A small particle has charge `-5.00muC` and mass `2.00 xx 10-4` kg. It moves from point A where the electric potential is `V_A = +200 V`. to point B, where the electric potential is `V_B=+ 800 V`. The electric force is the only force acting on the particle. The particle has speed `5.00 m/s` at point A. What is its speed at point B? is it moving faster or slower at B than at A. Explain,

Explanation

To solve the problem, we will use the principle of conservation of mechanical energy. The total mechanical energy (kinetic energy + potential energy) of the particle remains constant since the only force acting on it is the electric force.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Charge of the particle, \( Q = -5.00 \, \mu C = -5.00 \times 10^{-6} \, C \)
- Mass of the particle, \( m = 2.00 \times 10^{-4} \, kg \)
- Initial electric potential at point A, \( V_A = +200 \, V \)
- Final electric potential at point B, \( V_B = +800 \, V \)
- Initial speed at point A, \( v_A = 5.00 \, m/s \)

2. **Write the Conservation of Mechanical Energy Equation:**
The conservation of mechanical energy states that:
\[ U_A + K_A = U_B + K_B \]
Where:
- \( U \) is the electric potential energy (\( U = QV \))
- \( K \) is the kinetic energy (\( K = \frac{1}{2}mv^2 \))

3. **Calculate Initial Potential Energy and Kinetic Energy:**
- Initial potential energy at point A:
\[ U_A = Q \cdot V_A = (-5.00 \times 10^{-6} \, C)(200 \, V) = -1.00 \times 10^{-3} \, J \]
- Initial kinetic energy at point A:
\[ K_A = \frac{1}{2} m v_A^2 = \frac{1}{2} (2.00 \times 10^{-4} \, kg)(5.00 \, m/s)^2 = 2.50 \times 10^{-3} \, J \]

4. **Calculate Total Initial Energy:**
\[ E_{initial} = U_A + K_A = -1.00 \times 10^{-3} \, J + 2.50 \times 10^{-3} \, J = 1.50 \times 10^{-3} \, J \]

5. **Calculate Final Potential Energy at Point B:**
- Final potential energy at point B:
\[ U_B = Q \cdot V_B = (-5.00 \times 10^{-6} \, C)(800 \, V) = -4.00 \times 10^{-3} \, J \]

6. **Use Conservation of Energy to Find Final Kinetic Energy:**
\[ E_{initial} = U_B + K_B \]
Rearranging gives:
\[ K_B = E_{initial} - U_B = 1.50 \times 10^{-3} \, J - (-4.00 \times 10^{-3} \, J) = 1.50 \times 10^{-3} \, J + 4.00 \times 10^{-3} \, J = 5.50 \times 10^{-3} \, J \]

7. **Calculate Final Speed at Point B:**
\[ K_B = \frac{1}{2} m v_B^2 \]
Rearranging gives:
\[ v_B = \sqrt{\frac{2 K_B}{m}} = \sqrt{\frac{2 \times 5.50 \times 10^{-3} \, J}{2.00 \times 10^{-4} \, kg}} = \sqrt{55} \approx 7.42 \, m/s \]

8. **Conclusion:**
The speed of the particle at point B is approximately \( 7.42 \, m/s \). Since \( v_B > v_A \), the particle is moving faster at point B than at point A.