- #1
Sho Kano
- 372
- 3
Homework Statement
A charged particle (either an electron or a proton) is moving rightward between two parallel charged plates separated by distance d=2.87mm. The plate potentials are V1=-75.8V and V2=-49V. The particle is slowing from an initial speed of 90.0 km/s at the left plate. What is its speed just as it reaches plate 2?
Homework Equations
U = qV
W = ΔKE
KE = 1/2mv^2
The Attempt at a Solution
The particle is a proton because it is slowing down towards a plate of higher potential.
Attempt #1 using energy,
[itex] { U }_{ 0 }\quad +\quad { K }_{ 0 }\quad =\quad { U }_{ f }\quad +\quad { K }_{ f }\\ { U }_{ 0 }\quad -\quad { U }_{ f }\quad +\quad { K }_{ 0 }\quad =\quad { K }_{ f }\\ \frac { 2 }{ { m }_{ p } } { [q }_{ p }{ V }_{ 0 }\quad -\quad { q }_{ p }{ V }_{ f }\quad +\quad \frac { 1 }{ 2 } { m }_{ p }{ v }_{ 0 }^{ 2 }]\quad =\quad { v }_{ f }^{ 2 }\\ 1.2e27*[-1.2e-17\quad +\quad 7.84e-18\quad +\quad 6.76e-18]\quad =\quad { v }_{ f }^{ 2 }\\ { v }_{ f }\quad =\quad 54448.79\quad m/s[/itex]
Attempt #2 using energy
[itex]W\quad =\quad \Delta K\\ q\Delta V\quad =\quad Kf\quad -\quad Ki\\ q\Delta V\quad +\quad Ki\quad =\quad Kf\\ \frac { 2 }{ 1.67e-27 } [26.8*1.6e-19\quad +\quad 0.5*1.67e-27*{ 90000 }^{ 2 }]\quad =\quad { v }f^{ 2 }\\ vf\quad \neq \quad 54448.79[/itex]
Is 54448 m/s the correct answer? How come when I use the work energy theorem, the answer doesn't come out right? I noticed if I change the W done in attempt #2 to negative, the answer matches that of attempt #1... why?