An electron with speed 2.35×107m/s is traveling parallel to a uniform electric field of magnitude 1.16×104N/C .

Question

How far will the electron travel before it stops?

A=0.135m

How much time will elapse before it returns to its starting point?

I solved for the time it would take for the final velocity to reach zero (1.15x10^-8s). However, this answer is wrong so I'm assuming I misinterpreted the question?

Answer 1

F = m a

F = charge on electron * E
F = 1.6*10^-19 * 1.16*10^4 = 1.856*10^-15 N

m = mass of electron
a = F/m which is constant
a = 1.856 * 10^-15/9.1*10^-34
= .204 * 10^19 m/s^2

V = Vi - a t
solve for t, when it stops
0 = 2.35*10^7 - .204 *10^19 t
t = 11.52 * 10^-12 seconds
I disagree with you here

Answer 2

the first answer is somewhat wrong. the approach is almost correct, though.

Electric field = F/q. E*charge of electron = F
1.16E4 * 1.6E-19 = 1.856E-15 N

F=ma
mass of electron = 9.11E-31 kg
F/m = a
1.856E-15/9.11E-31 = 2.04E15 m/s2

0.5*(Vf-Vi)^2 = a(Xf-Xi)
0.5*(0-2.35E7)^2 = 2.04E15(Xf-0)
Xf=0.135 m
---------------------------------
Vf=Vi-at
0 = 2.35E7 - 2.04E15t
t=1.15E-8s

assuming that it will come back at the same acceleration as going forward...just double the time to get back to the beginning.

2t = 2.3E-8s