon_0} \frac{q \cdot q'}{r} \]
- Since there are two charges, the total initial potential energy \( PE_i \) is:
\[ PE_i = 2 \cdot \frac{1}{4 \pi \epsilon_0} \frac{q \cdot q'}{a} = \frac{q \cdot q'}{2 \pi \epsilon_0 a} \]

4. **Final Conditions**:
- At infinity, the potential energy \( PE_f \) is zero because the distance between charges becomes infinitely large.
- The final kinetic energy \( KE_f \) can be expressed as:
\[ KE_f = \frac{1}{2} m v^2 \]
- Here, \( v \) is the speed of the charge \( q' \) at infinity.

5. **Applying Conservation of Energy**:
- According to the conservation of mechanical energy:
\[ KE_i + PE_i = KE_f + PE_f \]
- Substituting the known values:
\[ 0 + \frac{q \cdot q'}{2 \pi \epsilon_0 a} = \frac{1}{2} m v^2 + 0 \]

6. **Solving for Speed \( v \)**:
- Rearranging the equation gives:
\[ \frac{1}{2} m v^2 = \frac{q \cdot q'}{2 \pi \epsilon_0 a} \]
- Multiplying both sides by 2:
\[ m v^2 = \frac{q \cdot q'}{\pi \epsilon_0 a} \]
- Dividing by \( m \):
\[ v^2 = \frac{q \cdot q'}{m \pi \epsilon_0 a} \]
- Taking the square root to find \( v \):
\[ v = \sqrt{\frac{q \cdot q'}{m \pi \epsilon_0 a}} \]

### Final Answer:
The speed of charge \( q' \) at infinity is:
\[ v = \sqrt{\frac{q \cdot q'}{m \pi \epsilon_0 a}} \]