To solve the problem, we will use the principle of conservation of mechanical energy. The initial potential energy of the charged particle will convert into kinetic energy as it moves closer to the two fixed charges.
### Step-by-Step Solution:
1. **Identify the Charges and Distances**:
- The two identical charges, \( Q_1 = Q_2 = 5 \, \mu C = 5 \times 10^{-6} \, C \), are fixed at a distance of \( 8 \, cm = 0.08 \, m \) apart.
- The charged particle, \( q = -10 \, \mu C = -10 \times 10^{-6} \, C \), is placed \( 5 \, cm = 0.05 \, m \) from each charge initially.
2. **Determine the Final Position**:
- When the particle is released and moves towards the two charges, it will be at the midpoint between them when it is nearest to both charges. The distance from each charge to the midpoint is \( 4 \, cm = 0.04 \, m \).
3. **Calculate Initial and Final Potential Energy**:
- The initial potential energy \( U_i \) when the particle is at \( 5 \, cm \) from each charge:
\[
U_i = k \left( \frac{Q_1 \cdot q}{r_1} + \frac{Q_2 \cdot q}{r_2} \right) = k \left( \frac{5 \times 10^{-6} \cdot (-10 \times 10^{-6})}{0.05} + \frac{5 \times 10^{-6} \cdot (-10 \times 10^{-6})}{0.05} \right)
\]
\[
U_i = 2 \cdot k \left( \frac{-50 \times 10^{-12}}{0.05} \right) = 2 \cdot k \cdot (-1 \times 10^{-9}) = -2k \times 10^{-9}
\]
- The final potential energy \( U_f \) when the particle is at \( 4 \, cm \) from each charge:
\[
U_f = k \left( \frac{Q_1 \cdot q}{r_1'} + \frac{Q_2 \cdot q}{r_2'} \right) = k \left( \frac{5 \times 10^{-6} \cdot (-10 \times 10^{-6})}{0.04} + \frac{5 \times 10^{-6} \cdot (-10 \times 10^{-6})}{0.04} \right)
\]
\[
U_f = 2 \cdot k \left( \frac{-50 \times 10^{-12}}{0.04} \right) = 2 \cdot k \cdot (-1.25 \times 10^{-9}) = -2.5k \times 10^{-9}
\]
4. **Apply Conservation of Energy**:
- According to the conservation of mechanical energy:
\[
K_i + U_i = K_f + U_f
\]
- Initially, the kinetic energy \( K_i = 0 \) (since the particle is released from rest), so:
\[
0 + U_i = K_f + U_f
\]
- Rearranging gives:
\[
K_f = U_i - U_f
\]
- Substituting the values:
\[
K_f = (-2k \times 10^{-9}) - (-2.5k \times 10^{-9}) = 0.5k \times 10^{-9}
\]
5. **Calculate the Kinetic Energy**:
- The kinetic energy \( K_f \) can also be expressed as:
\[
K_f = \frac{1}{2} mv^2
\]
- Setting the two expressions for kinetic energy equal:
\[
\frac{1}{2} mv^2 = 0.5k \times 10^{-9}
\]
- Solving for \( v^2 \):
\[
v^2 = \frac{k \times 10^{-9}}{m}
\]
6. **Substituting Values**:
- Using \( k = 9 \times 10^9 \, N \cdot m^2/C^2 \) and \( m = 9 \times 10^{-6} \, kg \):
\[
v^2 = \frac{9 \times 10^9 \times 10^{-9}}{9 \times 10^{-6}} = 10^3
\]
- Therefore, \( v = \sqrt{10^3} = 10^{1.5} = 31.62 \, m/s \).
### Final Answer:
The speed of the particle when it is nearest to the two charges is approximately \( 31.62 \, m/s \).
