(a) Find the speed of the particle at B by modeling it as a particle under constant acceleration.SOLUTIONConceptualize When the positive charge is placed at A, it experiences an electric force toward the right in the figure due to the electric field directed toward the right. As a result, it will accelerate to the---Select--- :) and arrive at B with some speed.Categorize Because the electric field is uniform, a constant electric force acts on the charge. Therefore, as suggested in the problem statement, the point charge can be modeled as a charged particle---Select---AnalyzeUse this equation to express the velocity of the particle as a function of position (Use the following as necessary: a, d, and q.):v? = v,? + 2a(xf– x;) = 0 + 2a(d – 0) =Using the equation for the force on a charge in an electric field, solve for vçand substitute for the magnitude of the acceleration (Use the following as necessary: E, m, d, and q.):Vf =(b) Find the speed of the particle at B by modeling it as a nonisolated system in terms of energy.SOLUTIONCategorize The problem statement tells us that the charge is a nonisolated system for energy. The electric force, like any force, can do work on a system. Energy is transferred to the system of the charge bywork done by the ( ---Select---force exerted on the charge. The initial configuration of the system is when the particle is at rest at A, and the final configuration is when it is moving with some speed at B.AnalyzeWrite the appropriate reduction of the conservation of energy equation, AK + AU + AEint= W + Q + TMw + TMT + TET + TER, for the system of the charged particle:ЕTW = AKReplace the work and kinetic energies with values appropriate for this situation. (Use the following as necessary: F, Ax, E, and m.)12FAx = Kg - KA→ Vf-mv- 0 → V.=Substitute for the electric force F and the displacement Ax. (Use the following as necessary: E, m, d, and q.)eFinalize The answer to part (b) is the same as that for part (a), as we expect. This problem can be solved with different approaches. We saw the same possibilities with mechanical problems. A uniform electric field E is directed along the x-axis between parallel plates of charge separated by a distanced as shown in the figure. A positive point charge q of mass m is released from rest at a point A next to thepositive plate and accelerates to a point B next to the negative plate.A positive point charge q in auniform electric field E undergoesconstant acceleration in thedirection of the field.++= 0Bd

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(a) Find the speed of the particle at B by modeling it as a particle under constant acceleration. SOLUTION Conceptualize When the positive charge is placed at A, it experiences an electric force toward the right in the figure due to the electric field directed toward the right. As a result, it will accelerate to the --Select-- ) and arrive at B with some speed. Categorize Because the electric field is uniform, a constant electric force acts on the charge. Therefore, as suggested in the problem statement, the point charge can be modeled as a charged particle -Select-- Analyze Use this equation to express the velocity of the particle as a function of position (Use the following as necessary: a, d, and q.): v? = v? + 2a(x,- x,) = 0 + 2a(d – 0) = Using the equation for the force on a charge in an electric field, solve for vand substitute for the magnitude of the acceleration (Use the following as necessary: E, m, d, and q.):

(a) Find the speed of the particle at B by modeling it as a particle under constant acceleration. SOLUTION Conceptualize When the positive charge is placed at A, it experiences an electric force toward the right in the figure due to the electric field directed toward the right. As a result, it will accelerate to the --Select-- ) and arrive at B with some speed. Categorize Because the electric field is uniform, a constant electric force acts on the charge. Therefore, as suggested in the problem statement, the point charge can be modeled as a charged particle -Select-- Analyze Use this equation to express the velocity of the particle as a function of position (Use the following as necessary: a, d, and q.): v? = v? + 2a(x,- x,) = 0 + 2a(d – 0) = Using the equation for the force on a charge in an electric field, solve for vand substitute for the magnitude of the acceleration (Use the following as necessary: E, m, d, and q.):

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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Figure 1 of 1 > A B -10.0cm
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yrom) Four point charges a1, a2, 93, and 94 are positioned at points (11,0),(0,12),(-15,0), and (0,-8), respectively. a) write the net electric field due to these charges at the origin in vector form using the unit vectors î and j. |- 4pC 5pC b) Calculate the magnitude and direction of the vector you found in part a. c) write the net electric force on 94 in vector form using î and J. d) calculate the magnitude and direction of +he vector you found in part c. Make sure you use the correct units in your calculations and final answers. Coordinates are in (x,4) form. 3M C 2) An electron's position and velocity at t = 0 are given by the vectors ŷ : - 4 x10 = 0. 50 m 一 4 x 1o" m/s There exists a uniform electric field given by E = - 1 00 N. i Ignoring gravity, find a) the acceleration vector a of the electron b) the position vector r when V= 4x 10° m/s î c) the velocity vector when the y-component of the position vector is 0.50 m again. (Express your answers in vector form using unit vectors…
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QUESTION 3 Problem A newly discovered light positively charged particle has a mass of m and charge q. Suppose it moves within the vicinity of an extremely heavy (fixed in place) particle with a positive charge Q and mass M. When the light particle is xi distance from the heavy particle, it is moving directly away from the heavy particle with a speed of vi. a) What is the lighter particle's speed when it is xf away from the heavy particle? (Consider the Newtonian Gravitation acting between the two charged particles. Ignore the effects of external forces) Solution: We may solve this using two approaches. One involves the Newton's Laws and the other involving Work-Energy theorem. To avoid the complexity of vector solution, we will instead employ the Work-Energy theorem, more specifically, the Conservation of Energy Principle. Let us first name the lighter particle as object 1 and the heavy particle as object 2. Through work-energy theorem, we will take into account all of the energy of…
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