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Movement of electron in an electric field

  1. Mar 24, 2016 #1
    1. The problem statement, all variables and given/known data
    An electron has an initial velocity of 5.00 x 10^6 m/s in a uniform 2.00 x 10^5 N/C strength electric field. The field accelerates the electron in the direction opposite to its initial velocity.
    (a) What is the direction of the electric field?
    (b) How far does the electron travel before coming to rest?
    (c) What is the electron's velocity when it returns to it's starting point?

    2. Relevant equations
    F= Eq
    F=ma
    Eq=ma

    3. The attempt at a solution
    F= Eq
    = (2.00 x 10^5)(-1..60x10^-19)
    = -3.2 x 10^-14 N
    F= ma
    -3.2x10^-14= (9.11x10^-31)a
    a= -3.51x10^16 m/s^2

    Taking downward direction as positive,

    v^2= u^2 + 2as
    0= 5.00 x 10^6 + 2(-3.51x10^16)s
    s= 3.56 x 10^-4 m

    Electron is accelerated in the direction towards the positive plate. It is stated that the direction of acceleration is opposite to that of its initial velocity. How is it possible?

    Can anyone helps me to figure out what has gone wrong in my equations? Thanks a lot.
     
  2. jcsd
  3. Mar 24, 2016 #2
    what is the basis of your observation?
    electric field force is defined by a test charge (unit +charge)
    so in which direction the field is operating?
     
  4. Mar 24, 2016 #3

    rude man

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    check this equation ...
     
  5. Mar 24, 2016 #4
  6. Mar 24, 2016 #5
    This is my observation. The electric field is directed from positive to negative. The electron will accelerate towards to positive side. But I don't get why the direction of initial velocity can be opposite to that of the acceleration,
     
  7. Mar 24, 2016 #6
    I still got s= 3.56 x10^-4m.
    May I know why s is positive as I have already taken downward direction as positive? In the sketch that I posted, the electric field is directed from positive to negative, that is downwards. Electron is directed towards the positive plate. Hence acceleration is in the direction towards the positive plate. If s is positive, won't it indicate a downward displacement? And how can the direction of initial velocity be opposite to that of acceleration?
     
  8. Mar 25, 2016 #7
    perhaps you have got an electron with initial velocity direction as given and -jy(0) is being decelerated .write the vector equation suppose your direction is -jv(o) and now you apply +j a
    your equation is
    (-jv(t))^2 = (-j v(0))^2 +2.(ja) (-) jS ; the final velocity v(t) =0
    so you get s= +ve as j.j=1 and in -j direction- i hope you can understand .
     
  9. Mar 25, 2016 #8

    rude man

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    You used u instead of u^2. Fix that & you're home.
    Your other questions don't make sense. OF COURSE the initial velocity HAS TO BE opposite to the acceleration. You are accelerating in a direction opposite to the initial velocity, otherwise the electron would just keep accelerating in the initial velocity direction, going to infinite speed instead of slowing down to zero.

    If you throw a ball straight up, what's the initial velocity direction? What's the acceleration direction?
    !
     
  10. Mar 25, 2016 #9
    If I throw a ball straight up, the direction of initial velocity will be upwards. Since the ball is slowing down, it's a deceleration and negative acceleration, so it's downwards. Thank you very much :D I now understood why the direction of acceleration is opposite to that of the initial velocity.

    But I have another question. If the electron is in a uniform electric field and the acceleration is upwards to the positive plate, what actually causes the electron to have an initial velocity that is downwards?
     
  11. Mar 25, 2016 #10

    cnh1995

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    Assuming ideal conditions, in the beginning, field direction was opposite, so the electron accelerated and when it reached the given speed, the direction of field was reversed. Reversing the field direction doesn't seem to be difficult.
     
  12. Mar 25, 2016 #11

    rude man

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    The initial velocity is whatever you give it!
     
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