A point charge `q_1=9.1muC` is held fixed at origin. A second point charge `q_2=-0.42muC` and `a` mass `3.2xx10^-4kg` is placed on the x-axis, `0.96m` from the origin. The second point charge is released at rest. What is its speed when it is `0.24m` from the origin?

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The initial potential energy of the system will be converted into kinetic energy as the second charge moves closer to the first charge.

### Step-by-step Solution:

1. **Identify the Initial and Final Positions:**
- The initial distance \( r_i = 0.96 \, \text{m} \) (where \( q_2 \) is initially placed).
- The final distance \( r_f = 0.24 \, \text{m} \) (where we want to find the speed of \( q_2 \)).

2. **Calculate Initial and Final Potential Energy:**
- The potential energy \( U \) between two point charges is given by:
\[ U = k \frac{q_1 q_2}{r} \]
- Where \( k \) is Coulomb's constant, \( k \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \).
- Calculate the initial potential energy \( U_i \):
\[ U_i = k \frac{q_1 q_2}{r_i} = k \frac{(9.1 \times 10^{-6} \, \text{C})(-0.42 \times 10^{-6} \, \text{C})}{0.96} \]
- Calculate the final potential energy \( U_f \):
\[ U_f = k \frac{q_1 q_2}{r_f} = k \frac{(9.1 \times 10^{-6} \, \text{C})(-0.42 \times 10^{-6} \, \text{C})}{0.24} \]

3. **Set Up the Energy Conservation Equation:**
- According to the conservation of energy:
\[ U_i + K_i = U_f + K_f \]
- Since the charge \( q_2 \) is released from rest, \( K_i = 0 \):
\[ U_i = U_f + K_f \]
- The kinetic energy \( K_f \) can be expressed as:
\[ K_f = \frac{1}{2} m v^2 \]

4. **Rearranging the Equation:**
- Rearranging gives:
\[ K_f = U_i - U_f \]
- Substitute for \( K_f \):
\[ \frac{1}{2} m v^2 = U_i - U_f \]

5. **Substituting Values:**
- Substitute the values of \( U_i \) and \( U_f \) into the equation:
\[ \frac{1}{2} m v^2 = k \frac{(9.1 \times 10^{-6})(-0.42 \times 10^{-6})}{0.96} - k \frac{(9.1 \times 10^{-6})(-0.42 \times 10^{-6})}{0.24} \]

6. **Calculate the Values:**
- Calculate \( U_i \) and \( U_f \):
- \( U_i = k \frac{(9.1 \times 10^{-6})(-0.42 \times 10^{-6})}{0.96} \)
- \( U_f = k \frac{(9.1 \times 10^{-6})(-0.42 \times 10^{-6})}{0.24} \)
- Find \( U_i - U_f \).

7. **Solve for \( v \):**
- Rearranging gives:
\[ v = \sqrt{\frac{2(U_i - U_f)}{m}} \]
- Substitute \( m = 3.2 \times 10^{-4} \, \text{kg} \) and compute \( v \).

8. **Final Calculation:**
- After performing the calculations, you should find that \( v \approx 26 \, \text{m/s} \).