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 will use the principle of conservation of energy. The total mechanical energy (potential energy + kinetic energy) of the system remains constant if no external work is done on it.

### Step-by-Step Solution:

1. **Identify the Initial Conditions:**
- We have two positive point charges \( q \) located at \( (0, a) \) and \( (0, -a) \).
- A third charge \( q' \) with mass \( m \) is placed at the origin \( (0, 0) \) and is displaced slightly in the positive x-direction.

2. **Calculate Initial Potential Energy (PE_initial):**
- The potential energy between the charge \( q' \) and each of the charges \( q \) is given by the formula:
\[ PE = k \frac{q \cdot q'}{r} \]
- The distance from the charge \( q' \) at the origin to each charge \( q \) is \( a \).
- Therefore, the potential energy due to both charges is:
\[ PE_{\text{initial}} = k \frac{q \cdot q'}{a} + k \frac{q \cdot q'}{a} = 2k \frac{q \cdot q'}{a} \]

3. **Calculate Initial Kinetic Energy (KE_initial):**
- Since the charge \( q' \) is initially at rest, its initial kinetic energy is:
\[ KE_{\text{initial}} = 0 \]

4. **Calculate Final Potential Energy (PE_final):**
- As the charge \( q' \) moves to infinity, the potential energy approaches zero:
\[ PE_{\text{final}} = 0 \]

5. **Calculate Final Kinetic Energy (KE_final):**
- At infinity, the charge \( q' \) will have some kinetic energy, which we denote as:
\[ KE_{\text{final}} = \frac{1}{2} mv^2 \]

6. **Apply Conservation of Energy:**
- According to the conservation of energy:
\[ PE_{\text{initial}} + KE_{\text{initial}} = PE_{\text{final}} + KE_{\text{final}} \]
- Plugging in the values we calculated:
\[ 2k \frac{q \cdot q'}{a} + 0 = 0 + \frac{1}{2} mv^2 \]
- Rearranging gives:
\[ 2k \frac{q \cdot q'}{a} = \frac{1}{2} mv^2 \]

7. **Solve for \( v \):**
- Multiplying both sides by 2:
\[ 4k \frac{q \cdot q'}{a} = mv^2 \]
- Dividing both sides by \( m \):
\[ v^2 = \frac{4k q q'}{ma} \]
- Taking the square root gives:
\[ v = \sqrt{\frac{4k q q'}{ma}} \]

8. **Substituting the value of \( k \):**
- Recall that \( k = \frac{1}{4\pi \epsilon_0} \):
\[ v = \sqrt{\frac{4 \cdot \frac{1}{4\pi \epsilon_0} \cdot q \cdot q'}{ma}} = \sqrt{\frac{q q'}{\pi \epsilon_0 ma}} \]

### Final Answer:
\[ v = \sqrt{\frac{q q'}{\pi \epsilon_0 ma}} \]