Question

An electron with speed 2.25×107m/s is traveling parallel to a uniform electric field of magnitude 1.10×104N/C .?
A) How far will the electron travel before it stops?
B) How much time will elapse before it returns to its starting point?

Answer 1

\(F=qE=ma\)

then use kinematic equations

Answer 2

How so?

Answer 3

I keep getting 2.4 x 10^-6 and it's wrong

Answer 4

the acceleration is \(a=\dfrac{qE}{m}=\dfrac{1.6*10^{-19}~C*1.10×10^4~N/C}{9.1*10^{-31}~kg}=1.93*10^{15}~m/s^2\)

Use this acceleration in the kinematic equation of motion under constant acceleration.

NOTE: it's negative because of the direction of the electron.

Answer 5

what is the kinematic equation of motion under constant acceleration?

Answer 6

Kinematic equations

\(\sf v_f=v_i+a\Delta t\)

\(\sf x_f=x_i+v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf v_f^2=v_i^2+2a_x\Delta x\)

Answer 7

so, what values would you use for delta x and for vf and vi

Answer 8

Have you done questions in mechanics where it asks "how high does an object go when you throw it upwards with a velocity of x m/s"?

It's the same thing.

Answer 9

I'll help you with the first one. Use: \(\sf v_f^2=v_i^2+2a_x\Delta x\)

\(\sf(0)^2=( 2.25*10^7~m/s )^2+2(-1.93∗10^{15} m/s^2)\Delta x\)

\(\sf \Delta x =0.1311528~m \approx 0.131 ~m\)

Answer 10

thank you so much! now to find out how much time is elapsed, do you just use vy=at?

Answer 11

not quite. It's asking for the time of the ROUND trip (there and back).

The accelerations are in different directions. First negative then positive

Answer 12

wait i got the electric field the opposite way, but it makes no difference.

Answer 13

I'm sorry, I'm confused so what number do you use for your v and will it be a positive number or a negative number?

Answer 14

i understand the two equations a=ve and a=-ve, but what values do you use for v

Answer 15

okay, so we got the distance 0.131 m, the initial speed 2.25×10^7m/s, and the acceleration 1.93*10^15 m/s^2. This is parabolic motion,

\(\sf x_f=x_i+v_o\Delta t +\dfrac{1}{2}a_x(\Delta t)^2\)

\(\sf 0 =(2.25×10^7m/s)\Delta t -\dfrac{1}{2}(1.93*10^{15} m/s^2)(\Delta t)^2\)

\(\sf \Delta t=0.0000000233160622~s=2.33*10^{-8}~s=23.2 ~ns\)