Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 23, Problem 23P
(a)
To determine
The final speed of the proton.
(b)
To determine
The accelerating electric field strength if the potential changed uniformly over a distance of
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionChapter 23 Solutions
Physics for Scientists and Engineers
Ch. 23 - Prob. 1PCh. 23 - Prob. 2PCh. 23 - Prob. 3PCh. 23 - Prob. 4PCh. 23 - Prob. 5PCh. 23 - Prob. 6PCh. 23 - Prob. 7PCh. 23 - Prob. 8PCh. 23 - Prob. 9PCh. 23 - Prob. 10P
Ch. 23 - Prob. 11PCh. 23 - Prob. 12PCh. 23 - Prob. 13PCh. 23 - Prob. 14PCh. 23 - Prob. 15PCh. 23 - Prob. 16PCh. 23 - Prob. 17PCh. 23 - Prob. 18PCh. 23 - Prob. 19PCh. 23 - Prob. 20PCh. 23 - Prob. 21PCh. 23 - Prob. 22PCh. 23 - Prob. 23PCh. 23 - Prob. 24PCh. 23 - Prob. 25PCh. 23 - Prob. 26PCh. 23 - Prob. 27PCh. 23 - Prob. 28PCh. 23 - Prob. 29PCh. 23 - Prob. 30PCh. 23 - Prob. 31PCh. 23 - Prob. 32PCh. 23 - Prob. 33PCh. 23 - Prob. 34PCh. 23 - Prob. 35PCh. 23 - Prob. 36PCh. 23 - Prob. 37PCh. 23 - Prob. 38PCh. 23 - Prob. 39PCh. 23 - Prob. 40PCh. 23 - Prob. 41PCh. 23 - Prob. 42PCh. 23 - Prob. 43PCh. 23 - Prob. 44PCh. 23 - Prob. 45PCh. 23 - Prob. 46PCh. 23 - Prob. 47PCh. 23 - Prob. 48PCh. 23 - Prob. 49PCh. 23 - Prob. 50PCh. 23 - Prob. 51PCh. 23 - Prob. 52PCh. 23 - Prob. 53PCh. 23 - Prob. 54PCh. 23 - Prob. 55PCh. 23 - Prob. 56PCh. 23 - Prob. 57PCh. 23 - Prob. 58PCh. 23 - Prob. 59PCh. 23 - Prob. 60PCh. 23 - Prob. 61PCh. 23 - Prob. 62PCh. 23 - Prob. 63PCh. 23 - Prob. 64PCh. 23 - Prob. 65PCh. 23 - Prob. 66PCh. 23 - Prob. 67PCh. 23 - Prob. 68PCh. 23 - Prob. 69PCh. 23 - Prob. 70PCh. 23 - Prob. 71PCh. 23 - Prob. 72PCh. 23 - Prob. 73PCh. 23 - Prob. 74PCh. 23 - Prob. 75PCh. 23 - Prob. 76PCh. 23 - Prob. 77PCh. 23 - Prob. 78PCh. 23 - Prob. 79PCh. 23 - Prob. 80PCh. 23 - Prob. 81PCh. 23 - Prob. 82PCh. 23 - Prob. 83PCh. 23 - Prob. 84PCh. 23 - Prob. 85PCh. 23 - Prob. 86PCh. 23 - Prob. 87PCh. 23 - Prob. 88PCh. 23 - Prob. 89PCh. 23 - Prob. 90PCh. 23 - Prob. 91PCh. 23 - Prob. 92PCh. 23 - Prob. 93PCh. 23 - Prob. 94P
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- At a certain distance from a charged particle, the magnitude of the electric field is 500 V/m and the electric potential is 3.00 kV. (a) What is the distance to the particle? (b) What is the magnitude of the charge?arrow_forwardA filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm, carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?arrow_forwardA proton is released from rest at the origin in a uniform electric field in the positive x direction with magnitude 850 N/C. What is the change in the electric potential energy of the protonfield system when the proton travels to x = 2.50 m? (a) 3.40 1016 J (b) 3.40 1016 J (c) 2.50 1016 J (d) 2.50 1016 J (e) 1.60 1019 Jarrow_forward
- The three charged particles in Figure P20.11 are at the vertices of an isosceles triangle (where d = 2.00 cm). Taking q = 7.00 C, calculate the electric potential at point A, the midpoint of the base. Figure P20.11arrow_forwardA filament running along the x axis from the origin to x = 80.0 cm carries electric charge with uniform density. At the point P with coordinates (x = 80.0 cm, y = 80.0 cm), this filament creates electric potential 100 V. Now we add another filament along the y axis, running from the origin to y = 80.0 cm. carrying the same amount of charge with the same uniform density. At the same point P, is the electric potential created by the pair of filaments (a) greater than 200 V, (b) 200 V, (c) 100 V, (d) between 0 and 200 V, or (e) 0?arrow_forwardFour particles are positioned on the rim of a circle. The charges on the particles are +0.500 C, +1.50 C, 1.00 C, and 0.500 C. If the electric potential at the center of the circle due to the +0.500 C charge alone is 4.50 104 V, what is the total electric potential at the center due to the four charges? (a) 18.0 104 V (b) 4.50 104 V (c) 0 (d) 4.50 104 V (e) 9.00 104 Varrow_forward
- If a proton is released from rest in an electric field, will it move in the direction of increasing or decreasing potential? Also answer this question for an electron and a neutron. Explain why.arrow_forwardFor the arrangement described in Problem 26, calculate the electric potential at point B, which lies on the perpendicular bisector of the rod a distance b above the x axis. Figure P20.26arrow_forwardTwo 5.00-nC charged particles are in a uniform electric field with a magnitude of 625 N/C. Each of the particles is moved from point A to point B along two different paths, labeled in Figure P26.65. a. Given the dimensions in the figure, what is the change in the electric potential experienced by the particle that is moved along path 1 (black)? b. What is the change in the electric potential experienced by the particle that is moved along path 2 (red)? c. Is there a path between the points A and B for which the change in the electric potential is different from your answers to parts (a) and (b)? Explain. FIGURE P26.65 Problems 65, 66, and 67.arrow_forward
- An electric potential exists in a region of space such that V = 8x4 2y2 + 9z3 and V is in units of volts, when x, y, and z are in meters. a. Find an expression for the electric field as a function of position. b. What is the electric field at (2.0 m, 4.5 m, 2.0 m)?arrow_forwardTwo particles each with charge +2.00 C are located on the x axis. One is at x = 1.00 m, and the other is at x = 1.00 m. (a) Determine the electric potential on the y axis at y = 0.500 m. (b) Calculate the change in electric potential energy of the system as a third charged particle of 3.00 C is brought from infinitely far away to a position on the y axis at y = 0.500 m.arrow_forwardA CD disk of radius (R = 3.0 cm) is sprayed with a charged paint so that the charge varies continually with radial distance r from the center in the following manner =(6.0C/m)r/R ?. Find the potential at a point 4 cm above the center.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Principles of Physics: A Calculus-Based TextPhysicsISBN:9781133104261Author:Raymond A. Serway, John W. JewettPublisher:Cengage LearningPhysics for Scientists and Engineers, Technology ...PhysicsISBN:9781305116399Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
- Physics for Scientists and Engineers: Foundations...PhysicsISBN:9781133939146Author:Katz, Debora M.Publisher:Cengage LearningCollege PhysicsPhysicsISBN:9781285737027Author:Raymond A. Serway, Chris VuillePublisher:Cengage LearningPhysics for Scientists and Engineers with Modern ...PhysicsISBN:9781337553292Author:Raymond A. Serway, John W. JewettPublisher:Cengage Learning
Principles of Physics: A Calculus-Based Text
Physics
ISBN:9781133104261
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers, Technology ...
Physics
ISBN:9781305116399
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Physics for Scientists and Engineers: Foundations...
Physics
ISBN:9781133939146
Author:Katz, Debora M.
Publisher:Cengage Learning
College Physics
Physics
ISBN:9781285737027
Author:Raymond A. Serway, Chris Vuille
Publisher:Cengage Learning
Physics for Scientists and Engineers with Modern ...
Physics
ISBN:9781337553292
Author:Raymond A. Serway, John W. Jewett
Publisher:Cengage Learning
Electric Fields: Crash Course Physics #26; Author: CrashCourse;https://www.youtube.com/watch?v=mdulzEfQXDE;License: Standard YouTube License, CC-BY