**Problem Statement:**A particle (charge = 3 μC) is released from rest at a point \( x = 11.4 \) cm. If a 61.1-μC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 94.1 cm?Round your answer to 2 decimal places.---**Explanation:**In this problem, a particle with a charge of \( 3 \) μC is initially at rest at \( x = 11.4 \) cm. A charge of \( 61.1 \) μC is fixed at the origin. We are to determine the kinetic energy of the particle after it has moved a distance of \( 94.1 \) cm.**Detailed Steps to Solve the Problem:**1. **Calculate the initial electric potential energy (U_initial):** The formula for electric potential energy between two point charges is: \[ U = \frac{k \cdot q_1 \cdot q_2}{r} \] where: - \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \)), - \( q_1 \) and \( q_2 \) are the charges, - \( r \) is the distance between the charges. For the initial position (\( x = 11.4 \) cm): \[ U_{\text{initial}} = \frac{(8.99 \times 10^9) \cdot (3 \times 10^{-6}) \cdot (61.1 \times 10^{-6})}{0.114 \text{ m}} \]2. **Calculate the final electric potential energy (U_final):** The particle moves an additional distance of \( 94.1 \) cm, so the final distance from the origin is: \[ r_f = 11.4 \text{ cm} + 94.1 \text{ cm} = 105.5 \text{ cm} = 1.055 \text{ m} \] Using the formula for electric potential energy: \[ U_{\text{final}} = \frac{(8.99 \times 10^9)

Given, Initially, charges are 3x10-6C at 11.4 cm (11x10-2 m) and 6.1x10-6 C at the origin. Then when…

A particle (charge = 3 µC) is released from rest at a point x %3D = 11.4 cm. If a 61.1-µC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 94.1 cm? Round your answer to 2 decimal places.

A particle (charge = 3 µC) is released from rest at a point x %3D = 11.4 cm. If a 61.1-µC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 94.1 cm? Round your answer to 2 decimal places.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

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Question

**Problem Statement:**

A particle (charge = 3 μC) is released from rest at a point \( x = 11.4 \) cm. If a 61.1-μC charge is held fixed at the origin, what is the kinetic energy of the particle after it has moved 94.1 cm?

Round your answer to 2 decimal places.

---

**Explanation:**

In this problem, a particle with a charge of \( 3 \) μC is initially at rest at \( x = 11.4 \) cm. A charge of \( 61.1 \) μC is fixed at the origin. We are to determine the kinetic energy of the particle after it has moved a distance of \( 94.1 \) cm.

**Detailed Steps to Solve the Problem:**

1. **Calculate the initial electric potential energy (U_initial):**

The formula for electric potential energy between two point charges is:
\[
U = \frac{k \cdot q_1 \cdot q_2}{r}
\]
where:
- \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \text{ N m}^2/\text{C}^2 \)),
- \( q_1 \) and \( q_2 \) are the charges,
- \( r \) is the distance between the charges.

For the initial position (\( x = 11.4 \) cm):
\[
U_{\text{initial}} = \frac{(8.99 \times 10^9) \cdot (3 \times 10^{-6}) \cdot (61.1 \times 10^{-6})}{0.114 \text{ m}}
\]

2. **Calculate the final electric potential energy (U_final):**

The particle moves an additional distance of \( 94.1 \) cm, so the final distance from the origin is:
\[
r_f = 11.4 \text{ cm} + 94.1 \text{ cm} = 105.5 \text{ cm} = 1.055 \text{ m}
\]

Using the formula for electric potential energy:
\[
U_{\text{final}} = \frac{(8.99 \times 10^9)

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