Motion of a charged particle in an electric field. Please help!?

Question

An electron with speed v0 = 29.5 106 m/s is traveling parallel to an electric field (v0||E) of magnitude E = 11.9 103 N/C.
(a) How far will it travel before it stops?
in cm
(b) How much time will elapse before it returns to its starting point?
in nano sec

Answer 1

Use final velocity squared minus initial velocity squared
equals
2 times a times change in position.

To get a, calculate the force, F = qE (look up the fundamental charge q in your book). F=ma, so a = F/m = qE / m.

You have final v and initial v (zero), so you can solve for the change in position.

To get the time, use a = change in velocity / time

time = change in velocity / a

Then double it for the round trop.

Good luck!

Answer 2

The force acting on the electron of charge q is qE. In the first case, the force is decelerating.

ma = qE., or a (deceleration) = qE/m ( look up q/m ratio of electron)

The rest is basic physics involving the 3 equations of motion

a) v^2 = vo^2 + 2 a s (s is distance and a is negative), find s.
b. v = vo + a t , find time it takes for the electron to come to a halt. The same of time is needed for the electron to be accerated in the opposite direction to reach the same speed and cover the same distance. This is by symmetry.

Answer 3

sensible value strikes contained contained in the route of the sphere. unfavorable value strikes opposite to the sphere. reason: enable field = E, value = Q Then rigidity = Q*E Above equation shows that i)If Q is unfavorable, then rigidity and field are in similar route. ii)If Q is unfavorable, then rigidity and field are in opposite training.