Textbook Question
FIGURE EX25.10 shows the potential energy of an electric dipole. Consider a dipole that oscillates between ±60°. What is the dipole's mechanical energy?
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Ch 25: The Electric Potential
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796
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Table of contents
- Ch 01: Concepts of Motion35
- Ch 02: Kinematics in One Dimension75
- Ch 03: Vectors and Coordinate Systems58
- Ch 04: Kinematics in Two Dimensions60
- Ch 05: Force and Motion23
- Ch 06: Dynamics I: Motion Along a Line53
- Ch 07: Newton's Third Law27
- Ch 08: Dynamics II: Motion in a Plane52
- Ch 09: Work and Kinetic Energy56
- Ch 10: Interactions and Potential Energy50
- Ch 11: Impulse and Momentum56
- Ch 12: Rotation of a Rigid Body71
- Ch 13: Newton's Theory of Gravity52
- Ch 14: Fluids and Elasticity44
- Ch 15: Oscillations55
- Ch 16: Traveling Waves37
- Ch 17: Superposition39
- Ch 18: A Macroscopic Description of Matter53
- Ch 19: Work, Heat, and the First Law of Thermodynamics60
- Ch 20: The Micro/Macro Connection53
- Ch 21: Heat Engines and Refrigerators34
- Ch 22: Electric Charges and Forces40
- Ch 23: The Electric Field39
- Ch 24: Gauss' Law32
- Ch 25: The Electric Potential63
- Ch 26: Potential and Field57
- Ch 27: Current and Resistance42
- Ch 28: Fundamentals of Circuits39
- Ch 29: The Magnetic Field47
- Ch 30: Electromagnetic Induction60
- Ch 31: Electromagnetic Fields and Waves32
- Ch 32: AC Circuits63
- Ch 33: Wave Optics32
- Ch 34: Ray Optics57
- Ch 35: Optical Instruments30
- Ch 36: Special Relativity38
- Ch 37: The Foundations of Modern Physics25
- Ch 38: Quantization49
- Ch 39: Wave Functions and Uncertainty31
- Ch 40: One-Dimensional Quantum Mechanics35
- Ch 41: Atomic Physics18
- Ch 42: Nuclear Physics27
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 25, Problem 14
What potential difference is needed to accelerate an electron from rest to a speed of 2.0×106 m/s?
Verified step by step guidance1
Identify the relationship between the kinetic energy gained by the electron and the work done by the electric field. The work done is equal to the change in kinetic energy: \( W = \Delta KE \).
Express the kinetic energy of the electron using the formula \( KE = \frac{1}{2}mv^2 \), where \( m \) is the mass of the electron (\( 9.11 \times 10^{-31} \; \text{kg} \)) and \( v \) is the final speed of the electron (\( 2.0 \times 10^6 \; \text{m/s} \)).
Relate the work done to the potential difference \( V \) using the equation \( W = qV \), where \( q \) is the charge of the electron (\( 1.6 \times 10^{-19} \; \text{C} \)).
Set the work done equal to the change in kinetic energy: \( qV = \frac{1}{2}mv^2 \). Solve for \( V \) by rearranging the equation: \( V = \frac{\frac{1}{2}mv^2}{q} \).
Substitute the known values for \( m \), \( v \), and \( q \) into the equation to calculate the potential difference \( V \).
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Kinetic Energy
Kinetic energy is the energy an object possesses due to its motion, calculated using the formula KE = 1/2 mv², where m is mass and v is velocity. For an electron accelerated from rest, its kinetic energy at a speed of 2.0×10⁶ m/s can be determined, which is essential for understanding the energy required for acceleration.
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Electric Potential Difference
Electric potential difference, or voltage, is the work done per unit charge to move a charge between two points in an electric field. It is directly related to the kinetic energy gained by a charged particle, such as an electron, when accelerated through a potential difference, allowing us to calculate the necessary voltage to achieve a specific speed.
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Charge of an Electron
The charge of an electron is a fundamental physical constant, approximately -1.6 × 10⁻¹⁹ coulombs. This negative charge is crucial in calculations involving electric fields and potential differences, as it determines the direction of force and energy transfer when an electron is accelerated by an electric field.
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Related Practice
Textbook Question
An electron with an initial speed of 500,000 m/s is brought to rest by an electric field. Did the electron move into a region of higher potential or lower potential?
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Textbook Question
Three charged particles are placed at the corners of an equilateral triangle that has edge length 2.0 cm. One particle has charge +3.0 nC and a second has charge +6.0 nC. What is the third charge if the electric potential energy of the three charged particles is zero?
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Textbook Question
In proton-beam therapy, a high-energy beam of protons is fired at a tumor. As the protons stop in the tumor, their kinetic energy breaks apart the tumor's DNA, thus killing the tumor cells. For one patient, it is desired to deposit 0.10 J of proton energy in the tumor. To create the proton beam, protons are accelerated from rest through a 10,000 kV potential difference. What is the total charge of the protons that must be fired at the tumor?
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Textbook Question
What is the speed of an electron that has been accelerated from rest through a potential difference of 1000 V?
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Textbook Question
A proton with an initial speed of 800,000 m/s is brought to rest by an electric field. What was the potential difference that stopped the proton?
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