1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    vx=4.0*10^6m/s
    ε=4.0*10^2 N/C
    Δdy=0.02m

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

    2. Relevant equations

    suvat
    electric field equation
    coulombs laws

    3. The attempt at a solution

    a)
    ma=qε
    (9.11*10^-31)(a)=(1.6*10^-19)(4*10^2)
    acceleration = 7.025246981*10^13 m/s/s [down]

    b)
    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

    c)
    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

    gneill

    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.

    Let
    and [down] be positive.

    c)
    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

    gneill

    User Avatar

    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.... :/

    Fixed:


    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]


    finally.

    look good?
     
  7. Jan 29, 2013 #6

    gneill

    User Avatar

    Staff: Mentor

    Much better. Be sure to round submitted results to appropriate significant figures.
     
  8. Jan 29, 2013 #7
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook