Return to Blackboard Halliday, Fundamentals of Physics, 10e Help I en Assignment

Question

Return to Blackboard Halliday, Fundamentals of Physics, 10e Help I en Assignment BACK NEXT Chapter 24, Problem 059 blem In the figure a charged particle (either an electron or a proton) is moving rightward between two parallel potentials are V1-74.0 V and V2-49. (b) What is its speed just as it reaches plate 27 charged plates separated by distance d 3.60 mm. The plate 0 V. The particle is slowing from an initial speed of 98.0 km/s at the left plate. (a) Is the particle an electron or a proton? blem blem blem blem blem lem blem blem (b) Number Units the tolerance is +/-296 lem By accessing this Question Assistance, you will learn while you earn points based on the Point Potential Policy set by your instructor study Question Attempts: 0 of 4 used SAVE FOR LATER SUBMET ANSWER Earn Maximum Points available only if you answer this question correctly in three m/edugenýshared/assignment/test/agist.uniid asnmt214

Explanation / Answer

Given  

Distance between the plates (d) = 3.60 mm

potential of first plate (v1) = - 74 v

potential of second plate (v9vi) = 2) = - 49 v

initial speed of the particle (vi) = 98 km /s = 98000 m/s

final speed of the particle (vf) = ?  

a) The particle is a proton as it is slowing down when moving from lower potential to higher potential.

b) According to work energy theorem work done on the charge in moving from - 74 v to - 49 v

W = q (v1-v2) = 1.6 *10^-19 * (-74 + 49) = -4 x 10^-18 J

This is the kinetic energy lost by the proton.

=> 1/2mp * vi^2 - 1/2mp *vf^2 = 4 * 10^-18 (mass of proton (mp)= 1.67 x 10^-27 kg)

=> 1/2 *1.67*10^-27 (98000^2 - vf^2 ) = 4* 10^-18

=> 98000 ^2 - vf^2 = 4*10^-18 *2 / (1.67*10^-27) = 4.97 * 10 ^9

vf = (98000^2 - 4.97 * 10^9)^0.5 = 68073.4 m/s = 68.07 km/s

Related Questions

Navigate

• Browse (All)
• Browse R
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.