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!

Charge to Mass ratio

  1. Sep 14, 2016 #1
    1. The problem statement, all variables and given/known data
    A small object with mass m, charge q, and initial speed v0 = 6.00×10^3 m/s is projected into a uniform electric field between two parallel metal plates of length 26.0 cm(Figure 1) . The electric field between the plates is directed downward and has magnitude E = 600 N/C . Assume that the field is zero outside the region between the plates. The separation between the plates is large enough for the object to pass between the plates without hitting the lower plate. After passing through the field region, the object is deflected downward a vertical distance d = 1.35 cm from its original direction of motion and reaches a collecting plate that is 56.0 cm from the edge of the parallel plates. Ignore gravity and air resistance.

    Part A
    Calculate the object's charge-to-mass ratio, q/m.

    fdU8nnS.jpg

    2. Relevant equations
    E = F/q
    E=kq/r^2
    kinematics

    3. The attempt at a solution
    I solved for q by using E = kq/r^2

    Er^2/k = q
    (600 N/C)(.26 m)^2 / 9x10^9 Nm^2/C^2 = q
    q = 4.51 * 10^-9 C

    My plan for mass is

    Eq = F = ma
    Eq/a = m

    Using Kinematics
    Solving for time t

    x - x0 =1/2(v0x + vx)t
    2(x-x0)/(v0x + vx) = t
    2(.56m)/(2 * 6 *10^3 m/s) = t
    t = 9.33*10^-5 s


    Solving for Acceleration y axis ay

    y = y0 + v0yt + (1/2)ayt^2
    2(y - y0 - v0yt)/t^2 = ay
    2(.56m)/t^2 = ay
    ay = 3127478 m/s


    Eq/a = m

    m = 8.668 * 10^-13 kg

    q/m = 5203 wrong answer


     
  2. jcsd
  3. Sep 14, 2016 #2

    haruspex

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member
    2016 Award

    The acceleration only applies while between the plates. Break the problem into two parts.
     
  4. Sep 14, 2016 #3

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    And while you do what haruspex suggested, consider that the E-field in which the particle travels is not kq/r^2, which is the field generated by a point charge, but a uniform field of 600 N/C generated by external charges on the plates.
     
  5. Sep 14, 2016 #4
    I tried solving for Vy(in plates) which is equal to V0y and Vy (outside the plates)
    y-y0 = 1/2(2Vy)t
    (-0.0135m)/t = Vy
    t = 5 * 10^-5 s (outside the from 26 to 56) calculated wrong in post
    Vy = (-0.0135m)/t = -270 m/s

    Now solving for t (x=0 to x=.26m)
    (x-x0) = (1/2)(v0x + v0x)t
    (.26m)/(6.00 * 10^3 m/s) = t
    t = 4.33*10^-5 s

    ay = (-270m/s)/ t
    ay = 6235565 m/s^2
     
    Last edited: Sep 14, 2016
  6. Sep 14, 2016 #5

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    A "previous question" is often an unreliable source. Here you don't have a line of charge, just a moving charge in a uniform electric field. The electric field in the line of charge question does not have the same dependence on space coordinates (it looks different if you draw field lines in space) as a uniform electric field, which is what you have here. Therefore, it is incorrect to use the line-of-charge of expression where it does not apply and expect to make sense out of it.

    I have not done the math yet to see if your number for ay is correct. Assuming that it is, can you find an expression for ay in terms of the charge q, the external electric field E and the mass m?
     
  7. Sep 14, 2016 #6
    ay = qE/m
     
  8. Sep 14, 2016 #7

    kuruman

    User Avatar
    Science Advisor
    Homework Helper
    Gold Member

    Right. Now look at the equation. What quantity are you trying to find and how can you write it in terms of known quantities?
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted



Similar Discussions: Charge to Mass ratio
  1. Charge to mass ratio (Replies: 3)

  2. Charge to mass ratio (Replies: 3)

  3. Charge-to mass ratio (Replies: 10)

Loading...