Proton electric potential/ final speed calculation

[physicsstar26]

Homework Statement

A proton (q=+e, m= 1.67*10^-27 kg) traveling at 300 m/s enters a uniform electric region opposite to the field direction.
a.) find the change in the energy of the proton as it moves 2 m across a potential difference of 100kV in this field. Express answer in eV.
b) find the final speed of the proton in km/s

Homework Equations

a) W=q*V
V= k*q/r I'm confused as to how to approach this problem.
b) once you know V or the above answer do you just plug it into V= (1/2) mv^2/q?
thanks

 

[dynamicsolo]

Homework Equations

a) W=q*V
V= k*q/r I'm confused as to how to approach this problem.
b) once you know V or the above answer do you just plug it into V= (1/2) mv^2/q?
thanks

What does your equation (a) represent? How will it relate to the change in (what?) energy of the proton?

As for the next equation, since the proton is in a *uniform* field, will that equation be useful? What does *it* represent?

 

[physicsstar26]

for (a) would you just use Va-Vb= E* dl to find the potential difference?

 

[dynamicsolo]

for (a) would you just use Va-Vb= E* dl to find the potential difference?

Well, you can take that route to work out the force on and acceleration of the proton. That will let you answer (b), after which you can back-figure part (a).

Alternatively, you can use the potential difference to figure out the kinetic energy change of the proton, then work out part (b). Either approach requires about the same amount of calculation.

 

[dynamicsolo]

Homework Statement

A proton (q=+e, m= 1.67*10^-27 kg) traveling at 300 m/s enters a uniform electric region opposite to the field direction.

Ah, be careful in handling either the kinematics or energetics for this situation. What is the proton going to do in this field?