#3 Electrostatic Potential Energy having a mass of 0.0550 kilograms, are separated by 0.250 meters. Two identical +3.00 µC charges, both a. Calculate the electrostatic potential energy of the pair of charges. 25 V= 0.324T b. Only one of the charges breaks free, while the other remains anchored and can- not move at all. Assuming no frictional losses, what is the kinetic energy of the departing charge as it approaches the point at infinity? 10.648J メ- Vん c. What is the magnitude of the velocity of the departing charge when it is very far away from the stationary charge? A55ン Ca
#3 Electrostatic Potential Energy having a mass of 0.0550 kilograms, are separated by 0.250 meters. Two identical +3.00 µC charges, both a. Calculate the electrostatic potential energy of the pair of charges. 25 V= 0.324T b. Only one of the charges breaks free, while the other remains anchored and can- not move at all. Assuming no frictional losses, what is the kinetic energy of the departing charge as it approaches the point at infinity? 10.648J メ- Vん c. What is the magnitude of the velocity of the departing charge when it is very far away from the stationary charge? A55ン Ca
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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b and c
![**Electrostatic Potential Energy**
In this exercise, we consider two identical charges of +3.00 μC, each having a mass of 0.0550 kilograms, separated by a distance of 0.250 meters.
**a. Calculation of Electrostatic Potential Energy**
The electrostatic potential energy (V) of the pair of charges is calculated using the formula:
\[ V = \frac{k \cdot q_1 \cdot q_2}{r} \]
Where:
- \( k \) is the electrostatic constant (\( \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( q_1 \) and \( q_2 \) are the charges (\( 3 \times 10^{-6} \, \text{C} \))
- \( r \) is the distance between the charges (0.25 meters)
The calculation yields:
\[ V = 0.324 \, \text{J} \]
**b. Kinetic Energy Calculation**
When one charge breaks free and the other remains anchored, the kinetic energy of the departing charge, assuming no frictional losses, is determined as it approaches infinity. The calculated kinetic energy is:
\[ 0.648 \, \text{J} \]
**c. Velocity of the Departing Charge**
To find the magnitude of the velocity (\( v \)) of the departing charge, when it is very far from the stationary charge, we use the kinetic energy formula and solve for \( v \):
\[ KE = \frac{1}{2} m v^2 \]
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
The specific values for this calculation are not completely shown in the image, but this is the procedure to follow.
This exercise illustrates concepts in electrostatics, particularly focusing on potential energy and kinetic energy transformations.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F15f07b92-5703-4de4-9426-36c09b5d48af%2F161e2a75-8443-433b-bcc6-a8d7cdde81ac%2Fgkkrskg_processed.png&w=3840&q=75)
Transcribed Image Text:**Electrostatic Potential Energy**
In this exercise, we consider two identical charges of +3.00 μC, each having a mass of 0.0550 kilograms, separated by a distance of 0.250 meters.
**a. Calculation of Electrostatic Potential Energy**
The electrostatic potential energy (V) of the pair of charges is calculated using the formula:
\[ V = \frac{k \cdot q_1 \cdot q_2}{r} \]
Where:
- \( k \) is the electrostatic constant (\( \approx 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \))
- \( q_1 \) and \( q_2 \) are the charges (\( 3 \times 10^{-6} \, \text{C} \))
- \( r \) is the distance between the charges (0.25 meters)
The calculation yields:
\[ V = 0.324 \, \text{J} \]
**b. Kinetic Energy Calculation**
When one charge breaks free and the other remains anchored, the kinetic energy of the departing charge, assuming no frictional losses, is determined as it approaches infinity. The calculated kinetic energy is:
\[ 0.648 \, \text{J} \]
**c. Velocity of the Departing Charge**
To find the magnitude of the velocity (\( v \)) of the departing charge, when it is very far from the stationary charge, we use the kinetic energy formula and solve for \( v \):
\[ KE = \frac{1}{2} m v^2 \]
\[ v = \sqrt{\frac{2 \times KE}{m}} \]
The specific values for this calculation are not completely shown in the image, but this is the procedure to follow.
This exercise illustrates concepts in electrostatics, particularly focusing on potential energy and kinetic energy transformations.
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