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!

Homework Help: Find charge q of particle passing through charged plates

  1. Apr 16, 2014 #1
    In an inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. The ink drops, which have a mass of 1.3 10-8 g each, leave the nozzle and travel toward the paper at 21 m/s, passing through a charging unit that gives each drop a positive charge q by removing some electrons from it. The drops then pass between parallel deflecting plates 2.0 cm long where there is a uniform vertical electric field with magnitude 7.7 104 N/C. If a drop is to be deflected 0.35 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop?

    So, I know that F= E*q=m*a
    I can solve for t, . I still don't know a, the acceleration, or vfinal of y either. I was thinking of solving for vfy by using dy=vi+vf/2 * t.

    Either way, I keep getting the wrong answers. They are supposed to be about ^-13.

    Any help is appreciated!!
  2. jcsd
  3. Apr 17, 2014 #2
    Some of your numbers and formulae are hard to understand with typos, I think. But you seem to be on the right track.

    You do know a in terms of q.

    What other kinematic formulae do you know?
  4. Apr 17, 2014 #3
    Here is what I can deduce the data from OP
    Mass of a drop ,m = 1.3e-8 g = 1.3e-11kg
    Deflection , y = .35mm = 3.5e-4 m
    Length of the plate ,l = 2cm = 2e-2m,
    Velocity of the drop = 21m/s
    E = 7.7e4.
    Here's the solution,
    [itex]a_{y} = \frac{qE}{m}[/itex] ( Since F = qE )
    Let t be the time required for the drop to pass the region and deflect,
    y = [itex]\frac{1}{2}a_{y}t^{2}[/itex] ... (1)
    l = vt
    ∴ t = [itex]\frac{l}{v}[/itex]
    Substituting for t and a in (1) , we get,
    y = [itex]\frac{qEl^{2} }{2mv^{2}}[/itex]
    Solving for q,
    q = 1.30e-13 = 1.3[itex]\times[/itex]10-13
    Note : the notation xey means x [itex]\times[/itex] 10 y
  5. Apr 17, 2014 #4
    Sorry about the typos guys!
    So, we can solve for t, even though the velocity is mostly in the x direction, because the time is the same for the x and y components of motion, right?
    ALso, eq(1) would have originally been y= initial y position + velocity in y * time + (acceleration stuff), but the y velocity component is 0 initially, right?

    Eventually, we have velocity in thr y direction, though, so why does eq(1) work? From what I see, it says the velocity in the y direction is always 0.

    The answer is absolutely correct; I;m just trying to understand the answers both of you gave. I appreciate this very much. This is better feedback than my prof gives.
    Last edited: Apr 17, 2014
  6. Apr 17, 2014 #5
    I did this one by accident, so I'll make a cat to not waste the post

  7. Apr 17, 2014 #6
    The velocity is in the x direction , this is anologous to projectile motion , if you fire in the horizontal direction , the horizontal velocity is unaffected by gravity but the vertical component is affected by gravity so in this case the drop moving across the plates in the x axis is deflected in the y axis due to the vertical field.
    The equation 1 works because acceleration is constant through out the motion.
    If you still have doubts about the equation recall how it is derived.
    ds = v.dt ,(now, since v - u= at and u = 0)
    ds = at.dt , integrating both sides we get our equation.
    Hope that clear things up.
    Last edited: Apr 17, 2014
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted