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Homework Help: Particle motion + electric fields

  1. Jan 27, 2013 #1
    1. The problem statement, all variables and given/known data

    see attachment for question wording.
    a) find acceleration
    b) find horizontal displacement
    c) find final velocity

    ε=4.0*10^2 N/C

    mass of electron = me = 9.11*10^-31kg
    charge of electron = qe = -1.6*10^-19 C

    2. Relevant equations

    electric field equation
    coulombs laws

    3. The attempt at a solution

    acceleration = 7.025246981*10^13 m/s/s [down]

    Δdy=vi yΔt+0.5(ay)Δt^2
    Δt = 2.4*10^-8 seconds

    Δdx = vΔt
    Δdx = (4*10^6)*2.4*10^-8)
    Δdx = 0.096m
    Δdx = 9.6cm

    vf y^2 = vi y^2 +2ayΔdy
    vf y = √(0+1.34884742*10^13)
    vf y = 3.672665817*10^6 m/s

    I'm pretty confident in parts A and B but I'm not sure about part C.
    I still have to find the direction of the final velocity; waiting to see if the magnitude is correct first.
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution

    Attached Files:

    Last edited: Jan 27, 2013
  2. jcsd
  3. Jan 27, 2013 #2


    User Avatar

    Staff: Mentor

    Your methods look okay. The results for a and b look good, but watch your significant figures.

    Something went wrong in the numerical calculation for part c; the value you obtained for the vertical velocity is not correct. Also, presumably the final velocity should include both the horizontal and vertical components.
  4. Jan 29, 2013 #3
    Ah. I used the horizontal displacement found in part b instead of the vertical displacement like I should've.

    and [down] be positive.

    vf y^2 = vi y^2 +2ayΔdy
    vf y = √(0+2((7.025246981)(0.02))
    vf y = 0.167633493 m/s [down]

    vf x = √(vi x^2 + 2(0)(0.096)
    vf x = vi x
    vf x = 4*10^6 m/s

    theta = tan^-1((4*10^6)/(0.167633493))
    theta = 89.9999976°
    so, im thinking this is negligible.

    vf = √(0.167633493^2 + (4*10^6)^2 )
    vf = 2000.00007 m/s

    is this correct?​
  5. Jan 29, 2013 #4


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    Staff: Mentor

    That seems way to small for the y-velocity. Check the acceleration value you've used.
    In other words the x velocity is unchanged because there is no acceleration in the x direction.
    Don't use angles to calculate the sum of the components, they are at 90° to each other. Use Pythagoras: square root of sum of squares.

    Nope. Fix the bits above and recalculate.​
  6. Jan 29, 2013 #5
    Ugh. Careless mistake. Seems I'm making a lot of those.... :/


    vf y^2 = vi y^2 +2ayΔdy
    vf y = √(0+2(7.025246981*10^13)(0.02)
    vf y = √(2.80098792*10^12)
    vf y = 1.676334928*10^6 m/s

    vf x = vi x = 4*10^6 m/s

    theta = tan^-1 (vfx/vfy) = 67.26 degrees

    vf = √((1.676334928*10^6)^2+(4*10^6)^2)
    vf = 4.337061078*10^5 m/s [down 67.26 degrees right]


    look good?
  7. Jan 29, 2013 #6


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    Staff: Mentor

    Much better. Be sure to round submitted results to appropriate significant figures.
  8. Jan 29, 2013 #7
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