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, likeany force, can do work on a system. Energy is transferred to the system of the charge by work 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,ΔΚ + Δυ + ΔΕ= W + Q + TMw + TMT + TET + TER, for the system of the charged particle:intW= ΔΚReplace the work and kinetic energies with values appropriate for this situation. (Use the following as necessary: F.,Ax, E, and m.)e'FAx = KB - KA2-mv? - 0 → v, =Substitute for the electric force F and the displacement Ax. (Use the following as necessary: E, m, d, and q.)Finalize The answer to part (b) is the same as that for part (a), as we expect. This problem can be solved withdifferent approaches. We saw the same possibilities with mechanical problems.EXERCISEIn an experiment similar to the setup described in the Example, a positively charged particle of mass m is released fromrest at the surface of the positive plate. The velocity of the particle at the negative plate is 1.30 x 105 m/s. If the electricfield magnitude is 6.00 x 103 N/C, the distance between the two plates is 5 cm, and the charge on the particle is 3e, whatis the particle's mass (in kg)? 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

Given,mass of the particle, mvelocity of the particle at the negative plate, v = 1.30×105…

A uniform electric field E is directed along the x-axis between parallel plates of charge separated by a distance d as shown in the figure. A positive point charge q of mass m is released from rest at a point A next to the positive plate and accelerates to a point B next to the negative plate. A positive point charge q in a uniform electric field E undergoes constant acceleration in the direction of the field. V= 0

A uniform electric field E is directed along the x-axis between parallel plates of charge separated by a distance d as shown in the figure. A positive point charge q of mass m is released from rest at a point A next to the positive plate and accelerates to a point B next to the negative plate. A positive point charge q in a uniform electric field E undergoes constant acceleration in the direction of the field. V= 0

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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