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.
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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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![**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)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa023f3c3-3186-4c61-aa74-30f5825e9d0f%2Fed8e77f8-e3e5-4a4a-bdbc-32487044ee9a%2Fk5940c.jpeg&w=3840&q=75)
Transcribed Image Text:**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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