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E Field Of A Ink Jet Printer

  1. Apr 16, 2007 #1
    1. One type of ink-jet printer, called an electrostatic ink-jet printer, forms the letters by using deflecting electrodes to steer charged ink drops up and down vertically as the ink jet sweeps horizontally across the page. The ink jet forms 30.0 (mu)m diameter drops of ink, charges them by spraying 800,000 electrons on the surface, and shoots them toward the page with a horizontal velocity of 16.0 m/s . Along the way, the drops pass through the long axis of two parallel electrodes that are 6.0 mm long, 4.0 wide, and spaced 1.0 mmapart. The distance from the center of the plates to the paper is 1.40 cm. To form the letters, which have a maximum height of 6.0 mm, the drops need to be deflected up or down a maximum of 3.0 mm. Ink, which consists of dye particles suspended in alcohol, has a density of 800 kg/m^3

    QUESTION: Estimate the maximum electric field strength needed in the space between the electrodes.

    What amount of charge is needed on each electrode to produce this electric field?

    2. Help me get started!
    I am given a velocity so I can solve for acceleration using F=EQ. That is the acceleration from the center of the electrodes to the paper. I am having trouble understanding the "format" of the electric field between the electrodes.
    The E field of a plane is N/(2*Epsilon). and N=Q/AREA???
    I know the E field between the electrodes must be strong enough to get the electrons to the paper. I just dont know how to relate the two. Help please. :bugeye:
  2. jcsd
  3. Apr 16, 2007 #2


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    The charge on the plates of the capacitor is related to the voltage and capacitance by Q=CV. That's what you use for the 2nd part of the questions (about the charge on the capacitor plates).
  4. Apr 16, 2007 #3
    Can someone please help with the first question?
  5. Apr 16, 2007 #4


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    Do you recognize that you have a projectile problem (from your first semester physics class)?
  6. Apr 16, 2007 #5
    Yes, I do, I can see that I can find acceleration and then I could find E, but its not the right E, so I am not sure how it relates. I need guidance.
  7. Apr 16, 2007 #6


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    It might be a good idea to show your attempt, including your calculations.
  8. Apr 16, 2007 #7
    You want me to show my attempt at finding accereration?

    -20^2=2(a)(.02+003), solve for a, a =8696 m/s^2,
    the .003 being the middle of plate

    if i was to go further and find E, I use the fact that f=eq, a=eq/m, and so e would be 4.95*10^-8
    Last edited: Apr 16, 2007
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