# Finding velocity given potentials of two parallel plates

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1. Apr 2, 2016

### Sho Kano

1. The problem statement, all variables and given/known data
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?

2. Relevant equations
U = qV
W = ΔKE
KE = 1/2mv^2

3. 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,
${ U }_{ 0 }\quad +\quad { K }_{ 0 }\quad =\quad { U }_{ f }\quad +\quad { K }_{ f }\\ { U }_{ 0 }\quad -\quad { U }_{ f }\quad +\quad { K }_{ 0 }\quad =\quad { K }_{ f }\\ \frac { 2 }{ { m }_{ p } } { [q }_{ p }{ V }_{ 0 }\quad -\quad { q }_{ p }{ V }_{ f }\quad +\quad \frac { 1 }{ 2 } { m }_{ p }{ v }_{ 0 }^{ 2 }]\quad =\quad { v }_{ f }^{ 2 }\\ 1.2e27*[-1.2e-17\quad +\quad 7.84e-18\quad +\quad 6.76e-18]\quad =\quad { v }_{ f }^{ 2 }\\ { v }_{ f }\quad =\quad 54448.79\quad m/s$

Attempt #2 using energy
$W\quad =\quad \Delta K\\ q\Delta V\quad =\quad Kf\quad -\quad Ki\\ q\Delta V\quad +\quad Ki\quad =\quad Kf\\ \frac { 2 }{ 1.67e-27 } [26.8*1.6e-19\quad +\quad 0.5*1.67e-27*{ 90000 }^{ 2 }]\quad =\quad { v }f^{ 2 }\\ vf\quad \neq \quad 54448.79$

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?

2. Apr 3, 2016

### Staff: Mentor

It should be $-q \Delta V$ for work, as the work done is the integral over the negative of the force.

3. Apr 3, 2016

### Sho Kano

I thought work was the positive integral of force, I've heard of voltage being the negative integral of electric field, but not this.

4. Apr 5, 2016

### Sho Kano

There is a negative in gravitational potential energy formula, but that is because negative work is done for an object to get closer to earth. So in this case, negative work is done to "slow" down the proton? (or get the proton closer to the less positive plate?)