A particle (charge = 5.5 µC) is released from rest at a point %3D x = 19.3 cm. If a 49.1-µC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 95.1 cm?

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### Physics Problem: Kinetic Energy of a Charged Particle

**Problem Statement:**
A particle (charge = \( 5.5 \, \mu C \)) is released from rest at a point \( x = 19.3 \, \text{cm} \). If a \( 49.1 \, \mu C \) charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved \( 95.1 \, \text{cm} \)?

**Instructions:**
- Round your answer to 2 decimal places.

**Solution Outline:**
1. Identify the initial and final positions of the particle.
2. Calculate the initial and final electric potential energies using Coulomb's law.
3. Determine the change in electric potential energy.
4. Apply the conservation of energy principle to find the kinetic energy.

### Detailed Explanation:
1. **Initial and Final Positions:**
   - Initial position of particle: \( x_i = 19.3 \, \text{cm} \)
   - Final position of particle: \( x_f = x_i + 95.1 \, \text{cm} = 114.4 \, \text{cm} \)

2. **Electric Potential Energy Calculation:**
   - \( q_1 = 5.5 \, \mu C = 5.5 \times 10^{-6} \, C \)
   - \( q_2 = 49.1 \, \mu C = 49.1 \times 10^{-6} \, C \)
   - Distance between charges initially: \( r_i = 19.3 \, \text{cm} = 0.193 \, \text{m} \)
   - Distance after moving: \( r_f = 114.4 \, \text{cm} = 1.144 \, \text{m} \)

   Using Coulomb's law for electric potential energy:
   - Initial potential energy: \( U_i = k \frac{q_1 q_2}{r_i} \)
   - Final potential energy: \( U_f = k \frac{q_1 q_2}{r_f} \)

   Where \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^
Transcribed Image Text:### Physics Problem: Kinetic Energy of a Charged Particle **Problem Statement:** A particle (charge = \( 5.5 \, \mu C \)) is released from rest at a point \( x = 19.3 \, \text{cm} \). If a \( 49.1 \, \mu C \) charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved \( 95.1 \, \text{cm} \)? **Instructions:** - Round your answer to 2 decimal places. **Solution Outline:** 1. Identify the initial and final positions of the particle. 2. Calculate the initial and final electric potential energies using Coulomb's law. 3. Determine the change in electric potential energy. 4. Apply the conservation of energy principle to find the kinetic energy. ### Detailed Explanation: 1. **Initial and Final Positions:** - Initial position of particle: \( x_i = 19.3 \, \text{cm} \) - Final position of particle: \( x_f = x_i + 95.1 \, \text{cm} = 114.4 \, \text{cm} \) 2. **Electric Potential Energy Calculation:** - \( q_1 = 5.5 \, \mu C = 5.5 \times 10^{-6} \, C \) - \( q_2 = 49.1 \, \mu C = 49.1 \times 10^{-6} \, C \) - Distance between charges initially: \( r_i = 19.3 \, \text{cm} = 0.193 \, \text{m} \) - Distance after moving: \( r_f = 114.4 \, \text{cm} = 1.144 \, \text{m} \) Using Coulomb's law for electric potential energy: - Initial potential energy: \( U_i = k \frac{q_1 q_2}{r_i} \) - Final potential energy: \( U_f = k \frac{q_1 q_2}{r_f} \) Where \( k = 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^
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