**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

Chapter16: Electrical Energy And Capacitance

Section: Chapter Questions

Problem 14P: Three charges are situated at corners of a rectangle as in Figure P16.13. How much work must an...

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Concept explainers

Electric Charges and Fields

One of the fundamental properties is the electromagnetic property. Charge cannot be destroyed by any process and this contributes formally to the law of charge conservation.

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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