Electron velocity in Electric Fields

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Homework Statement


An electron with a speed of 5.57 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.72 × 103 N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed?


Homework Equations


KE= .5mv^2
F=EQ
E=kq/r^2
W=FD


The Attempt at a Solution



For part A I got it correct which is 0.0325m.

I solved it by first calculating KE of the particle: .5 (9.11 E-31)(5.57 E 6) [converted velocity into m/s instead of cm/s]

KE= 1.413 E -17 J

Then I calculated F, F= (1.6 E-19)(2.72 E3) F=4.352 E-16

With those two values I plugged them into W=FD and got 0.0325m

Now to find the time I used all necessary information and applied t=d/v

t=0.0325m/5.57 E 6 t=5.83 E -9 s

On the homework I entered it both in scientific notation and by the complete decimal but it keeps marking it wrong. I've gone through it several times to make sure I did not do any calculations incorrectly but I cant seem to find any.


I would really appreciate it if someone could please review my work and see if it is correct. Thanks :smile:
 

Answers and Replies

  • #2
gneill
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Homework Statement


An electron with a speed of 5.57 × 108 cm/s in the positive direction of an x axis enters an electric field of magnitude 2.72 × 103 N/C, traveling along a field line in the direction that retards its motion. (a) How far will the electron travel in the field before stopping momentarily, and (b) how much time will have elapsed?


Homework Equations


KE= .5mv^2
F=EQ
E=kq/r^2
W=FD


The Attempt at a Solution



For part A I got it correct which is 0.0325m.

I solved it by first calculating KE of the particle: .5 (9.11 E-31)(5.57 E 6) [converted velocity into m/s instead of cm/s]

KE= 1.413 E -17 J

Then I calculated F, F= (1.6 E-19)(2.72 E3) F=4.352 E-16

With those two values I plugged them into W=FD and got 0.0325m

Now to find the time I used all necessary information and applied t=d/v

t=0.0325m/5.57 E 6 t=5.83 E -9 s

On the homework I entered it both in scientific notation and by the complete decimal but it keeps marking it wrong. I've gone through it several times to make sure I did not do any calculations incorrectly but I cant seem to find any.


I would really appreciate it if someone could please review my work and see if it is correct. Thanks :smile:

The electron will be subject to a constant force due to the electric field, so its motion will be that of an object undergoing acceleration (or deceleration if you wish). Because its velocity is changing, the formula t = d/v, which is for uniform motion, isn't going to work. Can you think of a formula that applies to accelerated motion? (hint: the velocity is going from its initial value to zero with constant acceleration)
 
  • #3
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Would it be Vf=Vi+at or d=Vi*t+1/2at^2?
 
  • #4
gneill
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Would it be Vf=Vi+at or d=Vi*t+1/2at^2?

It would. Choose the easier one to solve for t :wink:
 
  • #5
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I think I will as well. lol

a would just be 9.8, correct?

EDIT: should it be -9.8 to make t positive?
 
  • #6
gneill
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I think I will as well. lol

a would just be 9.8, correct?

EDIT: should it be -9.8 to make t positive?

No, no gravity here, just the electric field. The object being accelerated is an electron... you calculated the force earlier...
 
  • #7
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ahhh that makes more sense. lol

So applying F=ma..

4.352 E -16=9.11 E -31 [mass of the electron] *a

a=4.77 E14 m/s^2

Does that seem correct?

EDIT: Also, when using that value, t comes out negative...
 
Last edited:
  • #8
gneill
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ahhh that makes more sense. lol

So applying F=ma..

4.352 E -16=9.11 E -31 [mass of the electron] *a

a=4.77 E14 m/s^2

Does that seem correct?

It does.
 
  • #9
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and t being negative is also fine?
 
  • #10
gneill
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and t being negative is also fine?

No, not so fine. t should end up positive. Can you show your calculation?
 
  • #11
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using Vf=Vi+at

0=5.57E6+4.77 E14 t

t= -5.57 E6 / 4.77 E14

t= -1.25 E -8
 
  • #12
gneill
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using Vf=Vi+at

0=5.57E6+4.77 E14 t

t= -5.57 E6 / 4.77 E14

t= -1.25 E -8

Is electron speeding up or slowing down? That is, is the acceleration positive or negative?
 
  • #13
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It sounds like the electron is slowing down then speeding up in the opposite direction.
 
  • #14
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oh, would that mean that there is a negative acceleration?
 
  • #15
gneill
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It sounds like the electron is slowing down then speeding up in the opposite direction.

That's probably true, but you're only interested in the trajectory up to the point where it (momentarily) comes to a halt.
 
  • #16
gneill
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oh, would that mean that there is a negative acceleration?

The signs of displacements, velocities, and accelerations is determined by your choice of coordinate system. If your initial velocity is taken to be positive and it's slowing down, then the acceleration must be negative...
 
  • #17
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thats makes sense, so then my t value would be t= 1.25 E -8 but my online homework is saying that the answer is incorrect :/ Im rechecking my calculations and actually trying it with D=vi*t+1/2at^2 formula. Quadratic formula here we come!
 
  • #18
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ok so with the formula D=Vi*t+1/2at^2 I applied the quadratic formula and got..

0.0325=5.57 E6+1/2(4.77 E14)*t

t=4.84 E-9 or t=-2.82 E-8
 
  • #19
gneill
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thats makes sense, so then my t value would be t= 1.25 E -8 but my online homework is saying that the answer is incorrect :/ Im rechecking my calculations and actually trying it with D=vi*t+1/2at^2 formula. Quadratic formula here we come!

I think you'll want to check your arithmetic. 1.25 doesn't look like the result of 5.57/4.77.
 
  • #20
gneill
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ok so with the formula D=Vi*t+1/2at^2 I applied the quadratic formula and got..

0.0325=5.57 E6+1/2(4.77 E14)*t

t=4.84 E-9 or t=-2.82 E-8

That doesn't look right. There should be linear and quadratic terms for the equation. Also the acceleration still needs to have the appropriate sign!

0.0325 = 5.57x106 t - (1/2) 4.77x1014 t2
 
  • #21
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ohhhh thats what it was. ok so now after that and doing it correctly I get t= 1.197 E -8s for both values
 
  • #22
gneill
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ohhhh thats what it was. ok so now after that and doing it correctly I get t= 1.197 E -8s for both values

I think you'll want to keep several more decimal places in all intermediate results if you're going to use the quadratic formula method. They can be "sensitive" to round off errors. In fact, it's a good idea to always keep a few extra digits for all intermediate results, and only round final values.

Why don't you try the "simpler" method again now that you've sorted out the acceleration sign? vf = vi - a*t
 
  • #23
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with that formula I got that t= 1.16771 E -11

Should the answers be different?
 
  • #24
gneill
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with that formula I got that t= 1.16771 E -11

Should the answers be different?

No, the answers will not be different. Your power of ten looks to be off... did you convert cm/s to m/s before doing the calculation? You should also check that you kept sufficient decimals in your acceleration value.
 
  • #25
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I'll check it tomorrow morning because I am dead tired, but I just wanted to say that I EXTREMELY appreciate your help and time. :smile:
 

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