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Homework Help: Electric field and velocity

  1. Oct 20, 2008 #1
    1. The problem statement, all variables and given/known data

    Two 4.0 cm diameter disks face each other, 2.0 mm apart. They are charged to +-12 Nc.

    Part A
    What is the electric field strength between the disks?
    E = 1.1 * 10^6 N/C

    Part B
    A proton is shot from the negative disk toward the positive disk. What launch speed must the proton have to just barely reach the positive disk?
    2. Relevant equations

    E=(Q/(epsilon zero * A)

    3. The attempt at a solution

    I was able to figure out part A. I just need help with part B.
    V = ?
    x = 2mm
    t = ?

    Doesn't seem like this part should be difficult, so I must be missing something simple...

    Thanks in advance for any help.
  2. jcsd
  3. Oct 20, 2008 #2

    Doc Al

    User Avatar

    Staff: Mentor

    Two ways to attack this:

    (1) Using kinematics. Hint: What's the proton's acceleration?

    (2) Using energy. Hint: What's the potential difference between the plates?
  4. Oct 20, 2008 #3
    So...this is what I have so far.

    F = qE

    F = (12*10^-9)(1.1*10^6)
    F = (.0132N)

    F = ma

    mass of proton = 1.67*10^-27

    .0132 = (1.67*10^-27) (a)
  5. Oct 20, 2008 #4

    Doc Al

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    Staff: Mentor

    Since you need the force on the proton, use the charge of a proton. (Not the total charge on the plate!)
  6. Oct 20, 2008 #5
    F = 1.76*10^-13


    1.76*10^-13 = (1.67*10^-27)(a)
    1.05*10^14 = a
  7. Oct 20, 2008 #6

    Doc Al

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    Staff: Mentor

    Once you've found the acceleration, it's time for kinematics. You'll need a kinematic equation relating speed and distance.
  8. Oct 20, 2008 #7
    v = x/t
    f = m/a

    Those aren't it...hm...

    Kinetic energy = .5(mass)(velocity)^2

    so...(if I remember correctly) total energy = Kinetic energy + potential energy

    potential energy = (mgh)

    Sadly, if this is the correct approach I do not remember what variables are on the total energy side.

    O yeah...I forgot I was looking for an equation relating speed and distance. I am having trouble finding one.
  9. Oct 20, 2008 #8
    Hm...perhaps this equation
    v^2 = vo^2 + 2a(X - Xo)

    Yup...that would be the correct equation.

    Thanks for the help Doc Al.
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