Charged particle moving through an electric field

[TommyS123]

A proton moving to the right at 6.2 × 10^5 ms-1 enters a region where there is an electric field of 62 kNC-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field?

I approached this problem trying to solve for time.

Mass proton = 1.67 x 10^-27 kg
velocity of proton = 6.2 x 10^5 m/s
charge proton = 1.60 x 10^-19 C
t = time?

t = (electric field)/((Mass proton x velocity proton)/charge proton)

I was thinking that the units would cancel out and leave the answer in seconds.

 

[berkeman]

A proton moving to the right at 6.2 × 10^5 ms-1 enters a region where there is an electric field of 62 kNC-1 directed to the left. Describe qualitatively the motion of the proton in this filed. What is the time taken by the proton to come back to the point where it entered the field?

I approached this problem trying to solve for time.

Mass proton = 1.67 x 10^-27 kg
velocity of proton = 6.2 x 10^5 m/s
charge proton = 1.60 x 10^-19 C
t = time?

t = (electric field)/((Mass proton x velocity proton)/charge proton)

I was thinking that the units would cancel out and leave the answer in seconds.

What is the equation for the force on a charged particle due to an electric field E? What is the equation for the acceleration of a mass when subjected to a force? Is the acceleration in this problem constant? If so, use the kinematic equations of motion for constant acceleration...

 

[TommyS123]

F = qE
Acceleration in this case is negative
acceleration = (mass) (-force)
it would be a constant deceleration

 

[berkeman]

F = qE
Acceleration in this case is negative
acceleration = (mass) (-force)
it would be a constant deceleration

Correct. And what are the kinematic equations for position and velocity based on that acceleration and the initial velocity? They are very similar to the equations you use for gravity-type problems, right?

 

[TommyS123]

v=vi + a (delta t)

For this the final velocity is the same as the starting velocity and the starting position is the same as the starting position. Shouldn't the equation read

(v-vi)/a = delta t , for the above equation?

 

[berkeman]

v=vi + a (delta t)

For this the final velocity is the same as the starting velocity and the starting position is the same as the starting position. Shouldn't the equation read

(v-vi)/a = delta t , for the above equation?

Sure, but you will also need the position equation, in order to solve the equations.

 

[TommyS123]

Can I use the r^2 = kq/E as the position equation?

 

[berkeman]

Can I use the r^2 = kq/E as the position equation?

That looks more like for the electric field from a point charge. You have a point charge in a uniform electric field E. You already wrote the equation for the force on that point charge in a uniform E-field...