• Support PF! Buy your school textbooks, materials and every day products Here!

Electric Field

  • Thread starter JWHooper
  • Start date
JWHooper
1. Homework Statement
An electron has a mass of 9.11 x 10^(-31) kg. Its charge is -1.6 x 10^(-19) C. Suppose it is given a initial horizontal velocity of 8.0 x 10^7 m/s. It then travels through a vertical (upward pointing) electric field with a constant strength of 2.0 x 10^4 N/C. In a time T it has a horizontal displacement of 10 cm.
(a) How much force acts on the electron? In which direction is the force?
(b) What is the acceleration of the electron?
(c) What will ber the horizontal velocity of the electron at the end of time T?
(d) What will be the vertical velocity of the electron at the end of time T?
(e) At the end of time t, what angle will the electron's velocity make with its initial velocity?
(f) What will be the vertical displacement of the electron during time T?



2. Homework Equations
E = F_q / q = kQ / d^2
F = kq_1q_2 / d^2
x(t) = v_0t + 1/2at^2



3. The Attempt at a Solution
(a) Since E = F_q / q, this can arrange to F_q = Eq = (2.0 x 10^4 N/C, upward)(-1.6 x 10^(-19)) = 3.2 x 10^(-15) N, downward.
(b) Since v(t) = 8.0 x 10^7 m/s, v'(t) = a(t) = 0 m/s^2.
(c) Since x'(t) = v(t) = v_0 + at, v(T) = 8.0 x 10^7 m/s, right + (0 m/s^2)T = 8.0 x 10^7 m/s, right.
(d) Since no velocity is given in the problem, 0 m/s.

The last two questions are that I don't get. Please help!

Thanks,

J.
 

Answers and Replies

Doc Al
Mentor
44,827
1,083
(a) Since E = F_q / q, this can arrange to F_q = Eq = (2.0 x 10^4 N/C, upward)(-1.6 x 10^(-19)) = 3.2 x 10^(-15) N, downward.
Good.
(b) Since v(t) = 8.0 x 10^7 m/s, v'(t) = a(t) = 0 m/s^2.
Since there's a force on the electron, it had better have an acceleration. Use Newton's 2nd law.
(c) Since x'(t) = v(t) = v_0 + at, v(T) = 8.0 x 10^7 m/s, right + (0 m/s^2)T = 8.0 x 10^7 m/s, right.
OK
(d) Since no velocity is given in the problem, 0 m/s.
You'll need to calculate it based on the acceleration. (You know the initial velocity.)
 
JWHooper
For part (b), i got a = F/m = 3.51 x 10^15 m/s^2, downward.

Using this, for part (d), use the equation x'(t) = v(t) = v_0 + at =
v_0 + (3.51 x 10^15 m/s^2, downward)(t), but I can't figure out the initial velocity for vertical motion (although v_0 for horizontal motion is 8.0 x 10^7 m/s). Please help!

j.
 
Doc Al
Mentor
44,827
1,083
For part (b), i got a = F/m = 3.51 x 10^15 m/s^2, downward.
Good.
Using this, for part (d), use the equation x'(t) = v(t) = v_0 + at =
v_0 + (3.51 x 10^15 m/s^2, downward)(t), but I can't figure out the initial velocity for vertical motion (although v_0 for horizontal motion is 8.0 x 10^7 m/s).
The initial velocity is purely horizontal. So what must be the vertical component?
 
JWHooper
The initial velocity is purely horizontal. So what must be the vertical component?
Since it's purely horizontal, I will assume that the electron's vertical component is
0 m/s. Then, we get v(t) = v(T) = (3.51 x 10^15 m/s^2, downward)(T). Did I do this part right?
 
Doc Al
Mentor
44,827
1,083
Right!
 
JWHooper
Oh okay. Thanks for the help!

Btw, it would really help me if you give me tips on parts (e) and (f).
 
Doc Al
Mentor
44,827
1,083
Hint for (e): To find the angle of any vector, start by finding its components.

Hint for (f): This is a kinematics problem for accelerated motion, very similar to finding how far a ball would fall in a given time. Of course, here the acceleration is due to the electric force not gravity.
 
JWHooper
For part (e), since at end of time T, the horizontal velocity of the electron is
8.0 x 10^7 m/s right, and vertical velocity is (3.51 x 10^15 m/s^2, downward)(T), we use some trigonometry to find angle (use tan(x), where x is the angle).
tan(x) = (3.51 x 10^15 m/s^2)(T) / (8.0 x 10^7 m/s)
= (4.39 x 10^7 1/s)(T)
x = tan^-1(4.39 x 10^7 1/s)(T).
I don't know if I did it right; If I did it wrong, let me know.

For part (f), use antiderivative for velocity to get position function.
Integral of velocity = (1.76 x 10^15 m/s^2, downward)(t^2) = function of position = x(t)
Plug in t = T,
x(T) = (1.76 x 10^15 m/s^2, downward)(T^2).
I don't know if I did this part right either, but please let me know if I did some wrong calculation.
 
Doc Al
Mentor
44,827
1,083
Realize that the time "T" is not an unknown--it's the time it takes the electron to travel 10 cm horizontally. So find that time!

In part (d) and the following, be sure to use the actual value for the time to get your final answers.

For part (e), since at end of time T, the horizontal velocity of the electron is
8.0 x 10^7 m/s right, and vertical velocity is (3.51 x 10^15 m/s^2, downward)(T), we use some trigonometry to find angle (use tan(x), where x is the angle).
tan(x) = (3.51 x 10^15 m/s^2)(T) / (8.0 x 10^7 m/s)
= (4.39 x 10^7 1/s)(T)
x = tan^-1(4.39 x 10^7 1/s)(T).
I don't know if I did it right; If I did it wrong, let me know.
Looks good to me. That will be the angle below the horizontal. Be sure to plug in the time (after you figure it out) to get an actual angle.

For part (f), use antiderivative for velocity to get position function.
Integral of velocity = (1.76 x 10^15 m/s^2, downward)(t^2) = function of position = x(t)
Plug in t = T,
x(T) = (1.76 x 10^15 m/s^2, downward)(T^2).
I don't know if I did this part right either, but please let me know if I did some wrong calculation.
Good. You could also have just used the kinematic equation that you listed in your first post:
x(t) = v_0t + 1/2at^2
Again, get an actual distance by using the actual time.
 
JWHooper
Since the initial horizontal velocity is 8.0 x 10^7 m/s, we can convert 10 cm of horizontal displacement to meters.
(10cm)(1m / 100cm) = 0.1m
Now, divide displacement by velocity and get the time T,
T= d/v = (0.1m)/(8.0 x 10^7 m/s) = 1.25 x 10^9 s .
Hence, we have the time value.

I will assume that my calculations are correct, but just in case, I want you to check my calculation if I did any errors.
 
Doc Al
Mentor
44,827
1,083
JWHooper
Your method is correct, but that exponent should be minus 9.
Oh okay.
Thank you for helping me with the homework!

J.
 

Related Threads for: Electric Field

Replies
1
Views
7K
Replies
6
Views
458
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
5
Views
364
  • Last Post
Replies
10
Views
647
  • Last Post
Replies
2
Views
9K
  • Last Post
Replies
6
Views
8K
  • Last Post
Replies
2
Views
656
Top