Principles of Physics: A Calculus-Based Text
5th Edition
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Raymond A. Serway, John W. Jewett
Chapter20: Electric Potential And Capacitance
Section: Chapter Questions
Problem 8OQ: In a certain region of space, a uniform electric field is in the x direction. A particle with...
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![### Problem Statement:
Calculate the electric potential 0.85 m away from an electron.
(Give answer in volts but don't include the units.)
### Explanation:
**Electric Potential (V)** is defined as the work done per unit charge to bring a small positive test charge from infinity to a point in space. The formula to calculate the electric potential due to a point charge is:
\[ V = \dfrac{k_e \cdot q}{r} \]
Where:
- \( V \) is the electric potential.
- \( k_e \) (Coulomb's constant) is approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
- \( q \) is the charge of the electron, \( -1.6 \times 10^{-19} \, \text{C} \).
- \( r \) is the distance from the charge, in this case, 0.85 m.
Substitute the given values into the formula to compute the electric potential:
\[ V = \dfrac{(8.99 \times 10^9) \cdot (-1.6 \times 10^{-19})}{0.85} \]
By performing the calculation:
1. Multiply the constants:
\[ 8.99 \times 10^9 \times -1.6 \times 10^{-19} \approx -1.4384 \times 10^{-9} \]
2. Divide by the distance:
\[ V \approx \dfrac{-1.4384 \times 10^{-9}}{0.85} \approx -1.692 \times 10^{-9} \]
Thus, the electric potential at a distance of 0.85 meters from an electron is approximately \(-1.692 \, \text{volts}\).
**Note:** The negative sign indicates that the potential is negative due to the negative charge of the electron. However, as per the instructions, units should not be included in the final answer. Therefore, the answer is:
\[ \boxed{-1.692} \]](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F240c2e2e-41dc-444c-bfea-c5009c02e973%2F5b930b29-ace7-4f6b-b8b6-4770c9ee693b%2Fx9jvi0s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:### Problem Statement:
Calculate the electric potential 0.85 m away from an electron.
(Give answer in volts but don't include the units.)
### Explanation:
**Electric Potential (V)** is defined as the work done per unit charge to bring a small positive test charge from infinity to a point in space. The formula to calculate the electric potential due to a point charge is:
\[ V = \dfrac{k_e \cdot q}{r} \]
Where:
- \( V \) is the electric potential.
- \( k_e \) (Coulomb's constant) is approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
- \( q \) is the charge of the electron, \( -1.6 \times 10^{-19} \, \text{C} \).
- \( r \) is the distance from the charge, in this case, 0.85 m.
Substitute the given values into the formula to compute the electric potential:
\[ V = \dfrac{(8.99 \times 10^9) \cdot (-1.6 \times 10^{-19})}{0.85} \]
By performing the calculation:
1. Multiply the constants:
\[ 8.99 \times 10^9 \times -1.6 \times 10^{-19} \approx -1.4384 \times 10^{-9} \]
2. Divide by the distance:
\[ V \approx \dfrac{-1.4384 \times 10^{-9}}{0.85} \approx -1.692 \times 10^{-9} \]
Thus, the electric potential at a distance of 0.85 meters from an electron is approximately \(-1.692 \, \text{volts}\).
**Note:** The negative sign indicates that the potential is negative due to the negative charge of the electron. However, as per the instructions, units should not be included in the final answer. Therefore, the answer is:
\[ \boxed{-1.692} \]
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